SOLU Introducción a La Probabilidad y Estadística, 12ma Edición - W. Mendenhall, R. Beaver

STUDENT SOLUTIONS MANUAL FOR

M E N D E N HALL,

BEAVER,

I N T R O D U C T I O N

AND

BEAVER’S

TO

PROBABILITY STATISTICS

Duxbury

Student Solutions Manual for

M endenhall, Beaver, and Beaver's itro d u c tio n to P ro b a b ility and S tatistics Twelfth Edition

Barbara M. Beaver University of California, Riverside

T H O M

S O N

* — ------BR O O K S/C O L E

alia • Cariada • México • Singapore • Spain • United Kingdom • U nited States

© 2 0 0 6 D u x b u ry , a n im p rin t o f T h o m so n B ro o k s/C o le , a p a rt o f T h e T h o m so n C o rp o ra tio n . T h o m so n , th e S tar log o, an d B ro o k s/C o le are tra d e m a rk s u se d h erein u n d e r license.

T h o m so n H ig h er E d u c a tio n 10 D a v is D riv e B e lm o n t, C A 9 4 0 0 2 -3 0 9 8 U SA

A L L R IG H T S R E S E R V E D . N o p a rt o f th is w ork c o v e re d b y th e co p y rig h t h e re o n m ay be re p ro d u c ed

A sia (in c lu d in g In d ia ) T h o m so n L earn in g 5 S h e n to n W ay #01-01 U IC B u ild in g S in g a p o re 0 6 8 8 0 8

o r u sed in an y fo rm o r b y a n y m ean s— g rap h ic, e le c tro n ic , o r m ech a n ical, in c lu d in g p h o to co p y in g , re c o rd in g , tap in g , W eb d istrib u tio n , in fo rm atio n sto rag e an d retrie v al sy stem s, o r in a n y o th e r m an n er— w ith o u t the w ritten p e rm issio n o f the p u b lish er. P rin te d in C añada 1 2 3 4 5 6 7

09

08

07

06

05

P rin ter: W e b c o m L im ited IS B N : 0 -5 3 4 -4 6 3 2 5 -8

F o r m ore in fo rm a tio n a b o u t o u r p ro d u cts, co n ta c t u s at: T h o m so n L ea rn in g A c a d e m ic R e so u r c e C en ter 1-8 0 0 -4 2 3 -0 5 6 3 F o r p e rm issio n to use m ate ria l ffo m th is te x t or p ro d u ct, su b m it a re q u e s t o n lin e at h ttp ://w w w .th o m so n r ig h ts.c o m . A n y a d d itio n al q u e stio n s a b o u t p e rm issio n s ca n be su b m itted b y em ail to th o m so n r ig h ts@ th o m so n .c o m .

A u stra lia /N e w Z ea la n d T h o m so n L ea rn in g A u stra lia 102 D o d d s S treet S o u th b an k , V ic to ria 3 0 0 6 A u stralia C añada T h o m so n N e lso n 1120 B irch m o u n t R o ad T o ro n to , O n tario M 1 K 5G 4 C añ ad a U K /E u r o p e /M id d le E a st/A fr ic a T h o m so n L earn in g H igh H o lb o m H o u se 5 0 -5 1 B e d fo rd R o a d London W C 1R 4L R U n ite d K in g d o m L a tin A m erica T h o m so n L ea rn in g S en eca, 53 C o lo n ia P o lan co 11560 M éx ico D .F . M éx ico S p a in (in c lu d in g P o r tu g a l) T h o m so n P aran in fo C alle M a g a lla n es, 25 2 8 0 1 5 M a d rid , S p ain

Table of Contents

1: D escribing D ata w ith G raphs

1

2: D escribing D ata w ith N um erical M easures

9

3: D escribing B ivariate Data

23

4: P robability and Probability D istributions

31

5: Several U seful D iscrete D istributions

43

6: T he N orm al P robability D istribution

53

7: S am pling D istributions

65

8: L arge-Sam ple E stim ation

71

9: L arge-Sam ple T ests o f H ypotheses

81

10: Inference from Sm all Sam ples

91

11: T he A nalysis o f V ariance

107

12: L inear R egression and C orrelation

119

13: M últiple R egression A nalysis

135

14: A nalysis o f C ategorical Data

141

15: N onparam etric Statistics

151

1: Describing Data with Graphs a T h e ex p erim en tal is the student. b T h e ex p erim en tal c T h e ex p erim en tal d T h e ex p erim en tal e T h e ex p erim en tal

un it, the in d iv id u al o r o b je c t on w h ich a v a riab le is m easu red , u n it u n it u n it u n it

on w h ich the n u m b e r o f erro rs is m easu red is the exam . is the patient. is th e a z ale a plant. is th e car.

T h e p o p u latio n o f in terest co n sists o f v o ter o p in io n s (fo r o r ag a in st the c a n d id a te ) at the tim e o f the elec tio n fo r all p e rso n s v o tin g in th e e lec tio n . N o te th at w hen a sam ple is tak en (at so m e tim e p rio r o r the ele c tio n ), w e a re n o t actu a lly sa m p lin g fro m the p o p u latio n o f interest. A s tim e p asses, v o te r o p in io n s c h a n g e. H en ee, the p o p u latio n o f v o ter o p in io n s c h a n g e s w ith tim e, a n d the sam p le m ay n o t be re p rese n ta tiv e o f the p o p u latio n o f interest. a T h e v ariab le “re ad in g sc o re ” is a q u a n tita tiv e v aria b le , w h ich is p ro b ab ly in teg er-v alu ed an d h en ee d iscrete . b T h e in d iv id u al on w hich the v ariab le is m easu red is the student. c T h e p o p u latio n is h y p o th e tica l - it d o es n o t e x ist in fact - b u t co n sists o f the read in g sco res for all stu d en ts w h o co u ld p o ssib ly be tau g h t by th is m ethod. 13

a T h e p ercen tag es g iv en in the ex e rc ise o n ly a d d to 9 4 % . W e sh o u ld add an o th er c a te g o ry called “O th er” , w h ich w ill ac co u n t fo r the o th e r 6 % o f the resp o n ses. b E ith e r ty pe o f c h a rt is a p p ro p ria te . S in ce the d ata is alread y p resen ted as p e rc en tag es o f the w h o le g ro u p . w e c h o o se to use a p ie ch a rt, show n in the figure below .

c

A n sw ers w ill vary. 1

1.15

a

T h e total p e rc e n ta g e o f re sp o n se s g iv en in the table is o nly

(4 0 + 34 + 19)% = 93% . H en ee th ere are 7% o f the o p in io n s n o t re c o rd e d . w h ich sh o u ld g o into a c a te g o ry c a lle d “O th e r” o r “M o re than a few d a y s” . b Y es. T h e b ars are very c ió se to the c o rre c t prop o rtion s. c S im ilar to p rev io u s ex erc ise s. T he pie ch art is show n b elo w . T h e b a r c h a rt is

1.21

a S in ce th e v ariab le o f in terest c a n o nly take the v alú es 0 , 1, o r 2, th e c lasse s can be c h o sen as the in teg er v alú es 0 , 1, and 2. T he tab le b elo w sh o w s the c la sse s, th eir c o rre sp o n d in g freq u e n cie s and th eir relativ e freq u en cies. T h e relativ e freq u en ey h isto g ram is sh o w n below .___________ _______________________ V alu é

F req u en ey

R ela tiv e F req u en ey

0 1

5 9

2

6

.25 .45 .30

0.5

b U sin g th e tab le in p art a, the p ro p o rtio n o f m easu re m en ts g re a te r then 1 is the sam e as th e p ro p o rtio n o f “2 ” s, or 0.3 0 . 2

c T h e p ro p o rtio n o f m e a su re m e n ts less th an 2 is th e sam e as th e p ro p o rtio n o f “0 ” s a n d “ l ”s, o r 0 .2 5 + 0 .4 5 = .70 . d

T h e p ro b ab ility o f sele c tin g a “2 ” in a ra n d o m selectio n fro m th ese tw en ty

m e a su re m e n ts is 6 / 2 0 = 3 0 . e

T h e re are no o u tliers in th is re la tiv e ly sy m m e tric , m o u n d -sh ap e d d istrib u tio n .

a

T h e test sco res are g ra p h ed u sin g a stem an d le a f p lo t g e n e ra te d by Mi ni t ab .

Stem-and-Leaf Display: S co res Stem-and-leaf of Scores Leaf Unit = 1.0 2 5 8 9 (2) 9 7 3

5 6 6 7 7 8 8 9

N

=20

57 123 578 2 56 24 6679 134

b -c T h e d istrib u tio n is not m o u n d -sh a p e d , b u t is ra th e r b im o d al w ith tw o peaks c e n te re d aro u n d th e sco res 65 and 85. T h is m ig h t in d ícate th a t the stu d e n ts are d iv id ed in to tw o g ro u p s - th o se w h o u n d e rsta n d the m aterial a n d d o w ell on exam s, a n d th o se w ho d o not h av e a th o ro u g h co m m a n d o f the m aterial. a T h e d a ta ran g es fro m .2 to 5 .2 , o r 5.0 units. S in ce the n u m b e r o f class in terv als sh o u ld be b etw een five an d tw en ty , w e ch o o se to use e lev e n c la ss in terv als, w ith each c lass in terv al hav in g len g th 0 .5 0 (5 .0 /1 1 = .4 5 , w h ich , ro u n d e d to th e nearest c o n v e n ie n t fractio n , is .50). W e m u st n o w sele ct in te rv al b o u n d a ries su c h th a t no m easu rem en t c an fall o n a b o u n d a ry p o in t. T h e su b in te rv a ls .1 to < .6 , .6 to < 1.1, an d so on, are c o n v en ie n t a n d a ta lly is co n stru c te d . ^ ^ _______________________ R ela tiv e freq u en cy , C la ss T a lly C la ss í B o u n d a r ie s i fjn 1 2

0.1 to < 0.6 0 .6 t o < 1.1

3

1.1 to < 1.6

4

1.6 to < 2.1 2.1 to < 2.6

5 6 7 8

2 .6 to < 3.1 3.1 to < 3.6 3.6 to < 4.1

11111 11111 11111 11111 11111 11111 1111

11111 11111

10 15

.167 .250

11111

15

.250

11111

10 4

.167 .067 .017 .033 .017

1

1

11

2

1

1

9 10

4.1 to < 4 .6 1 4 .6 to < 5.1 5.1 to < 5.6 1 11 T h e re lativ e freq u en cy h isto g ra m is sh o w n below .

3

1

.017

0

.0 0 0

1

.017

15/60

10/60

IV I

5/60

0

a T h e d is trib u tio n is sk ew ed to th e rig h t, w ith several u n u su ally large o b serv atio n s. b F o r so m e reaso n , one p erso n had to w ait 5 .2 m inutes. P e rh ap s the su p e rm a rk e t w as u n d e rsta ffe d th at day , o r th ere m ay have been an u n u su ally larg e n u m b e r o f c u sto m e rs in the store. c T h e tw o g rap h s co n v e y the sam e in fo rm atio n. T h e stem and le a f p lo t allo w s us to actu a lly re c re a te the actual d a ta set, w hile the h isto g ram d o e s not. 1.35

a H isto g ram s w ill v a ry fro m stu d e n t to student. A ty p ical h isto g ram , g e n e ra te d by M i n i í a b is sh o w n on th e n e x t p ag e. ^

b

S in ce 2 o f the 2 0 p lay ers h av e av e ra g es a b o v e 0 .4 0 0 , the c h an ce is 2 o u t o f 2 0 or

2 /2 0 = 0 .1 . 1.39

T o d e te rm in e w h e th e r a d istrib u tio n is lik ely to be skew ed, lo o k fo r the lik e lih o o d o f o b serv in g e x trem e ly large o r e x trem e ly sm all v alú es o f the v ariab le o f in terest. 4

a T h e d istrib u tio n o f n o n -se c u re d loan sizes m ig h t be sk e w e d (a few ex trem ely larg e lo an s are p o ssib le). b T h e d istrib u tio n o f sec u re d lo an siz es is n o t lik ely to co n ta in u n u su ally larg e o r sm all valúes. c N o t likely to be skew ed. d N o t likely to be skew ed. e I f a p ack ag e is d ro p p e d , it is lik ely th at all th e sh e lls w ill be b ro k en . H en ee, a few larg e n u m b e r o f b ro k en sh ells is p o ssib le. T h e d istrib u tio n w ill be skew ed. f If an an im al h as o n e tick , he is lik ely to have m ore th an on e. T h ere w ill be som e “0 ”s w ith u n in fected rab b its, an d th en a la rg e r n u m b e r o f larg e valúes. T he d istrib u tio n w ill n o t be sy m m etric. 1.43

a S tem an d le a f d isp la y s m ay v ary fro m stu d en t to stu d en t. T h e m o st o b v io u s ch o ice is to use the ten s d ig it as th e stem and the o n es d ig it a s the leaf. 7 | 8 9 8 | 0 1 7 9| 0 1 2 4 4 5 6 6 6 8 8 101 1 7 9 11

|

2

T h e d isp lay is fairly m o u n d -sh a p e d , w ith a larg e p e a k in the m iddle. 1.47

A n sw e rs w ill vary fro m stu d en t to stu d e n t. T h e stu d en ts sh o u ld notice th at the d istrib u tio n is sk ew ed to th e rig h t w ith a few p re sid e n ts (T ru m an , C le v e la n d , and F.D . R o o sev elt) e a stin g an u n u su ally larg e n u m b e r o f vetoes._________________

Vetoes

1.51

a T h e p o p u la r v o te w ith in e a c h state sh o u ld vary d e p e n d in g on th e size o f the state. S ince th ere are several v ery larg e States (in p o p u la tio n ) in the U n ite d S tates, the d istrib u tio n sh o u ld be sk ew ed to the right. b -c H isto g ram s w ill v ary fro m stu d e n t to stu d en t, b u l sh o u ld rese m b le the h isto g ram g en erated b y M i n i t a b in the figure b elo w . T h e d istrib u tio n is in d eed sk ew ed to the rig h t, w ith tw o o u tliers - C a lifo rn ia and N ew Y ork.

5

O

1000

2000

3000

4000

5000

Popubr Vote

1.55

a-b A n sw ers w ill vary fro m stu d e n t to student. T h e line c h art sh o u ld lo o k sim ila r to the o n e sh o w n b elo w .

c T h e p erce n ta g e o f p eo p le w ho w ere n o t w o rrie d w as risin g at a slo w rate until S e p te m b e r 11, 2 0 0 1 , at w h ich tim e the p erc e n ta g e s re v erse d th em selv es d ram atica lly . d T h e h o rizo n ta l ax is o n the w w w .g a llu p .c o m ch a rt is n o t an actu a l tim e line, so t h a tth e tim e fram e in w h ich th ese c h a n g e s o c c u r m ay be d isto rted . 1.59

a -b A n sw e rs w ill vary. A ty p ical h isto g ram is sh o w n b elo w . N o tic e the g a p s an d the b im o d al n atu re o f th e h isto g ram , p ro b ab ly due to the fact th at the sa m p le s w ere c o lle c te d at d ifferen t lo catio n s.

6

1.63

a -b T h e M i n i t a b stem a n d le a f p lo t is sh o w n b elo w . T h e d istrib u tio n is slig h tly sk ew ed to th e left.

Stem-and-Leaf Display: Percent Stem-and-leaf of Percent Leaf Unit = 1.0 1 2 3 4 6 12 20 (11) 20 12 4 1

0 0 1 1 1 1 1 2 2 2 2 2

N

=51

7 8 0 3 45 666777 88888999 00000001111 22222333 44444555 677 9

7

c G e o rg ia (7 .5 ) and A rk a n sas (8 .0 ) have g aso lin e tax es th a t a re so m e w h a t sm a lle r than m o st, b u t th ey m ay n o t be “o u tlie rs” in the sen se th at th ey lie fa r aw ay fro m the rest o f the m e a su re m e n ts in the d a ta set. 1.67

a-b T h e d istrib u tio n is a p p ro x im a tely m o u n d -sh ap ed . w ith o n e unusual m e a su re m e n t, in the c la ss w ith m id p o in t at 100.8°. P e rh a p s the perso n w hose tem p e ra tu re w as 1 0 0 .8 has so m e so rt o f illn ess co m in g on? c T h e v alu é 9 8 .6 ° is slig h tly to the rig h t o f center.

1.69

a T h e d istrib u tio n is so m e w h a t m o u n d -sh a p e d (as m u ch as a sm all set c a n be); there a re no outliers. b 1.73

2 /1 0 = 0.2

a T h e re a re a few e x tre m ely large n u m b ers, in d icatin g th at th e d istrib u tio n is p ro b a b ly sk ew ed to the right. b-c T h e d istrib u tio n is in d eed sk ew ed right w ith three p o ssib le o u tlie rs - Y ah o o !, T im e W a rn e r and M S N -M icro so ft.

2: Describing Data with Numerical Measures 2.1

a T h e d o tp lo t sho w n b elo w p lo ts the five m e a su re m e n ts a lo n g the h o riz o n tal axis. S in ce th ere are tw o “ 1”s, the co rre sp o n d in g d o ts are p lac e d one abo ve the o th er. T he a p p ro x im ate ce n te r o f the d a ta a p p e a rs to be a ro u n d 1 .

b T h e m ean is th e su m o f the m e a su re m e n ts d iv id ed b y the n u m b e r o f m e a su rem en ts, or _ = ^

= 0 + 5 + 1+ U 3 = m = 2

x ~ n 5 ~ 5 ~ T o c a lc ú la te the m ed ian , the o b se rv a tio n s are first ran k ed fro m sm a lle st to largest: 0, 1, 1, 3, 5. T h en since n = 5 ,th e p o sitio n o f th e m ed ian is 0 .5 (n + l) = 3 , and the m ed ian is th e 3rd ran k ed m e a su re m e n t, o r m = 1 .T h e m o d e is the m e asu re m e n t o c c u rrin g m o st freq u en tly , or m ode = 1. c T h e three m easu res in p art b are lo cated o n the d o tp lo t. S in ce the m e d ia n and m o d e are to the left o f th e m ean, w e co n c lu d e th a t th e m e asu re m e n ts are sk ew ed to the right. 2.5

a A lth o u g h th ere m ay be a few h o u se h o ld s w h o o w n m o re th an one D V D p lay er, th e m a jo rity sh o u ld o w n e ith e r 0 o r 1. T h e d istrib u tio n sh o u ld be slig h tly sk ew ed to th e right. b Since m o st h o u se h o ld s w ill h av e o n ly o n e D V D p la y e r, w e g u ess th at the m o d e is 1 . c T h e m ean is Z jL = 1 + 0 + - + , a 27 n

25

25

T o c a lc ú late the m ed ian , the o b se rv a tio n s are first ra n k ed fro m sm a lle st to largest: T h e re are six Os, th írteen ls , fo u r 2s, and tw o 3s. T h en sin ce n = 25 , the p o sitio n o f 9

th e m ed ia n is 0 .5 (n + 1 ) = 13, w h ich is the 13th ran k ed m ea su re m en t, o r m = 1 . T h e m o d e is the m ea su rem e n t o c c u rrin g m o st freq u en tly , or m ode = 1. d T h e relativ e freq u e n cy h isto g ram is sh o w n b elo w , w ith th e th ree m easu res su p e rim p o se d . N o tic e th at th e m ean falls slig h tly to the rig h t o f the m ed ian an d m o d e , in d ic a tin g th at th e m e a su re m e n ts are slig h tly sk ew ed to the right.

0

2 .9

1 m edian m ean m o d e.

2

3

VCRs

T h e d istrib u tio n o f sp o rts sa la rie s w ill be sk ew ed to the rig h t, b e c a u se o f the v ery h igh salarie s o f so m e sp o rts fig u res. H en ee, the m ed ian salary w o u ld be a b e tte r m easu re o f c e n te r th an the m ean.

n b

5

C reate a tab le o f d iffe re n c e s, ( x ¡ - x ) and th e ir sq u a res, (*. - T ) 2 . X¡

x¡ - x

(* /-* )

2

- 0 .4 - 1 .4 - 1 .4

0 .1 6 1.96 1.96

3 5

0 .6 2 .6

0.36 6.76

T otal

0

1 1 .2 0

1 1

Then^ j 3 _ ! ( * , - ? ) • _ ( 2 - 2 A ) 2 -i------H 5 - 2 .4 ): .

n -1 c

4

11.20

. „

4

T h e sam p le sta n d a rd d ev ia tio n is the p o sitiv e sq u are ro o t o f the v arian c e or s = V 7 = V l 8 = 1 .6 7 3

C alcú late ' Z x f = 2 2 + 1 2 + --- + 5 2 = 4 0 . T h en

10

_

n n- 1

1 - = — 4

4

= 2.8 a n d s = y f ? = - J l .8 = 1.673 .

T h e resu lts o f p arts a and b are identical. 2.17

a

T h e ran g e is R = 2 .3 9 - 1 .2 8 = 1 .1 1 .

b

C a lcú late I *,2 = 1,282 + 2 .3 9 2 + ••• + 1 . 5 l 2 = 15.415 . T h en ^ , ( I x )2 (8 .5 6 )2 Z*?-15.451 —-— —1— 7. n?8 — ---------"— = ------------------- 5— = ^ 6 0 2 8 _ 1 9 0 0 7 n —1 4 4

an d 5 = 7 7 = V. 19007 = .436 c

T h e ran g e, R = 1 .1 1 , is 1 .1 1/.436 = 2.5 stan d ard d ev iatio n s. T h e range o f the d a ta is R = 6 - 1 = 5 a n d the ra n g e ap p ro x im a tio n w ith n = 10 is

2 .1 9

s ~ — = 1.67 3 T h e stan d a rd d ev iatio n o f the sam p le is

U

-

& * ,f n

(3 2 ) 130 —

_

= 7 3 .0 6 6 7 = 1 .7 5 1 n- 1 w h ic h is v ery c ió se to the estim ate fo r p art a. c -e F ro m the d o tp lo t on the next p ag e , y o u c a n see th a t th e d a ta set is n o t m o u n d sh ap ed . H enee y o u can use T c h e b y s h e ff s T h e o re m , b u t n o t the E m p iric a l R u le to d e scrib e the data. =V 7 =

2.21

a

T h e interv al from 4 0 to 6 0 rep re se n ts f i ± g = 5 0 ± 10. S in ce the d istrib u tio n is

re la tiv e ly m o u n d -sh ap e d , the p ro p o rtio n o f m e a su re m e n ts b e tw een 4 0 a n d 6 0 is 6 8 % acc o rd in g to the E m p irical R ule and is sh o w n on th e next page 11

b

A gain, u sin g the E m p iric a l R ule, the in terv al / / ± 2 o = 5 0 ± 2(10) o r b e tw een 30

a n d 7 0 c o n tain s ap p ro x im ate ly 9 5 % o f the m e asu rem en ts. c R efer to the figure below .

S ince a p p ro x im ate ly 6 8 % o f the m easu re m e n ts are b e tw een 4 0 a n d 60, the sy m m etry o f the d is trib u tio n im p lies th a t 34% o f the m e a su rem en ts are b e tw e e n 5 0 a n d 60. S im ilarly , sin ce 9 5 % o f the m easu re m e n ts are b e tw een 3 0 a n d 70, a p p ro x im ately 4 7 .5 % are b etw een 3 0 and 50. T h u s, the p ro p o rtio n o f m e asu re m e n ts b e tw e e n 3 0 and 6 0 is 0 .3 4 + 0 .4 7 5 = 0 .8 1 5 d F ro m the fig u re in p art a , the p ro p o rtio n o f the m e a su re m e n ts b e tw een 5 0 an d 60 is 0 .3 4 an d the p ro p o rtio n o f the m easu re m e n ts w h ich are g re a te r th an 5 0 is 0.50. T h e re fo re , the p ro p o rtio n th at are g rea te r than 6 0 m u st be 0 . 5 - 0 . 3 4 = 0.16 2 .2 5 A cco rd in g to the E m p irical R u le, if a d istrib u tio n o f m e a su re m e n ts is ap p ro x im a te ly m o u n d -sh ap ed . 12

a ap p ro x im a te ly 6 8 % o r 0 .6 8 o f the m easu re m e n ts fall in the interval H ± o = 1 2 ± 2 .3 o r 9.7 to 14.3 b ap p ro x im a te ly 9 5 % o r 0 .9 5 o f the m e a su rem en ts fall in the interval / / ± 2 a = 12 ± 4 .6 o r 7 .4 to 16.6 c ap p ro x im a te ly 9 9 .7 % o r 0 .9 9 7 o f the m easu re m e n ts fall in the interval /y ± 3< r = 1 2 ± 6 .9 o r5 .1 to 18.9 T h e re fo re , ap p ro x im a te ly 0 .3 % o r 0 .0 0 3 w ill fall o u tsid e this interval. 2.31

a ------------C la ss i

W e ch o o se to use 12 c la sse s o f len g th 1.0. T h e tally and the relativ e freq u en cy C la ss B o u n d a r ie s

1 2

3 4 5 6

7

T ally 1

1

1

1

5

111

3

1/70 1/70 3 /70

4 to < 5 to < 6 to < 7 to < 8 to <

6

11111

7

11111

5 5

5 /70 5 /70

8

1 1 1 1 1 1 1 1 1 1 11

12

9

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 111

18 15

12/70 18/70

8

9 to < 10

11111 11111 11111

9

1 0 to < 11

11111 1

10

11 to < 12

111

11

12 to < 13 13 to < 14

12

R ela tiv e fr e q u e n c y ,///!

fi

2 to < 3 3 to < 4

15/70 6 /70 3 /70

6

3

1

0

0

1

1/70

20/70

L «

■E

10/70

1

5

10

15

TRH3

b

V* y 1 C a lc ú la te n = 70, X x = 5 4 1 and X x f = 4 4 5 3 . T h en x = — — = ------ = 7 .7 2 9 is n 70

an e stím ate o f / / . c

T h e sam p le sta n d a rd d ev ia tio n is

13

U - í 541): J =V \

n- 1

—

= \ -------------- ^ — = V 3.9398 = 1.985 \ 69

T h e th ree in terv als, T ± ^ f o r k = 1,2 ,3 are c a lc u la te d b elo w . T h e ta b le sh o w s the actu al p erce n ta g e o f m easu re m e n ts fallin g in a p a rtic u la r in terv al as w ell as the p ercen tag e p re d ic te d by T c h e b y sh e ff’s T h e o re m a n d the E m p iric a l R u le. N o te that the E m p iric a l R u le sh o u ld be fairly accu rate, as in d icated by the m o u n d -sh a p e o f the h isto g ram in p a rt a. ______________________________________________________________ k

2.35

x±ks

Interval

F ractio n in Interval

T c h e b y sh e ff

E m p irical R ule

1

7 .7 2 9 ± 1 .9 8 5

5 .7 4 4 to 9.714

5 0 /7 0 = 0.71

at le a st 0

- 0 .6 8

2

7 .7 2 9 ± 3 .9 7 0

6 7 /7 0 = 0 .9 6

at least 0 .7 5

-0 .9 5

3

7 .7 2 9 ± 5 .9 5 5

3 .7 5 9 to 11.699 1.774 to 13.684

7 0 /7 0 = 1.00

at le ast 0 .8 9

- 0 .9 9 7

a

C a lcú la te R = 2 .3 9 - 1 .2 8 = 1.11 so th a t s - R / 2 . 5 = 1.11 /2 .5 = .444 .

b

In E x ercise 2 .1 7 , w e calc u la te d X * . = 8 . 5 6 and

I *,2 = 1.282 + 2 .3 9 2 + --- + 1.512 = 1 5 .4 1 5 . T h en

■> (

Z

I xf-±

,2

* (8. 56)" 15.451 - L

n- 1

.76028

4

4

= .1 9007

an d 5 = V 7 = V -19007 = .4 3 6 , w h ich is v ery ció se to o u r e stím a te in p a rt a. 2 .3 9

a

T h e d a ta in this ex ercise have b ee n arran g e d in a fre q u en cy table. 0 1 2 3 4 5 6 7 8 9 10 Xi 1 0 5 3 2 1 1 1 0 0 1 1 fi

U sin g the freq u en cy tab le and the g ro u p e d fo rm u las, c a lc ú la te ' L x if = 0(10) + 1 (5 ) H------1-10(1) = 51 X * , 2/; = 0 2(10) + 12( 5 ) h------1-10 2 (1) = 2 93

T hen I = I iA n U

f -

= 51=204 25

(Z xií)2

293-

(5 1 )1

s~ = n- 1

24

^ - = 7 .8 7 3 and

s = y ¡ l .873 = 2 .8 0 6 . b-c

T h e three in terv als x ± k s fo r k = 1,2,3 are ca lc u la te d in th e tab le a lo n g w ith the

actual p ro p o rtio n o f m e a su re m e n ts fa llin g in th e in terv als. T c h e b y s h e ff s T h e o re m is

14

sa tisficd an d th e a p p ro x im a tio n g iv en b y the E m p irica l R ule are fairly ció se fo r k = 2 an d k = 3 . k

2.41

x±ks

In terv al

F rac tio n in In terv al

T c h e b y sh e ff

E m p irical R ule = 0 .6 8

1

2 .0 4 ± 2 .8 0 6

- 0 .7 6 6 to 4 .8 4 6

2 1/25 = 0 .8 4

at least 0

2

2 .0 4 ± 5 .6 1 2

- 3 .5 7 2 to 7 .6 5 2

2 3 /2 5 = 0 .9 2

at least 0.75

= 0.95

3

2 .0 4 + 8.418

- 6 .3 7 8 to 10.458

2 5 /2 5 = 1.00

at least 0.89

- 0 .9 9 7

T h e d ata h av e alre a d y b e e n so rte d . F in d the p o sitio n s o f the q u artiles. a n d the m ea su re m e n ts th a t are ju s t ab o v e and b e lo w those p o sitio n s. T h en find the q u artiles by in terp o latio n . S o rted D a ta Set

Qi

P o sitio n o f

A bove

Q.

an d b elow

1, 1.5, 2 , 2 , 2.2

.2 5 ( 6 ) = 1.5

1 an d 1.5

1.25

0, 1.7, 1.8, 3.1, 3.2,

.2 5 (1 2 ) = 3

N one

1.8

.2 5 (9 ) = 2.25

.30 an d .35

.30 + .2 5 (.0 5 ) = .3125

7, 8 , 8 .8 , 8.9, 9, 10 .23, .30, .35, .41, .56, .58, .76, .80

P o sitio n o f Q 3

q3

2 an d 2 .2

2.1

.7 5 (1 2 ) = 9

N one

8.9

.7 5 (9 ) = 6.75

.58 an d .76

.58 + .7 5 (.1 8 ) = .7150

.7 5 (6 ) = 4.5

2 .4 5

A b o v e an d b elow

T h e o rd e re d d a ta are: 2, 3 ,4 , 5 ,6 , 6 , 6 , 7 , 8 , 9 , 9, 10, 22 F o r « = 1 3 , the p o sitio n o f th e m e d ia n is 0 .5 (« + 1 ) = 0.5(13 + 1 ) = 7 and m = 6 . T h e p o sitio n s o f th e q u a rtile s are 0 .2 5 (n + l) = 3.5 and 0 .7 5 (n + l) = 10.5 , so that Q¡ = 4.5, Q y = 9, and I Q R = 9 - 4.5 = 4.5 . T h e l owe r a n d u p p e r f e n c e s are: Qx - 1 . 5 I Q R = 4.5 - 6.75 = -2 .2 5 Q, + 1 .5 /0 /? = 9 + 6.75 = 15.75 T h e valu é x = 2 2 lies o u tsid e th e u p p e r fen ce and is an outlier. T h e b o x p lo t is show n b elow . T h e lo w e r w h isk e r co n n e c ts th e b o x to the sm allest valué th at is n o t an o u tlier, w hich h ap p en s to be the m ín im u m valu é, x = 2 . T h e u p p e r w h isk er co n n ec ts the box to the la rg e st v alu é th a t is n o t an o u tlie r o r x = 1 0 .

15

2 .4 9

a

F o r n = 18 , the p o sitio n o f the m edian is 0 .5 (n + 1 ) = 9.5 a n d the p o sitio n s o f the

q u a rtiles are 0 .2 5 (n + 1) = 4 .7 5 an d 0 .7 5 (n + l) = 14.25 . T h e lo w er q u a rtile is Va the w ay b e tw een the 4 lh an d 5 111 m easu rem e n ts an d th e u p p e r q u artile is Va th e w ay b etw een the 14Ih and 15,h m easu rem en ts. T h e so rte d m ea su re m e n ts are sh o w n below . F avre: 10, 12, 13, 14, 15, 15, 18, 19, 21, 22, 22, 23, 23, 23, 23, 25, 25, 26 M cN a b b : 9, 10, 11, 15, 15, 16. 16, 17, 18, 18, 18, 18, 19, 21, 21, 2 3 , 24, 27 F o r B re tt F av re, m = (21 + 2 2 ) / 2 = 2 1 . 5 , 0 , = 1 4 + 0 .7 5 ( 1 5 - 1 4 ) = 14.75 and 0 3 = 23 + 0 .2 5 ( 2 3 - 2 3 ) = 2 3 . F or D o n o v an M cN ab b , m = (18 + 1 8 ) / 2 = 18 , 0 , = 15 + 0 .7 5 ( 1 5 - 1 5 ) = 15 and 0 3 = 2 1 + 0 .2 5 ( 2 1 - 2 1 ) = 2 1 . T h en the fiv e -n u m b e r su m m arie s are M in F av re M cN abb b

10

9

Q. 14.75 15

M edian 21.5 18

03 23 21

M ax 26 27

F o r B re tt F av re, calc ú late I Q R = 0 , - 0 , = 2 3 - 1 4 .7 5 = 8.25 . T h e n th e l o w e r

a n d u p p e r f e n e e s are: 0 , - 1 .5I Q R = 14.75 - 1 2 .3 7 5 = 2.375 0 , + 1 . 5 / 0 / ? = 2 3 + 1 2 .3 7 5 = 3 5 .375 F o r D o n o v an M cN a b b , calc ú late I Q R = 0 , - 0 , = 2 1 - 1 5 = 6 . T h e n the l o w e r a n d u p p e r f e n c e s are: 0 ,- 1 .5 /0 /? = 1 5 -9 = 6 0 , + 1 .5 I QR = 21 + 9 = 30 T h e re are no o u tlie rs, and the box p lo ts are sh o w n on th e n ex t p ag e.

16

c A n sw e rs w ill vary. T h e F av re d istrib u tio n is sk ew ed left, w h ile the D o n o v an d istrib u tio n is ro u g h ly sy m m e tric , p ro b ab ly m o u n d -sh ap ed . T he M cN a b b d istrib u tio n is slig h tly m o re v ariab le ; F av re h as a h ig h e r m ed ian n u m b e r o f co m p le te d passes. .53

A n sw e rs w ill vary. T h e stu d en t sh o u ld no tice the o u tliers in the fem ale g ro u p , th at the m ed ian fem ale te m p e ra tu re is h ig h e r th an th e m ed ian m ale tem p eratu re.

.55 24 26 27

T h e o rd e re d G en eric 25 25 26 26 28 28

sets are sh o w n b elow : 25 26 28

26 27

22

25 28

24 25 28

S u n m a id 24 24 27 28 29 30

24 28

F o r n = 1 4 , the p o sitio n o f the m e d ia n is 0 .5 (/i + l) = 0.5(14 + 1) = 7.5 and the p o sitio n s o f the q u a rtile s are 0 .2 5 (n + 1 ) = 3.75 a n d 0 .7 5 (n + 1 ) = 11.25 , so that G en eric:

m = 26, Qx= 25, Qy = 27.25, and I Q R = 27.25 - 25 = 2.25

S u n m a id :

m = 26, (2, = 24, ’(2 = 1 9 8 7 6 ; I * , ? , = 3 2 1 3 6

T h en the c o v a rian c e is

sn = *

—

^

= 220.78571

n —1

T h e sam p le sta n d a rd d e v ia tio n s are sx = 17.010501 a n d $v = 1 3 .3 7 1 0 7 7 5 so that r = 0 .9 7 1 . b S in ce the c o rre la tio n c o e ffic ien t is so ció se to 1, the stro n g c o rre la tio n in d icates that th e se co n d an d q u ic k e r test c o u ld be used in p lace o f th e lo n g er test-in te rv ie w . 3 .4 3

a

C a lcú late

, 2 = 8 ; I * . = 4 5 1 ; L y , = 5 5 5 ; X x 2 = 2 9 ,6 1 9 ; X y f = 4 3 ,2 0 5 ; I x , ^ = 3 5 , 0 8 2 . T h en the c o v a ria n c e is

=

= 541.9821 a; - 1

T h e sam p le sta n d a rd d e v ia tio n s are s x = 2 4 .4 7 7 0 and s y = 25.9171 so that r = 0 .8 5 4 4 . b-c T h e sc a tte rp lo t sh o u ld lo o k lik e the one sh o w n on th e next p a g e . T h e co rre la tio n c o e fficie n t sh o u ld be c ió se to r = 0 .8 5 . T h e re is a stro n g p o sitiv e trend.

28

29

4: Probability and Probability Distributions 4.1

a T h is e x p e rim e n t sam p le sp ace fo r this E\ . O b serv e E 2: O b serv e £ 3: O b serv e b E v en ts m anner: A: B: C: c

in v o lv e s to ssin g a sin g le die and o b serv in g the o u tco m e. e x p e rim e n t c o n sists o f th e fo llo w in g sim p le eyents: a 1 £ 4: O b se rv e a 4 a 2 £ 5: O b serv e a 5 a 3 £ 6: O b serv e a 6

T he

A th ro u g h F are c o m p o u n d e v e n ts a n d are c o m p o se d in the fo llo w in g ( E 2) ( E 2, E 4 , E b) ( £ 3, £4, £5, £ 6)

D: ( E 2) E: ( E 2, E ^ E 6) F: c o n ta in s no sim p le e v en ts

S in ce the sim p le e v e n ts £ ,- ,/ = 1, 2, 3,

6 are eq u a lly likely, £ ( £ , ) = 1/6 .

d T o find the p ro b a b ility o f an e v en t, w e sum the p ro b a b ilitie s a ssig n ed to the sim ple ev e n ts in th at even t. F o r ex am p le,

6

/>( A) = /> (£ ,) = j

S im ilarly , P ( D ) = \ / 6 : P ( B ) = P ( E ) = £ ( £ 2) + £ ( £ 4) + £ ( £ 6) = - = - ; and

6

2

4 2 P ( C ) = — = — . S in ce e v e n t £ c o n tain s no sim p le e v en ts, P ( £ ) = 0 .

4 .5

a T h e e x p e rim e n t c o n sists o f c h o o sin g th ree c o in s at ra n d o m fro m fo u r. T h e o rd er in w h ich th e co in s are d raw n is u n im p o rta n t. H en ee, e ac h sim p le e v e n t c o n sists o f a trip let, in d icatin g the th ree c o in s d ra w n . U sin g the letters N , D, Q , a n d H to rep resen ! the n ickel, d im e , q u arter, a n d h alf-d o lla r, resp ec tiv e ly , the fo u r p o ssib le sim p le ev en ts are listed below . £ , : (N D Q ) £ 2: (N D H ) £ 3: (N Q H ) £ 4: (D Q H ) b T h e ev en t th at a h a lf-d o lla r is c h o sen is a sso c ia te d w ith th e sim p le e v e n ts £ 2, £3, and £4. H en ee, P lc h o o se a h alf-d o llar] = £ ( £ , ) + P ( E , ) + P ( £ 4) = - + - + - = 4 4 4 4 since e ach sim p le e v en t is eq u a lly likely. c

T h e sim p le e v en ts a lo n g w ith th e ir m o n e ta ry v a lú es follow : E^ N D Q $ 0 .4 0 0.65 e 2 NDH 0 .8 0 Ei NQH 0.85 £4 D QH 31

H en ee, P [to tal am o u n t is $ 0 .6 0 o r m ore] = P ( E 2) + P ( E y) + P ( E 4 ) = 3 /4 . 4.9

T h e four p o ssib le o u teo m e s o f the e x p e rim e n t, or sim p le ev en ts, a re re p re sen te d as the c ells o f a 2 x 2 table, and have p ro b a b ilitie s as g iv en in the table. a R [adult ju d g e d to need g lasses] = .44 + .14 = .58 b P [a d u lt need s g la sse s b u t d o e s not u se them ] = . 14 c F [a d u lt u se s g lasses] = .44 + .02 = .46

4 .1 3

a E x p e r i m e n t : A ta s te r tastes a n d ra n k s th re e v arieties o f tea A , B , and C, a cco rd in g to p refe re n ce . b S im p le ev en ts in S are in trip let form . E l : (1 ,2 ,3 )

E 4 : (2 ,3 ,1 )

E 2 : (1 ,3 ,2 )

£ 5 : (3 ,2 ,1 )

E y : (2 ,1 ,3 )

£ 6 : (3 ,1 ,2 )

H ere 1 is assig n ed to the m o st d e sira b le , 2 to the n ex t m o st d e sirab le , an d 3 to the least d esirable. c D efin e the ev en ts D: v ariety A is ran k e d first F: v ariety A is ra n k e d third T hen P ( D ) = /> (£ ,) + P ( E 2) = 1/6 + 1 /6 = 1/3 T h e p ro b ab ility th at A is least d e sira b le is £ ( £ ) = P ( E 5) + P Í E , ) = 1/6 + 1 /6 = 1/3 4.1 7

U se th e m n R ule. T h e re a re 10(8) = 8 0 p o ssib le pairs.

4.21

8' S in ce o rd e r is im p o rtan t, you use p e r m u t a t i o n s and p \ = —j = 8 (7 )(6 )(5 )(4 ) = 6 7 2 0 .

4.2 5

S in ce o rd e r is u n im p o rtan t, y o u use c o mb i n a t i on s a n d C í° = 3! 17!

4 .2 9

^ = 120. 3(2)(1)

a E ach stu d en t has a c h o ic e o f 52 card s, sin ce the c a rd s are re p la c e d b etw een selectio n s. T h e m n R ule allo w s y o u to fin d the to ta l n u m b e r o f co n fig u ra tio n s fo r th ree stu d en ts as 5 2 (5 2 )(5 2 ) = 1 4 0 ,6 0 8 . b N o w e a c h stu d en t m u st p ic k a d iffe re n t card . T h a t is, the first stu d en t h as 52 ch o ice s, but the se co n d and th ird stu d en ts h av e o n ly 51 and 5 0 ch o ic e s, re sp ectiv ely . T h e total n u m b e r o f c o n fig u ra tio n s is fo u n d u sin g th e m n R u le o r th e rule for p erm u tatio n s: m n t = 5 2 (5 1 )(5 0 ) = 1 3 2 ,6 0 0

or

£ 5,2 = — = 1 3 2 ,6 0 0 . 49!

c L et A be the ev en t o f in terest. S in ce th e re are 52 d iffe re n t c a rd s in the deck, there are 52 co n fig u ra tio n s in w h ich all th ree stu d e n ts p ic k the sa m e c ard (o n e fo r

32

e ach c a rd ). T h a t is, th ere a re nA = 5 2 w ay s fo r the ev en t A to o cc u r, o u t o f a total o f N = 1 4 0 ,6 0 8 p o ssib le c o n fig u ra tio n s from p a rt a. T h e p ro b a h ility o f in te re st is P ( A ) = ^ - = — — — = .00 0 3 7 N 1 4 0,608 d

A g ain , let A be th e e v e n t o f in terest. T h e re are n A = 1 3 2 ,6 0 0 w ay s (fro m p a rt b)

fo r th e e v e n t A to o ccu r, o u t o f a total o f N = 1 4 0 ,6 0 8 p o ssib le c o n fig u ra tio n s froTn p art a, a n d th e p ro b ah ility o f in te re st is

^9 4 3

= ^ = 1 3 2 ,6 0 0 N .33

N o tic e that a sa m p le o f 10 n u rses w ill be the sam e n o m atter in w h ich o rd e r th e y w ere selected . H en ee, o rd e r is u n im p o rta n t an d c o m b in a tio n s are used. T h e n u m b e r o f sa m p les o f 10 se lected from a total o f 9 0 is Cm =

1.37

1 40,608

90i 2 .0 7 5 9 0 7 6 Í1 0 19) - = -------------- = 5 .7 2 0 6 4 5 (10 ) 1 0 !8 0 ! 3 .6 2 8 8 (1 0 )

. V

’

T h e situ atio n p re se n te d h ere is an a lo g o u s to d ra w in g 5 item s fro m a j a r (the five m em b ers v o tin g in fa v o r o f th e p lain tiff). I f the ja r c o n tain s 5 re d and 3 w hite item s (5 w o m en an d 3 m en ), w h at is the p ro b a h ility th at all five item s are re d ? T h a t is, if th ere is no sex b ias, five o f the eig h t m e m b ers are ran d o m ly c h o sen to be th o se vo tin g fo r the plain tiff. W h a t is the p ro b a h ility th at all five are w o m en ? T h ere are 8' N = C \ = — — = 56 5!3! sim p le ev e n ts in the e x p e rim e n t, o n ly o n e o f w h ich re su lts in ch o o sin g 5 w o m en . H enee, P (fiv e w o m en ) = ^

4.41

4 .4 7

.

F o llo w th e in stru c tio n s g iv en in the M y P erso n al T ra in e r sectio n . T h e a n sw e rs are g iv en in th e table. n B)

P (A |B )

P (A )

P (B )

C o n d itio n s f o r e v e n ts A a n d B

.3

.4

M u tu ally e x clu siv e

0

.3

.4

In dep en d en t

.3 ( 4 ) = .12

.3 + .4 - ( .3 ) ( .4 ) = .58

.3

.1

.5

In dep en d en t

.1 ( 5 ) = .05

, l + . 5 - ( . l ) ( . 5 ) = .55

.1

.2

.5

M u tu ally ex clu siv e

0

.2 + .5 = .7

0

P (A

P (A u B ) .3

+ .4 = .7

0

R e fe r to the so lu tio n to E x e rc ise 4.1 w h ere the six sim p le e v en ts in the e x p e rim e n t are giv en , w ith P { E j ) = 1 /6 . a

5 = { E l , E 2, E i , E A, E 5, E 6} and P ( S ) = 6 /6 = 1 33

4 .4 8

b

P (A \B )= « A n B ) = yi= x v ' P(B) 1/3

c

B = { £ „ £ , } a n d P ( B ) = 2 /6 = 1/3

d

A n f i n C co n tain s no sim p le e v e n ts, and P ( A n B n C ) = 0

e

P ( A n B ) = P ( A \ B ) P ( B ) = 1(1/3) = 1/3

f

A n C co n tain s no sim p le e v en ts, a n d P ( A n C ) = 0

g

B n C c o n t a i n s n o sim p le e v e n ts, an d P ( f i n C ) = 0

h

A k j C = S an d P ( A u C ) = l

¡

8 u C = { £ l , £ , , £ 4, £ s , £ j an d P ( P u C ) = 5 /6

a

F ro m E x e rc ise 4 .4 7 , P (A n B ) = 1/3 , P ( A | B ) = 1 , P (A ) = 1/2 ,

P ( A | B ) * P ( A ) , so th at A a n d B are n o t in d ep en d e n t.

P ( A n B ) ^ 0 , so th a t A and

B are not m u tu ally ex clu siv e. b

P ( A |C ) = P ( A n C ) / P ( C ) = 0 , P (A ) = 1 /2 , P ( A n C ) = 0 . Since

P ( A | C ) * P ( A ) , A an d C a re d e p e n d e n t. S in ce P ( A r \ C ) = 0 , A and C are m u tu ally exclu siv e. 4 .4 9

4.5 5

a

S ince A an d B are in d ep en d e n t, P (A r \ B ) = P { A ) P { B ) = .4 (.2 ) = .08 .

b

P ( A u B ) = P ( A ) + P ( P ) - P ( A n f i ) = .4 + .2 - (.4 )(.2 ) = .52

D efin e th e fo llo w in g events: A: p ro je c t is ap p ro v ed fo r fu n d in g D: p ro jec t is d isa p p ro v e d fo r fu n d in g F o r the first g ro u p , P { \ ) = .2 and P ( D ,) = .8 . F o r the se co n d g ro u p , P [sam e d ecisió n as first g ro u p ] = .7 and P [rev ersal] = .3 . T h a t is, P ( A 2 \ A t ) = P ( D 1 \ D t ) = . l and

P ( \ \ D , ) = P ( D 7 \ A ) = .3 .

a

P ( A n 4 ) = P ( A ) í >(A 2 | A ) = -2(.7) = .14

b

P(D¡ n D 2 ) = P ( D ¡)Pi,D2 \ £),) = . 8(.7 ) = .56

c P (D , n A 2) + P ( A i n D 2) = P ( D ,)P (A , | D ,) + P ( A ,) P ( D 2 1A ,) = ,8(.3) + ,2(.3) = .30 4.5 9

F ix th e b irth events: A 2: A 3: A 4:

d a te o f th e first p e rso n en te rin g th e ro o m . T h e n d e fin e th e fo llo w in g se co n d p e rs o n ’s b irth d a y d iffe rs fro m th e first th ird p e rs o n ’s b irth d ay d iffe rs fro m the first a n d seco n d fo u rth p e rs o n ’s b irth d a y d iffe rs fro m all p rec e d in g

A n: nlh p e rs o n ’s b irth d a y d iffe rs fro m all p rece d in g 34

T hen P ( A ) = P ( A 2) P ( A ) - P ( A ii) =

' 364^1^ 363 p 65-/i + n v365J ,365J l 365 )

since at e a c h step, o n e less b irth d ate is a v a ila b le fo r selectio n . S in ce ev en t B is the c o m p le m e n t o f e v e n t A, P(B) = \ - P ( A )

4 .6 3

a

F o r n = 3 , P ( A ) = (3 6 4 )(3 6 3 ) = 9 9 l8 and P ( B ) = 1 - .9 9 1 8 (365)

b

For n = 4 ,

D e fin e

= .0082

P (A ) = (3 6 4 X3 6 3 X3 6 2 ) = 9 8 3 6 a n d P ( B ) = 1 - .9 8 3 6 = .0164 (3 6 5 ) A: sm o k e is d e te c te d by d e v ice A B : sm o ke is d e tec te d b y d e v ic e B

I f it is g iv e n th a t P ( A ) = .95, P ( B ) = .98, a n d P ( A n B ) = .94 .

4.69

a

P { A k j B ) = P { A ) + P ( B ) - P( A n B ) = .95 + .98 - .94 = .99

b

P ( A C n B c ) = l - P ( A k j B ) = 1 - .99 = .01

a

U se the L a w o f T o ta l P ro b a b ility , w ritin g

b

U se the re su lts o f p art a in the fo rm o f B a y e s’ R ule:

P ( A ) = P (5 , ) P ( A 15 ,) + P ( S 2 ) P ( A | S 2) = ,7 (.2 ) + .3(.3) = .23

P ( S | M ) = -------------W , ) P ( » | 5 . ) P ^P iA lS ^+ P iS ^P iA lS ,) For i = l ,

P(S, | A) = ----------------1 ,7 (.2 ) + .3(.3)

For

P(5, |A) = --------------

í

= 2,

2 L73

D efin e T h en

,7 (.2 ) + .3(.3)

= — = .6087 .23

=^

= .3913

.23

A: m a ch in e p ro d u c e s a d e fe c tiv e item B: w o rk e r fo llo w s in stru ctio n s P ( A \ B ) = .0\, P ( B ) = .90, P ( A \ B c ) = .03, P ( B C) = .1 0 . T h e p ro b a b ility o f

in terest is P(A) = P ( A r ^ B ) + P ( A n B c ) = P { A | B ) P ( B ) + P ( A | B C) P ( B C) = ,01(.90) + .0 3 (.1 0 ) = .012 1.77

T h e p ro b a b ility o f in te re st is P ( A \ H ) w h ich can be calc u la ted using B a y e s’ R ule a n d th e p ro b a b ilitie s g iv en in th e ex ercise.

35

P (A \H ) =

________________ P ( A ) P ( H \ A ) ________________ P ( A ) P ( H | A ) + P ( B ) P ( H | B ) + P ( C ) P ( H \ C) __________ .01(.90)____________

.009

= .3 1 3 0

.01(.90) + .0 05(.95) + .0 2 (.7 5 ) ” .028 75

4 .8 3

a

S ince o n e o f the re q u ire m e n ts o f a p ro b a b ility d istrib u tio n is th a t ^

p(x) = 1

X

w e need p (3 ) = 1 - (. 1 + .3 + .3 + . 1) = 1 - .8 = .2 b

T h e p ro b ab ility h isto g ra m is sh o w n below .

c

F o r th e ran d o m v ariab le x g iv e n h ere, H = E ( x ) = Z x p ( x ) = 0 (. 1) + 1(.3) H— + 4 (. 1) = 1.9

T h e v arian ce o f x is d efin ed as ( j 1 ^ E ^ x - p f Y ^ x - p ) 1 p ( x ) = (0 - 1.9)2(.1) + (1 - 1 .9 ) 2(.3) + --- + ( 4 - 1.9)2(.l) = 1.29 an d í t = V T 2 9 = 1 .1 3 6 . d U sin g th e tab le fo rm o f the p ro b a b ility d istrib u tio n g iv en in the ex ercise, P ( x > 2) = .2 + . l = . 3 . e 4 .8 7

P ( x < 3) = 1 - P ( x = 4) = 1 - . 1 = .9 .

a-b O n the first try, th e p ro b a b ility o f se lec tin g th e p ro p e r key is 1/4. I f th e key is not found oír the first try, th e p ro b a b ility c h a n g e s o n th e se co n d try. L et F d e n o te a failu re to find th e key a n d S d e n o te a su ccess. T h e ra n d o m v a ria b le is x, th e n u m b er o f keys tried b efo re the c o rre c t key is fo u n d. T h e fo u r a sso c ia te d sim p le ev e n ts are show n below . E ,:S

( x = \)

E 3: F F S

( x = 3)

E 2:F S

(^ = 2)

E 4: F F F S

(* = 4)

36

c -d

T hen

P ( l) = P ( x = 1) = P ( S ) = 1/4 p{ 2) = P(jc = 2) = P ( F 5 ) = P ( F ) P ( S ) = ( 3 /4 ) (1/3) = 1/4 p ( 3) = P ( * = 3) = P ( F F S ) = P ( F ) P ( F ) P ( S ) = ( 3 /4 ) ( 2 /3 ) (1 /2 ) = 1/4 p (4 ) = F(jc = 4 ) = P ( F F F S ) = P ( F ) P ( F ) P ( F ) P ( S ) = ( 3 / 4 ) ( 2 / 3 ) ( l / 2 ) ( l ) = 1/4 T h e p ro b a b ility d istrib u tio n and p robability h isto g ram follow . JC 2 4 1 3 1/4 1/4 1/4 1/4 p( x)

,91

L et x be the n u m b e r o f d rillin g s until the first su ccess (oil is stru ck ). It is g iv en that th e p ro b ab ility o f strik in g oil is P ( O ) = .1 , so that th e p ro b a b ility o f n o oil is P ( N ) = .9 a

p{ 1) = F [ o il stru ck on first d rillin g ] = P ( 0 ) = .1 p ( 2) = P [ o il stru ck on seco n d d r illin g ] . T h is is th e p ro b a b ility th at oil is not

fo u n d on th e first d rillin g , b u t is fo u n d on th e seco n d d rillin g . U sin g the M u ltip lic a tio n L aw , p ( 2 ) = P ( N O ) = (.9)(. 1) = .0 9 . F in ally , p ( 3) = P ( N N O ) = (.9 )(.9 )(. 1) = .081 . b -c F o r th e first su ccess to o c c u r on trial x, (jc - 1) failu res m ust o c c u r b e fo re the first success. T hus, p ( x ) = P ( N N N ... N N O ) = (.9)*"' (. 1) sin ce th ere are (jc - 1) N ’s in th e seq u en ce. T h e p ro b ab ility h isto g ram is sh o w n on the n ex t p age.

37

1

3

5

7

9

11

13

15

17

x

4 .9 5

T h e ran d o m v ariab le G , to tal g ain to th e in su ra n c e co m p a n y , w ill be D if th ere is no th eft, but D - 5 0 ,0 0 0 if th ere is a th eft d u rin g a g iv en y ear. T h e se tw o ev e n ts w ill o c c u r w ith p ro b a b ility .99 an d .01, resp ec tiv e ly . H e n ee, th e p ro b a b ility d istrib u tio n for G is g iv en below . T h e e x p e cte d g ain is G_____________ p ( G) D .99 E { G ) = l G p ( G ) = .99 D + .0 1 (D - 5 0 ,0 0 0 ) = D - 5 0 ,0 0 0 D - 5 0 ,0 0 0

.01

In o rd e r that E ( G ) = 1000 , it is n e cessary to h av e 1000 = D - 5 0 0 o r D = $ 1 5 0 0 . 4 .9 7

a S im ila r to E x e rc ise 4 .9 1. F o r the fírst n o n -b e lie v e r to be fo u n d on ca li jc, (x - 1) p eo p le w ho d o b eliev e in h eav en m u st be c a lle d b e fo re the first n o n -b e lie v e r is found. T hus, p ( x ) = P ( N N N . .. N N Y ) = ( .8 1)*"1(. 19) b A s w ith o th e r p h o n e su rv ey s, th ere is a lw ay s a p ro b le m o f n o n -re sp o n se p eo p le w ho d o not an sw er th e te le p h o n e o r d e c lin e to p a rtic ip a te in th e su rv ey . A lso, th ere is a p ro b lem o f tru th fu ln ess o f th e re sp o n se fo r a q u e stio n su c h as th is w h ich m ay be a sen sitiv e su b je ct fo r so m e p eo p le.

4 .101

D efin e the fo llo w in g events: A: w o rk er fails to re p o rt fraud B: w o rk er su ffers rep risal It is g iv en th at P ( B | A c ) = .23 and P ( A ) = .69 . T h e p ro b a b ility o f in te rest is P ( A C r \ B ) = P ( B | Ac ) P ( A r ) = .23(.31) = .0713

4 .1 0 5

T w o sy stcm s are se lected from sev en , th ree o f w h ich a re d efec tiv e . D en o te the seven sy stem s as G 1( G 2, G 3, G 4, D (, D 2, D 3 a c co rd in g to w h eth e r th e y are g o o d o r d efectiv e. 38

E ach sim p le e v en t w ill re p re se n t a p a rtic u la r p a ir o f sy stem s c h o sen fo r testin g , and th e sa m p le sp ace, c o n sistin g o f 21 p airs, is show n below .

G1G2 G jD | G,G4 G ,D 3 G2G4 G2D2

G2D3 g]d2

G4D1

G4D2 G1G3 D ,D 2 G 2G 3 D2D3 G3G4

G1G2 G3D1 G 2D, G3D3

G4D3 D ,D 3

N o te that th e tw o sy ste m s are d raw n sim u ltan e o u sly and th at o rd e r is u n im p o rtan t in id e n tify in g a sim p le e v en t. H e n ee, the p airs G |G 2 a n d G 2G! are not c o n sid e re d to re p re se n t tw o d iffe re n t sim p le e v en ts. T h e e v e n t A , “no d e fe c tiv e s are se lec ted ” , co n sists o f th e sim p le ev e n ts G |G 2, G1G3, G1G4, G 2G 3, G2G4, G3G4. S in ce the sy stem s are selected at ra n d o m , a n y p a ir has an eq u al p ro b a b ility o f b ein g selected . H en ee, the p ro b a b ility a ssig n e d to e a ch sim p le e v e n t is 1/21 and P ( A ) = 6 /2 1 = 2 /7 .

4 .1 0 9

4 9 + 43 + 34

126

= .4565 276 276 F: p e rso n has fo u r o r fiv e re latio n sh ip s D efine: S: p erso n h a s six o r m o re re latio n sh ip s T h en for the tw o p e o p le ch o sen fro m th e total 276,

a

P (c o ld ) =

P ( o n e F and one S ) = P ( F n S ) + P ( S n F ) 100 276

( 96 275 y

+

r 96 ^ í k x T

= .2530

^ 2 7 6 , 1 ,2 7 5 ,

. P ít h r e e o r f e w e r n c o l d ) 4 9 /2 7 6 49 P Í T h r e e o r fe w e r c o id ) = = — = ----- = . 3889 v 1 ’ P ( c o ld ) 1 2 6 /2 7 6 126 4 .1 1 3

D efin e th e fo llo w in g ev en ts: B j: clien t b u y s o n first c o n ta c t B 2: c lie n t b u y s o n se co n d co n tac t Since the c lie n t m ay b u y on e ith e r the first o f the seco n d co n tac t, the d e sired p ro b a b ility is

a

P [c lie n t w ill b uy ] = P [c lie n t b u y s o n first co n tac t] + P [clien t d o esn 't b u y on first, b u t buys on seco n d ] = P ( P , ) + ( l - P ( f í 1) ) P ( f i 2) = .4 + ( l - .4 ) ( .5 5 ) = .73 b T h e p ro b ab ility th a t th e c lie n t w ill n o t buy is o n e m in u s the p ro b a b ility th at the clien t w ill buy, o r 1 - .73 = .27 . 4 .1 1 7

E a c h hall can be c h o sen fro m the se t (4, 6 ) and there are th ree su ch b alls. H enee, th e re are a to tal o f 2 (2 )(2 ) = 8 p o te n tial w in n in g num b ers.

4 .121

a C o n sid e r a sin g le trial w h ich co n sists o f to ssin g tw o co ins. A m atch o ccu rs w h en e ith e r H H o r T T is o b serv ed . H enee, the p ro b a b ility o f a m atch on a sin g le trial

39

is P ( H H ) + P ( T T ) = 1 /4 + 1/4 = 1 /2 . L et M M M d e n o te the e v e n t “m atch on triáis 1, 2, and 3” . T h en P ( M M M ) = P ( M ) P ( M ) P ( M ) = (1 /2 )3 = 1/8. b

O n a sin g le trial the e v e n t A , “tw o tra ils are o b se rv e d ” h as p ro b a b ility

P ( A ) = P ( T T ) = 1/4 . H enee, in th ree triá is P ( A A A ) = P ( A ) P ( A ) P ( A ) = (1 /4 )' = 1/64 c T h is lo w p ro b a b ility w o u ld n o t su g g e st c o llu sio n , sin ce th e p ro b a b ility o f th ree m atch es is lo w o nly if w e assu m e th at e a c h stu d e n t is m erely g u e ssin g at e ac h an sw er. If the stu d en ts h av e stu d ied to g e th e r o r if th ey b o th k n o w th e c o rre c t an sw er, the p ro b ab ility o f a m atch on a sin g le trial is no lo n g er 1/2, b u t is su b sta n tia lly h igher. H enee, the o ccu rren c e o f th ree m atch es is no t u n u su al. 4 .1 2 5

D efin e the ev en ts:

A: the m an w aits five m in u te s o r lo n g er B : the w o m an w aits five m in u te s o r lo n g er

T h e tw o ev en ts are in d ep en d e n t, and P ( A ) = P ( B ) = .2 . a

P ( A r ) = l - / > M ) = .8

b

P ( A r f l r ) = / > ( / l c ) / > ( £ c ) = (.8 )(.8 ) = .64

c

P [ a t least one w aits five m in u te s o r lo n g er] = 1 - P [n e ith e r w a its five m in u te s o r lo n g er] = 1 - P ( A ( B c ) = 1 - . 6 4 = .36

4 .1 2 9

S ince th e first p o o le d test is p o sitiv e, w e a re in tereste d in th e p ro b a b ility o f req u irin g five single tests to d ete c t the d isease in the sin g le a ffe c te d p erso n . T h e re are (5 )(4 )(3 )(2 )(1 ) w ays o f o rd e rin g th e five tests, a n d th ere are 4 (3 )(2 )(1 ) w ay s o f o rd erin g the tests so th at the d ise a se d p erso n is g iv e n the fin al test. H en ee, the d e sired 4! 1 p ro b ab ility is — = - . If tw o p e o p le are d iseased , six tests a re n e e d e d if the la st tw o tests are g iv en to the d iseased p eo p le. T h ere are 3 (2 )(1 ) w ay s o f o rd e rin g the tests o f the o th e r th ree p eo p le and 2(1) w ays o f o rd e rin g the tests o f th e tw o d ise a se d p eo p le. H en e e, the p ro b a b ility 2 !3! 1 that six tests w ill be n eed ed is ------ = — . 5! 10

4 .1 3 3

a D efin e P: sh o p p e r p refers P epsi a n d C: sh o p p e r p refers C o k e . T h e n if th ere is actu a lly no d ifferen c e in th e taste, P (P ) = P (C ) = 1/2 and

b P (e x a c tly one p refers P ep si) = P ( P C C C ) + P ( C P C C ) + P ( C C P C ) + P ( C C C P )

40

4 .1 3 7

R efer to the T o ssin g D ice ap p le t, in w h ich the sim p le e v e n ts fo r th is e x p e rim e n t are d isp lay ed . E ach sim p le e v e n t h as a p artic u la r v alué o f T a sso c ia te d w ith it, and by su m m in g the p ro b a b ilitie s o f all sim p le e v e n ts p ro d u cin g a p artic u la r v alu é o f T , the fo llo w in g p ro b a h ility d istrib u tio n is o b tain ed . T h e d istrib u tio n is m o u n d -sh ap ed . a-b T 2

p(T) 1/36

T 8

3 4

2 /3 6 3 /36 4 /3 6 5 /36

9 10

5 6 7

2

4

11 12

p( T) 5/36 4 /3 6 3 /36 2 /36 1/36

6 /3 6

6

8 X

-______________

41

10

12

5: Several Useful Discrete Distributions F o llo w the in stru c tio n s in th e P erso n al T ra in e r sectio n . T h e an sw ers are sh o w n in the tab les below .

5.1

k

0

1

2

3

4

5

6

7

8

P ( x < k)

.000

.001

.011

.058

.194

.448

.745

.942

1.000

T h e P r o b le m

T hree o r less

L is t th e

W r ite th e

R e w r ite th e

F in d th e

V a lú e s o f x

p r o b a b ility

p ro b a b ility

p ro b a b ility

0, 1 , 2 ,3

P ( x < 3)

.058

3, 4 . 5, 6. 7, 8

P (x

> 3)

1 - P (x

< 2)

1 —.011 = .989

M ore th a n three

4 , 5, 6, 7, 8

P (x

> 3)

1 - P (x

4 ) = 1 - .382 = .6 1 8 . c

U se the resu lts o f p arts a and b. P ( x > 4 ) = 1 - P ( x < 4) = 1 - ( . 3 8 2 + .2 5 1) = .367 43

5.1 5

5 .1 9

d

F ro m p a rt c, P { x < 4 ) = P ( x < 3) + P ( x = 4 ) = .382 + .251 = .6 3 3 .

e

n = n p = 1 0 (.4 ) = 4

f

o = yj ñpq = 7 l0 ( .4 ) ( .6 ) = y ¡ 2 A = 1.549

a

P f j t < 12] = P [ x < 1 1] = .748

b

P [ x < 6] = .610

c

/>[ jc > 4] = 1 - ^ [ jc < 4] = 1 - .6 3 3 = .367

d

P [ x > 6] = 1 - P [ x ^ 5] = 1 —.034 = .966

e

P [ 3 < J t < 7 ] = P [ x < 6 ] - P [ ; t < 3 ] = .8 2 8 - .1 7 2 = .656

a

p (0 ) = C o °(-l)° (-9 )20 = .1 2 1 5 7 6 7

p (3 ) = C f ( - l ) ' (-9 )17 = .1 9 0 1 1 9 9

M D = C ?0(.1 )1(.9 )19 = .2 7 0 1 7 0 3

p (4 ) = C 4 ,( . l ) 4 ( .9 ) '6 = .0 8 9 7 7 8 8

^ ( 2 ) = C l° (-1 )2 (-9 )'8 = .2 8 5 1 7 9 8 so that

P [ x < 4] = p ( 0 ) + p ( 1) + p { 2) + p ( 3) + p ( 4 ) = .9 5 6 8 2 5 5 P [ x < 4] is re a d d ire c tly as .957.

b

U sin g T a b le 1, A p p e n d ix I,

c

A d d in g th e e n trie s fo r x = 0 ,1 ,2 ,3 ,4 , we h av e P [x < 4] = .9 5 6 8 2 6 .

d

p = n p = 2 0(. 1) = 2 a n d O = yj npq = V E 8 = 1.3416

e

F o r ¿ = 1, / / ± c r = 2 ± l .342 o r .658 to 3 .3 4 2 so that P [.658 < jc < 3 .3 4 2 ] = P [ 1 < x < 3] = .2702 + .2 8 5 2 + . 1901 = .7455 F or k = 2 , /j ± 2 ct = 2 ± 2 .6 8 3 or - .6 8 3 to 4 .6 8 3 so that P [ - . 6 8 3 < x < 4 .6 8 3 ] = P [ 0 < * < 4 ] = .9569 F o r k = 3 , p ± 2 o = 2 ± 4 .0 2 5 o r -2 .0 2 5 to 6 .0 2 5 so that P [- 2 .0 2 5 < jc < 6 .0 2 5 ] = P [ 0 < x < 6] = .9977

f 5.23

T h e resu lts are c o n siste n t w ith T c h e b y s h e ff s T h e o re m a n d th e E m p irical R ule.

D efin e x to be the n u m b er o f a la rm sy ste m s th at a re trig g e re d . T h en p = P [alarm is trig g e re d ] = .99 a n d n = 9 . S in ce th ere is a tab le av a ila b le in A p p en d ix I fo r n = 9 and p = .99 , y o u sh o u ld use it rath e r than th e b in o m ial fo rm u la to c a lc ú late the n e cessary p ro b ab ilities.

5 .2 5

a

P [ a t least o n e alarm is trig g e re d ] = P ( x > 1) = 1 - P ( x = 0 ) = 1 - . 0 0 0 = 1 .0 0 0 .

b

P [m o re th an se v en ] = P ( x > 7 ) = \ - P ( x < 7 ) = 1 - . 0 0 3 = .997

c

P [e ig h t o r fe w e r] = P ( x < 8) = .086

D efin e x to be the n u m b er o f c a rs th at are black. T h en p = P [b la c k ] = .1 an d ti = 25. U se T ab le 1 in A p p e n d ix I. 44

5 .2 9

a

/ >( jc > 5 ) = 1 - / >( jc < 4 ) = 1 - .902 = .098

b

P ( x < 6 ) = .991

c

P ( x > 4 ) = l - P ( x < 4 ) = 1 - .9 0 2 = .098

d

P ( x = 4 ) = P ( x < 4 ) - P ( x < 3 ) = .9 0 2 - .7 6 4 = .138

e

P ( 3 < jc < 5 ) = P ( x < 5 ) - P ( x < 2 ) = .9 6 7 - .5 3 7 = .430

f

P ( m o r e than 20 n ot b la c k ) = P ( le s s th an 5 b la c k ) = P ( x < 4 ) = .902

D efin e x to be the n u m b e r o f field s in fested w ith w hitefly. T h en p = P [in fe c te d field ] = .1 an d n = 100. a

p = n p = 1 0 0 (.l) = 10

b S in ce n is larg e, th is b in o m ia l d istrib u tio n sh o u ld be fairly m o u n d -sh a p e d , ev en th o u g h p = .1 . H en ee y o u w o u ld e x p e c t ap p ro x im a tely 9 5 % o f th e m e a su re m e n ts to lie w ithin tw o stan d a rd d ev iatio n o f the m e a n w ith o = y j n p q = ^ 1 0 0 (.1 )(.9 ) = 3. T h e lim its are calc u la te d as / i ± 2 í t => 10 ± 6 o r from 4 to 16 c

F ro m p a rt b, a valué o f x = 25 w o u ld be very u n lik ely , a ssu m in g th a t the

ch ara c te ristic s o f th e b in o m ial e x p erim en t are m et an d th at p = . 1 . I f this v a lu é w ere actu a lly o b serv ed , it m ig h t be p o ssib le th at the triáis (fie ld s) are n o t in d ep en d e n t. T h is co u ld easily be th e case, sin ce an in festatio n in o n e field m ig h t q u ic k ly sp re a d to a n e ig h b o rin g field. T h is is e v id e n c e o f c ontagi ón. 5 .3 3

D efin e x to be the n u m b e r o f A m e ric an s w h o a re “ta s te rs” . T h e n , n = 2 0 and p = .7 . U sin g th e b in o m ial ta b le s in A p p en d ix I,

5 .3 5

a

P ( x > 17) = 1 - P ( x < 16) = 1 - . 8 9 3 = .107

b

P ( x < 15) = .762

F o llo w the in stru c tio n s in the M y P erso n al T ra in e r sectio n . T h e a n sw ers are sh o w n in the tab le below . P ro b a b ility P (x = 0)

F o rm u la

C a lc u la te d v a lu é .0821

2 . 5 V 25 0!

P (* = l)

.2 0 5 2

2 .5 1e~25 1!

P (* = 2)

.2565

2 . 5 V 25 2!

P (2 o r few er su ccesses)

P (x = 0 ) + P (x = 1) + P (x = 2)

45

.5438

5.3 9

U sin g p ( x ) =

X

p x e~^

2* e

x\

2°e~2 p [jc = 0 l = ---------- = .135335

b

/> [* = l] = —— = .27 0 6 7

c

P [ x > 1] = 1 - P [ x < 1] = 1 - .1 3 5 3 3 5 - .2 7 0 6 7 = .5 9 3 9 9 4

1

J

O! 2V 2

9-V2

P [ * = 5] = --------- = .0 3 6 0 8 9 1 J 5!

L et x be the n u m b e r o f m isse s d u rin g a g iv en m o n th . T h en x has a P o isso n d istrib u tio n w ith p = 5. p(

c 5.47

X!

a

d

5.43

-2

5V 5

0) = e ' 5 = .0067

p ( 5) = ---------= .1755 5!

P [ x > 5] = 1 - />[* < 4] = 1 - . 4 4 0 = .5 6 0 fro m T a b le 2.

T h e ran d o m v ariab le x, n u m b e r o f b a cte ria, has a P o isso n d istrib u tio n w ith p = 2 . T h e p ro b ab ility o f in te re st is P [ x ex c e ed s m áx im u m c o u n t] = P [ x > 5] U sin g th e fact th a t p = 2 an d < j = 1.414 fro m E x e rc ise 5 .4 7 , m o st o f the o b se rv a tio n s sh o u ld fall w ith in p ± 2cr o r 0 to 4 . H en ee, it is u n lik e ly th a t x w ill e x c ee d 5. In fact, the e x a c t P o isso n p ro b a b ility is P [ x > 5] = .017.

5.51

T h e fo rm u la fo r p { x ) is p ( x ) =

b

c 4c "

165

C j5

455

p ( 0)

=

^

= £ ^ = « - = .15 C j5 455

= .36

C 4C " x_ ¿ x fo r x = 0 ,1 ,2 ,3

p(\) =

C 4r “

220

= — — = .48 C j5 455

C 4C “ 4 P( 3) = - b r - = — - . 0 1 c 15 45 5

T h e p ro b ab ility h isto g ra m is s h o w n below .

U sin g th e fo rm u la s g iv en in S ectio n 5.4. (

b

ii ’j j

c

í 4^ , 15]

( N - M >Í N - n >

d

i * C alcú late th e in terv als

o í 4 "l f 15 —4 ^ i

U - l y

,5

= .5 0 2 8 6 .

) I . 5 - 1)

p ± 2 a = .8 ± 2 > /.5 0 2 8 6 = .8 + 1.418 o r - . 6 1 8 to 2 .2 1 8 / / ± 3cr = .8 ± 3-V-50286 = .8 ± 1.418 or - 1 .3 2 7 to 2 .9 2 7 T h en , P [- .6 1 8 < jc < 2 .2 1 8 ] = p ( 0 ) + p ( 1) + p ( 2 ) = .99 P [- 1 .3 2 7 < x < 2 .9 2 7 ] = p ( 0) + p ( 1) + p { 2) = .99 T h e se resu lts a g ree w ith T c h e b y sh e ff’s T h eo rem . 5.55

a

T h e ran d o m v a riab le x has a h y p e rg e o m e tric d istrib u tio n w ith

N = 8 ,M = 5 an d rc = 3. T h en -.5^-3 p(x) =

* fo r x = 0 ,1 ,2 ,3 C

5 .6 1

b

P ( x = 3) =

d

P ( x < 1) = V '

C ¡C ¡ _ 10 = — = .1786 56 C* r 5r ? C¡

= — = .01786 56

r 5r 3 1 + 15 + - 4 - ^ - = I j l i l = .2857 C3* 56

R e fe r to E x ercise 5 .6 0 an d assu m e th at p = . 1 in stead o f p = .5 . a

P [ .t = 0] = p ( 0 ) = C o (-l)° (-9 )' = .729

P [ j : = 1]= /> (1) = C ¡'(-1 )‘ (-9 )2 = .2 4 3 />[.v = 2] = p ( 2 ) = C 5 ( . l ) 2 (.9 )' = .027 P [ .i = 3] = p (3 ) = C ! ( . l ) '( . 9 ) “ = .0 0 1 b N o te that th e p ro b a b ility d istrib u tio n is n o lo n g e r sy m m etric ; th a t is, sin ce the p ro b ab ility o f o b serv in g a h ead is so sm all, th e p ro b a b ility o f o b se rv in g a sm all n u m b e r o f h ead s on th ree flips is in creased (see th e fig u re on the n ex t p ag e).

47

0

1

2

3

c

f j = n p = 3 (. 1) = .3 an d (T = yj npq = ^ 3 ( . l ) ( . 9 ) = .520

d

T h e d esired in terv als are // + ít = .3 ± .5 2 0 /y ± 2 (7 = .3 ± 1 .0 4

o r - . 2 2 0 to .820 o r - . 7 4 0 to 1.34

T h e o n ly valu é o f x w h ich falls in this first in terv al is

jc =

0 , and the fractio n o f

m e asu rem en ts in th is in terv al w ill be .729. T h e v alú es o f * = 0 and

jc =

1 a re en clo sed

by th e seco n d in terv al, so th at .7 2 9 + .243 = .972 o f the m e a su re m e n ts fall w ith in tw o sta n d a rd d e v ia tio n s o f th e m ean , c o n siste n t w ith b o th T c h e b y sh e fP s T h e o rem an d the E m p irical R ule. 5.6 5

R efer to E x ercise 5 .6 4 . R e d e fin e x to be the n u m b e r o f p e o p le w ho c h o o se an in te rio r n u m b er in the sa m p le o f n = 20. T h en x has a b in o m ial d istrib u tio n w ith p = .3 . a

/> [jc > 8] = 1 - / > [ .* < 7 ] = 1 - 7 7 2 = .228

b O b serv in g e ig h t o r m o re p e o p le ch o o sin g an in te rio r n u m b e r is n o t an u n lik ely event, a ssu m in g th a t th e in te g e rs are all e q u a lly likely. T h e re fo re , th ere is no e v id en c e to in d íc a te th at p e o p le are m o re lik ely to c h o o se th e in te rio r n u m b ers than an y o thers. 5 .6 9

It is g iv en that

jc =

n u m b e r o f p a tie n ts w ith a p sy c h o so m atic p ro b lem , n = 2 5 , and

p = f ’fp atien t has p sy ch o so m a tic p ro b le m ]. A p sy c h ia trist w ish es to d eterm in e w h eth er o r not p = .8 . a jc

A ssu m in g th at th e p sy c h ia trist is c o rre c t (th at is, p = .8 ), th e e x p e cte d v alu é o f is E ( x ) = n p = 2 5 (.8 ) = 20.

b

( f L = n p q = 2 5 (.8 )(.2 ) = 4

c

G iv e n th at p = .8 , P [ x < 14] = .006 fro m T a b le 1 in A p p e n d ix I.

d

A ssu m in g th a t the p sy c h ia trist is co rrec t, the p ro b ab ility o f o b se rv in g

jc =

14 or

the m ore u n lik ely v alú es, x - 0 , 1, 2 , .. . , 13 is v ery u n lik ely . H en ee, o n e o f tw o c o n c lu sio n s can be d raw n . E ith e r w e have o b se rv e d a v ery u n lik e ly e v e n t, o r the p sy ch iatrist is in c o rre ct and p is ac tu a lly less th an .8. W e w o u ld p ro b a b ly c o n c lu d e that the p sy c h ia trist is in co rre c t. T h e p ro b a b ility th at w e h av e m ad e an in co rrect d e cisió n is 48

P [ x < 14 g iv en p = .8] = .006 w hich is q u ite sm all. 5.71

D efin e x to be the n u m b er o f stu d en ts 3 0 y ears o r o ld er, w ith ti = 2 0 0 and p = ^ [ s tu d e n t is 3 0 + years] = .25. a

S in ce x h as a b in o m ial d istrib u tio n , p = n p = 2 0 0 (.2 5 ) = 5 0 and

G - yfñ p q - yj2 0 0 ( . 2 5 ) ( . 1 5 ) = 6.124. b

T h e o b se rv e d valué, x = 35 , lies ^ « = - 2 .4 5 6 .1 2 4

stan d ard d ev ia tio n s b e lo w the m ean. It is u n lik ely th a t p = .25. 5 .7 5

a

T h e ran d o m v ariab le x, the n u m b er o f p la n ts w ith re d p etá is, has a b in o m ial

d is trib u tio n w ith n = 10 and p = P [r e d p etáis] = .75. b

S in ce th e v alu é p = .75 is n o t g iv en in T a b le 1, you m u st use th e b in o m ia l

fo rm u la to calc ú late

P ( x > 9 ) = C¿° (.75 f (.2 5 )' + C¡¡¡ (.7 5 )'° (.2 5 )° = . 1877 + .0 5 6 3 = .2440 c

P ( x < 1) = C

d

R efer to p art c. T h e p ro b a b ility o f o b se rv in g x = 1 o r so m eth in g

(.7 5 )° (.2 5 )'° + C,10 (.7 5 )' (,2 5 )9 = .0 0 0 0 2 9 6 . e v en m ore

u n lik ely ( x = 0 ) is very sm all - .0 0 0 0 2 9 6 . T h is is a hig h ly u n lik e ly e v e n t if in fact p = .75 . P erh ap s th ere has been a n o n ran d o m c h o ice o f seed s, o r the 7 5 % figure is not c o r r e d for this p a rtic u la r g en etic cross. 5.7 9

a

T h e d is trib u tio n o f x is actu a lly h y p erg e o m e tric , w ith N = 1200, n = 20 and M =

n u m b e r o f d e fe c tiv e s in th e lot. H o w ev er, sin ce N is so larg e in co m p a riso n to n , the d istrib u tio n o f * can be clo se ly a p p ro x im a te d by th e b in o m ia l d istrib u tio n w ith // = 2 0 a n d p = P [d efectiv e]. b

If p is sm all, w ith n p < 7, the P o isso n a p p ro x im a tio n c a n be used.

c

If there are 10 d efectiv es in the lot, then p = 10/12 00 = .0 0 8 3 3 3 an d p = .1667.

T h e p ro b ab ility th a t the lot is sh ip p ed is , ( . 1 6 6 7 ) % - 1667 ---------= .85 p(x = 0 W

0!

If th ere are 2 0 d efe ctiv e s, p = 2 0 /1 2 0 0 and p = .3333. T h en x ( . 3 3 3 3 Í V 3333 />(jc = 0 ) * = .72

V

0!

’

If th ere are 3 0 d efe c tiv es, p = 3 0 /1 2 0 0 and p = . 5. T h en ( 5)° é f 5 /> (* = 0 ) = ^ = .61 ’ 0! 49

5 .8 3

a T h e ran d o m v ariab le x, th e n u m b e r o f tasters w h o p ick the c o rrec t sam p le, h a s a b in o m ial d istrib u tio n w ith n = 5 an d , if th ere is no d ifferen c e in th e taste o f the th rce sam ples, p = P{ ta ste r p ic k s th e c o rre ct sam p le) = ^ b T h e p ro b a b ility th at e x a c tly o n e o f th e five ta ste rs c h o o se s th e late st b atc h as d ifferen t fro m the o th e rs is

5.8 7

T h e ran d o m v ariab le x has a P o isso n d istrib u tio n w ith p = 2. U se T a b le 2 in A p p en d ix I or the P o isso n fo rm u la to fin d the fo llo w in g p ro b a b ilitie s. a

P(jc = 0 ) = ----------= e~2 = .1 35335

0!

= . 135335 + .270671 + .270671 = .6 7 6 6 7 6 5.91

T h e ran d o m v a riab le x , the n u m b e r o f C alifo rn ia h o m e o w n e rs w ith earth q u a k e insurance, h as a b in o m ial d istrib u tio n w ith n = 15 and p = . 1. a

P ( x > 1) = 1 - P ( x = 0) = 1 - .2 0 6 = .794

b

P ( x > 4) = 1 - P (x < 3) = 1 - .944 = .056

c

C alcú late p =

np = 15(. 1) = 1.5 an d a = yfñpq = > /l5 (.l)(.9 )

= 1 .1 6 1 9 . T h en

ap p ro x im a tely 9 5 % o f th e v a lú e s o f x sh o u ld lie in the interv al //± 2 c r = > 1 .5 ± 2 ( 1 .1 6 1 9 )

= > - .8 2 to 3.82.

o r b etw een 0 and 3. 5 .9 5

5 .9 9

U se the C a lc u la tin g B in o m ia l P r o b a b ilitie s ap p let. T h e c o rre c t a n sw ers are giv en below . a

P ( x < 6 ) = 6 .0 (10)"5 = 0 .0 0 0 0 6

d

P ( 2 < x < 6 ) = .5948

b

P ( x = 8) = .042

e

P(*>6)= 1

c

P ( x > 14) = .0207

D efin e x to be the n u m b e r o f y o u n g ad u lts w h o p re fe r M c D o n a ld ’s. T h en x has a b inom ial d istrib u tio n w ith n = 100 an d p = .5. U se the C a lc u la tin g B in o m ia l P r o b a b ilitie s ap p let. a

P ( 6 l < x < 100) = .0176 50

b

P ( 4 0 < x < 6 0 ) = .9648

c I f 4 0 p re fe r B u rg er K ing, th en 6 0 p re fer M cD o n ald s, and vice v ersa. T he p ro b ab ility is th e sam e as th at ca lc u la te d in p a rt b . sin ce p = .5.

51

6: The Normal Probability Distribution 6 .3

T h e first few e x e rc ise s are d e sig n e d to p ro v id e p ra c tic e fo r th e stu d en t in ev alu a tin g areas u n d er the n orm al cu rv e . T h e fo llo w in g n o tes m ay be o f so m e assistan ce. 1 T ab le 3, A p p en d ix I ta b u la te s th e c u m u la tiv e a re a u n d e r a stan d ard n o rm al cu rv e to the left o f a sp e cifie d v alu é o f z. 2 S in ce th e total a rea u n d e r the cu rv e is on e, the total a re a ly in g to the rig h t o f a sp ecifie d valu é o f z a n d the total a re a to its left m u st a d d to 1. T h u s, in o rd e r to calc ú late a “tail a re a ” , su ch as the o n e sh o w n in F ig u re 6 .1 , the v a lu é o f z = z 0 w ill be in d ex ed in T a b le 3, a n d the a rea th at is o b ta in e d w ill be su b tra c te d fro m 1. D eno te the area o b tain ed by in d e x in g z = z0 in T a b le 3 b y /\( z 0) and th e d e sired a re a by A. T h en , in th e ab o v e e x a m p le , A = 1 - A (z0) .

3

T o find the a re a u n d e r th e sta n d a rd n o rm al c u rv e b etw ee n tw o v alú es, z¡ and z2, c a lc ú late the d iffe re n c e in th e ir c u m u la tiv e a reas, A = A { z 2 ) - A ( z x).

4

N o te that z, sim ila r to x, is a c tu a lly a ra n d o m v ariab le w h ich m ay tak e o n an infinite n u m b e r o f v alú es, b o th p o sitiv e and n eg a tiv e. N e g a tiv e v a lú e s o f z lie to th e le ft o f th e m ean , z = 0 , a n d p o sitiv e v alú es lie to the right.

a

It is n ecessa ry to fin d th e a re a to th e left o f z = 1 .6 . T h a t is, A = A (1.6) = .9452.

b

T h e area to th e left o f z = 1.83 is A = /f tl .83) = .9664.

c

A = A (.90) = .8159

d

A = A (4.58 ) = 1. N o tic e th at th e v alú es in T a b le 3 a p p ro a c h 1 as the v alu é o f z

increases. W h en th e v alu é o f z is la rg e r th an z = 3 .4 9 (th e la rg e st v a lu é in th e table), w e can assu m e th at th e area to its left is a p p ro x im a tely 1. 6.7

N o w w e are ask e d to fin d the z-v a lu e co rre sp o n d in g to a p a rtic u la r area.

53

a

W e need to find a zo su c h th a t P ( z > z 0 ) = .025. T h is is e q u iv a le n t to fin d in g an

in d ex ed a re a o f 1 - .025 = .975 . S earch th e in te rio r o f T a b le 3 until y o u fin d the fo u rd ig it n u m b er .9 7 5 0 . T h e c o rre sp o n d in g z-v alu e is 1.96; th at is, A ( \ .9 6 ) = .9 7 5 0 . T h e re fo re , z0 = 1.96 is th e d e sire d z-v a lu e (see the fig u re b elow ).

b

W e need to find a zo su ch th a t P { z < z Q) = .9251 (see b elo w ). U sin g T ab le 3, w e

find a v alu é such that the in d e x ed a re a is .9251. T h e c o rresp o n d in g z-v a lu e is * 0 = 1 .4 4 .

T h e p th p ercen tile ,o f th e sta n d a rd n o rm al d is trib u tio n is a v alu é o f z w h ich has a rea p / 100 to its left. Since all fo u r p erc e n tile s in this ex e rc ise are g re a te r th an th e 5 0 ,h percen tile, th e valu é o f z w ill all lie to the rig h t o f z = 0 , as sh o w n fo r the 9 0 th percen tile in th e fig u re on th e n ex t p ag e.

54

a F ro m the fig u re, the a re a to the left o f th e 9 0 lh p e rc en tile is .9 0 0 0 . F ro m T a b le 3, the ap p ro p ria te v alu é o f z is c lo se st to z = 1.28 w ith a re a .8 9 9 7 . H en ee th e 9 0 lh p e rc e n tile is ap p ro x im ate ly z = 1 .2 8 . b A s in p art a , th e a rea to the left o f th e 9 5 th p e rc e n tile is .9500. F ro m T a b le 3, the ap p ro p ria te v a lu é o f z is fo u n d u sin g lin ea r in terp o latio n (see E x e rc ise 6 .9 b ) as z = 1 .6 4 5 . H en ee the 9 5 lh p erce n tile is z = 1 .6 4 5 . c T h e area to the left o f th e 9 8 th p erc en tile is .9800. F ro m T a b le 3, the a p p ro p ria te v alu é o f z is c lo se st to z = 2.05 w ith a rea .9798. H en ee the 9 8 lh p e rc e n tile is a p p ro x im ately z = 2 .0 5 . d T h e a re a to th e left o f the 9 9 lh p erc en tile is .9900. F ro m T a b le 3, th e ap p ro p riate v alu é o f z is c lo se st to z = 2.33 w ith a re a .9901. H en ee th e 9 9 ,h p e rc e n tile is a p p ro x im a te ly z = 2 .3 3 . 6 .1 5

T h e 9 9 ,h p e rc e n tile o f the stan d a rd norm al d istrib u tio n w as fo u n d in E x ercise 6.1 Id to be z = 2 .3 3 . S in ce the re la tio n sh ip b e tw een the g en eral n o rm al ra n d o m v a riab le * an d the stan d a rd n orm al z is z =

x —u — , the c o rre sp o n d in g p e rc e n tile fo r th is g en eral o

no rm al ran d o m v a riab le is fo u n d b y so lv in g fo r x = f j + z(J \

10 jc-

6 .1 9

35 = 2 3 .3 or

x = 58.3

T h e ra n d o m v ariab le jc, the h e ig h t o f a m ale h u m an , h as a n o rm al d istrib u tio n w ith fJ = 6 9 and o = 3.5 . a

A h eig h t o f 6 ’0 ” rep rese n ts 6(12) = 72 in ch es, so that P (x > 72) = p Í z > —

b

= p { z > -86) = 1 -.8 0 5 1 = .1949

H eig h ts o f 5 ’8” a n d 6 ’ 1” re p rese n t 5(12) -t- 8 = 68 and 6(12) +1 = 73 inches,

resp ectiv ely . T hen

55

6 8 -6 9 7 3 -6 9 ^ 1 P ( 6 8 < jc < 7 3 ) = p \ —— — < z < — j = ^ ( - - 2 9 < z < 1.14) = .8 7 2 9 - .3 8 5 9 = .4870 c

A h eig h t o f 6 ’0 ” rep re sen ts 6(12) = 72 in ch es, w h ich h as a z-v alu e o f

3.5 T h is w ould not be c o n sid e re d an u n u su ally larg e valué, sin ce it is less th an tw o stan d ard d e v iatio n s fro m the m ean. d T he p ro b ab ility th at a m an is 6 ’0 ” o r ta lle r w as fo u n d in p art a to be .1 9 4 9 , w hich is not an u nusual o cc u rre n ce . H o w e v er, if y o u d efin e y to be th e n u m b e r o f m en in a ran d o m sam p le o f size « = 36 w h o are 6 ’0 ” o r taller, then y h as a b in o m ial d istrib u tio n w ith m ean f i = n p = 36(. 1949) = 7 .0 2 a n d sta n d a rd d ev iatio n a = y¡ñpq = y l 3 6 ( . \ 9 4 9 ) ( M 5 \ ) = 2 . 3 8 . T h e valué y = 17 lies

o 2.38 stan d ard d e v ia tio n s from th e m ean , a n d w o u ld be c o n sid e re d an u n u su al o ccu rre n c e for the g en eral p o p u la tio n o f m ale h u m an s. P erh a p s o u r p re sid e n ts d o n o t re p re se n t a ran do m sam p le fro m th is p o p u latio n . 6 .2 3

T h e ran d o m v ariab le jc, total w eig h t o f 8 p eo p le , has a m ean o f / r = 1200 an d a varian ce o 1 = 9 8 0 0 . It is n e cessary to fin d P ( x > 1300) a n d P ( x > 1500) if the d istrib u tio n o f jc is a p p ro x im a te ly n o rm al. R e fe r to th e n ext figurc-

1200

1300

1300

-p 1 3 0 0 - 1 2 0 0 _ 100 -4 -= — . = ------------ = 1.01 o V 9800 9 8 .995 P ( x > 13 0 0 ) = P ( z > 1.01) = 1 —A(1.01) = 1 —.8 4 3 8 = .1 5 6 2 .

56

S im ilarly , th e z-v alu e c o rre sp o n d in g to x 2 = 1 5 0 0 is _ *2 - / / _ 1 5 0 0 - 1 2 0 0 Zi — — .------(7 >/9800

.Vj j .

P ( x > 1500) = P ( z > 3 .0 3 ) = 1 - ¿ (3 .0 3 ) = 1 - .9 9 8 8 = .0 0 1 2 .

and 6.2 7

^ j

a It is g iv en th a t the p rim e in te re st rate fo recasts, x, are a p p ro x im a te ly n o rm al w ith m ean /y = 4.5 an d stan d a rd d e v ia tio n o - 0 .1 . It is n e cessary to d e te rm in e the p ro b a h ility that x e x c e e d s 4 .7 5 . C alcú late x -n 4 .7 5 -4 .5 — = --------------- = 2 .5 . T h en

z=

o

0.1

P ( x > 4 .7 5 ) = P ( z > 2 .5 ) = 1 - .9 9 3 8 = .0 0 6 2 . b

C a lcú late z =

x —u

4 3 7 5 -4 5 ----- -- = - 1 -25 . T h en

o

0.1

P ( x < 4 .3 7 5 ) = P ( z < - 1 .2 5 ) = .1 0 5 6 . 6.31

L et w be the n u m b e r o f w o rd s sp ecifie d in the co n trac t. T h e n jc, th e n u m b e r o f w ords in th e m a n u scrip t, is n o rm ally d is trib u te d w ith /y = w + 2 0 ,0 0 0 an d cr = 1 0 ,0 0 0 . T he p u b lish e r w ould like to sp ecify w so that P ( x < 1 0 0 ,0 0 0 ) = .9 5 . A s in E x ercise 6 .3 0 , calc ú late 1 0 0 ,0 0 0 0 - ( w + 2 0 ,0 0 0 ) Z--

T h en

P ( x < 1 0 0 ,0 0 0 ) = P z <

K

10,000 8 0 ,0 0 0 -w }

’

10,000

)

8 0 ,0 0 0 - w

10,000 —_ T , = .95 . It is n e c e ssa ry th at

z0 = ( 8 0 , 0 0 0 - w ) /l 0 ,0 0 0 be su ch th at /, ( z < z 0 ) = -95

=>

A ( z 0 ) = .9 5 0 0

or

z0 = 1 .6 4 5 .

H enee, 8 0 ,0 0 0 - w

= 1.645

or

w = 6 3 ,5 5 0 .

10,000 6 .3 7

a T h e norm al a p p ro x im a tio n w ill be a p p ro p ria te if b o th n p and n q are g rea te r than 5. F o r th is b in o m ial ex p e rim en t, n p = 2 5 ( 3 ) = 1.5

and

n q = 2 5 (.7 ) = 17.5

an d the no rm al a p p ro x im atio n is ap p ro p riate. b F o r the b in o m ia l ra n d o m variable, p = n p = 7.5 c

an d

o = yjnp q = -v/2 5 (.3 )(.7 ) = 2 .2 9 1 .

T h e p ro b ah ility o f in te re st is the area u n d er the b in o m ial p ro b a h ility h isto g ram

c o rre sp o n d in g to the re c ta n g le s x = 6 ,7 ,8 and 9 in th e fig u re on the n ex t p ag e.

57

T o ap p ro x im ate th is are a . u se the “c o rre c tio n fo r c o n tin u ity ” and fin d the a re a u n d e r a no rm al cu rv e w ith m ean fu = 7.5 and o = 2.291 b e tw e e n x { = 5.5 a n d x 2 = 9.5 . T he Z-values co rre sp o n d in g to the tw o v a lú e s o f x are ^ = - . 87 an d 2.291 T h e a p p ro x im a tin g p ro b a b ility is z, =

9 .5 - 7 .5 ^ z , = ------------- = .87 2 2.291

P ( 5.5 < * < 9 .5 ) = P ( - . 8 7 < z < .8 7 ) = .8 0 7 8 - .1 9 2 2 = .6 1 5 6 . d

F rom T ab le 1, A p p en d ix I, / >( 6 < j c < 9 ) = / >( j c < 9 ) —/ >( j c < 5 ) = .811 —. 193 = .618

w h ich is not too fa r fro m the a p p ro x im a te p ro b ab ility c a lc u la te d in p art c. 6.41

6.4 5

U sing th e b in o m ia l tab les fo r n = 2 0 a n d p = .3 , y o u ca n v erify th at a

P ( .r = 5) = P ( . x < 5 ) - P ( j t < 4 ) = .4 1 6 - .2 3 8 = .178

b

P ( x > 7 ) = 1 - P ( a : < 6 ) = 1 - .608 = .392

a

T h e a p p ro x im a tin g p ro b a b ility w ill be P ( x > 2 0 .5 ) w h ere x h as a norm al

d istrib u tio n w ith ¡a = 5 0 (.3 2 ) = 16 a n d o = >/5 0 (.3 2 )(.6 8 ) = 3 .2 9 8 . T h en 2 0 .5 -1 6 ' = P ( Z > 1.36) = 1 —.9131 = .0869 P ( x > 2 0 .5 ) = P z > ■ 3 .2 9 8 \ b

T h e ap p ro x im a tin g p ro b a b ility is P ( x < 14.5) = P

z <

1 4 .5 - 1 6

= P ( z < - .4 5 ) = .3264

3 .2 9 8 c If few er than 28 stu d e n ts d o n o t p re fe r ch erry , th en m o re th an 5 0 - 28 = 22 do p refer cherry. T h e a p p ro x im a tin g p ro b a b ility is P ( x > 2 2 .5 ) = p ( z > 22-5 ~ .16. = P ( z > 1.97) = 1 - .9 7 5 6 = .0244 v ’ 1, 3.298 d A s lo n g as y o u r c la ss c a n be a ssu m ed to be a re p re se n ta tiv e sam p le o f all A m erican s, the p ro b a b ilitie s in p arts a -c w ill be accu rate.

58

6 .4 9

D efin e x to be the n u m b e r o f e le c tio n s in w h ich the ta lle r ca n d id a te w on. If A m e ric a n s are not b iased by h eig h t, th en the ra n d o m v ariab le x h as a b in o m ial d istrib u tio n w ith n = 31 a n d p = .5 . C alcú late p = n p = 31(.5) = 1 5 .5 and G = y¡3\(.5)(.5) = V 7/75 = 2 .7 8 4 a U sin g the n o rm al a p p ro x im a tio n w ith co rre ctio n fo r c o n tin u ity , w e fin d the a rea to the rig h t o f x = 16.5 : P ( x > 16.5) = P \^z > 1625 ~g45 '5 ) = P ( z > -3 6 ) = 1 - 6 4 0 6 = .3 5 9 4 b S in ce the o c c u rren c e o f 17 o u t o f 31 ta lle r c h o ic e s is n o t u n u su a l, b a se d o n the resu lts o f p art a, it a p p e a rs th at A m e ric a n s d o not c o n sid e r h e ig h t w h en c a stin g a vote fo r a can d id ate.

6 .5 3

R e fe r to E x ercise 6 .5 2 , a n d le t x be the n u m b e r o f w o rk in g w o m e n w h o put in m o re than 4 0 h o u rs p e r w eek on the jo b . T h e n x has a b in o m ial d istrib u tio n w ith n = 50 an d p = .62 . a

T h e a v erag e v alu é o f * is (i = np = 5 0 (.6 2 ) = 31.

b

T h e stan d ard d ev ia tio n o f x is G = yjnp q = >/5 0 (.6 2 )(.3 8 ) = 3.432.

T h e z-sco re fo r x = 25 is z = —— — = — — — = - 1 .7 5 w h ich is w ith in tw o G 3.432 stan d ard d e v ia tio n s o f the m ean. T h is is n o t co n sid e re d an u n u su al o ccu rren ce.

c

6 .5 5

a T h e d esired are A |, as sh o w n in the fig u re on the n ext p ag e , is fo u n d by su b tra c tin g the cu m u la tiv e a re as c o rresp o n d in g to z = 1.56 an d z = 0 .3 , resp ectiv ely .

A, = A ( 1.56) - A (.3) = .9406 - .6 1 7 9 = .3 2 2 7 . b

T h e d esired a re a is sh o w n o n the n e x t page: A, + A, = A(. 2) - A { - . 2 ) = .5793 - .4 2 0 7 = .1 5 8 6

59

6.5 9 d esired valu é, zo, will be b e tw e e n z, = -67 and z 2 = .68 w ith a sso c ia te d p ro b a b ilitie s Px = .2 5 1 4 and P2 = .2483 . S in ce the d esire d tail area , .2500, is c lo se r to P[ = .2 5 1 4 , w e ap p ro x im a te zo as z0 = .67 . T h e v alú es z = - . 6 7 and z = .67 re p re se n t the 2 5 1*1 and 7 5 th p erce n tiles o f the sta n d a rd n orm al d istrib u tio n . 6.6 3

It is giv en th at x is n o rm ally d is trib u te d w ith fu = 10 and 8) = P ^ z >

d istrib u tio n ) o f .01 is z0 = 2 .3 3 . Flence, the v alu é o f /y can be o b ta in e d b y so lv in g fo r/y in th e fo llo w in g equ atio n :

6.7 5

D efin e

2 .3 3 = - ^ ^ o r / i = 7.301 .3 x = n u m b er o f in c o m in g c a lis th at are lo n g d istan ce p = P [in c o m in g ca li is lo n g d istan ce] = .3

61

n = 200 T he d esired p ro b ab ility is P ( x > 5 0 ) , w h ere x is a b in o m ial ra n d o m v ariab le w ith

H = n p = 2 0 0 (.3 ) = 6 0

and

^

= V 2 Ó 0 (3 )C 7 ) = >/42 = 6.481

A co rre ctio n for co n tin u ity is m ad e to in clu d e the e n tire a rea u n d e r the rectan g le co rre sp o n d in g to x = 5 0 and h en ee the a p p ro x im a tio n w ill be

6 .7 9

T h e ra n d o m v ariab le jc, the g e sta tio n tim e fo r a h u m an b aby is n o rm ally d istrib u te d w ith p = 278 and 6 8 0 ) = P ( z > 1.27) = 1 - . 8 9 8 0 = .1 0 2 0 T h u s, a p p ro x im ate ly 10.2% o f the p eo p le w ho to o k the te st sc o re d h ig h e r th a n 680. (T h e ap p let u ses three d ecim al p lace a c c u racy and sh o w s z = 1 .2 7 3 w ith P ro b = 0 .1 0 1 6 .)

63

7: Sampling Distributions Y o u can select a sim p le ra n d o m sam p le o f size n = 2 0 u s in g T a b le 10 in A p p e n d ix I.

7.1

F irst ch o o se a startin g p o in t an d c o n sid e r the first th ree d ig its in e ac h n u m b er. S in ce the e x p erim en tal u n its have alre a d y been n u m b ered from 0 0 0 to 9 9 9 , th e first 2 0 ca n be used. T h e three d ig its O R the (th ree d ig its - 5 0 0 ) w ill id en tify th e p ro p e r ex p e rim e n ta l unit. F o r ex am p le , if th e th ree d ig its are 7 4 2 , you sh o u ld select the e x p e rim e n ta l u nit n u m b ered 7 4 2 - 5 0 0 = 2 4 2 . T he p ro b a h ility th at any th ree d ig it n u m b e r is se lected is 2 /1 0 0 0 = 1/500 . O ne p o ssib le selectio n for the sam p le size n = 2 0 is 134 412 244 156

242 056 469 399

173 188 332 028

128 255 439 238

399 388 101 231

7 .5

If all o f th e to w n c itiz e n ry is lik ely to p ass th is c o m e r, a sa m p le o b ta in e d by selectin g e v e ry ten th p erso n is p ro b a b ly a fa irly ra n d o m sam ple.

7 .9

U se a ra n d o m izatio n sc h e m e sim ila r to th at u sed in E x ercise 7 . 1. N u m b e r e a c h o f the 5 0 rats from 01 to 50. T o ch o o se the 25 rats w h o w ill re c eiv e the d o se o f M X , se lec t 25 tw o -d ig it ra n d o m n u m b e rs fro m T a b le 10. E ach tw o -d ig it n u m b e r O R the (tw o d ig its - 5 0 ) w ill id en tify th e p ro p e r e x p erim en tal unit.

7 .1 3

a T h e first q u estio n is m o re unbiased . b N o tice that th e p erce n ta g e fav o rin g the new space p ro g ram d ro p s d ram atica lly w h en the p h rase “ sp en d in g b illio n s o f d o lla rs” is ad d ed to the q u estio n .

7 .1 9

R e g ard le ss o f the sh ap e o f the p o p u la tio n from w h ich w e are sa m p lin g , th e sa m p lin g d istrib u tio n o f the sam p le m e a n will have a m ean / / eq u al to the m ean o f the p o p u latio n from w h ich w e a re sam p lin g , a n d a stan d ard d ev ia tio n e q u a l to o ¡ \ f ñ .

a

/y = 10;

( j / y f ñ = 3/> /36 =.5

b

ju = 5;

( j / y f ñ = 2 / -n/ToO = .2

c

n = 120;

cr/V ñ = l/>/8 = .3536

7 .2 2 -2 3 F o r a p o p u latio n w ith o = 1 , the sta n d a rd e rro r o f th e m ean is .0848 .

c

A n erro n e o u s c o n c lu sió n w ill h av e o cc u rre d if in fact p < .0 8 4 8 and th e sa m p le has

p ro d u c e d p = .15 by ch an ce. O n e can o b tain an u p p e r b o u n d on th e p ro b a b ility o f this p a rtic u la r type o f e rro r by c a lc u la tin g P ( p > .15 w h en p = .0 8 4 8 ). 7 .7 9

A n sw e rs w ill vary fro m stu d en t to stu d en t. P ay in g c a sh fo r o p in io n s w ill not n ec e ssa rily p ro d u ce a ran d o m sam p le o f o p in io n s o f all P ep si and C o k e d rin k ers.

7 .8 3

a

T h e av erag e p ro p o rtio n o f in o p e ra b le c o m p o n e n ts is _ _ 6+ 7h P ~

h5 _ 75 _ ]0

50(15)

_ 7 5 0 ~~

a n d the c o n tro l lim its are

and I f su b se q u e n t sam p les d o no t stay w ith in the lim its, U CL = .2273 a n d L C L = 0 , the p ro cess sh o u ld be ch eck ed . 7 .8 7

a

T h e th eo retical m ean and sta n d ard d ev ia tio n o f the sa m p lin g d istrib u tio n o f x w h en

ti = 4 are p = 3.5

and

a / y [ ñ = 1.708/ \[Á = .854

b-c A n sw ers w ill vary fro m stu d en t to stu d en t. T h e d istrib u tio n sh o u ld be re lativ ely u n ifo rm w ith m ean a n d sta n d a rd d e v ia tio n ció se to those g iv en in p art a.

8: Large-Sample Estimation T h e m arg in o f e rro r in e stim a tio n p ro v id e s a p ractical u p p e r b o u n d to the d ifferen ce b etw een a p a rtic u la r e stím ate a n d the p a ra m e te r w h ich it e stim a te s. In th is e h ap te r, the m arg in o f e rro r is 1 .9 6 x (sta n d a rd e rro r o f the estim a to r).

T h e m arg in o f erro r is 1.96 S E = 1.96-^ L , w h ere o can be e stim a te d b y th e sam p le yjn stan d ard d ev iatio n s fo r larg e v alú es o f n. a

1 . 9 6 , — = .5 5 4 V 50

b

1.9 6 , — = .1 7 5 y 500

c

1 .9 6 ./—^ — = .055 V 5000

F o r the estím ate o f p g iv en as p = x /n , the m arg in o f e rro r is 1.96 S E = 1.96. — V n

.

U se the e stim a te d v alu é g iv en in the ex e rc ise fo r p. a

d f

1.96

= .0588 \

100

V

100

b

1

.

9 V

1.96

= .0898

6 = 100

e

.0898

c

1.

9 V

6 100

= .098

1 .9 6 J — ^ — = .0 5 8 8 V 100

T h e larg est m argin o f e rro r o c c u rs w h en p = .5 .

T h e p o in t estím ate o f p is x = 39.8° an d the m arg in o f e rro r w ith 5 = 17.2 and n = 50 is 1.96

a

S E = 1 . 9 6 = 1 .9 6 -4 = = 1, 9 6 - 4 ¿ = 4 .7 6 8 yfn yjn V50 X

T h e p o in t estím ate fo r p is g iv en as p = — = .lS and th e m a rg in o f e rro r is

n

ap p ro x im ately 1.96

= 1 .9 6 ,f 7 8 ^ 2 2 ^ = .026 V 1000 b T h e p o ll’s m arg in o f e rro r d o es n o t ag ree w ith the re su lts o f p art a, b e c a u se the sam p lin g e rro r w as re p o rted u sin g the m áx im u m m a rg in o f e rro r u sin g p = .5 : 1.96

J—

n

E = 1.96 = .0 3 1 \ n \ 1000

o r ± 3 .1 %

A po in t estím ate fo r th e m ean len g th o f tim e is x = 1 9 .3 , w ith m arg in o f e rro r

1.96

8 .2 3

S E = 1 .9 6 - ^ r « 1 . 9 6 - ^ = l . 9 6 - 5 ¿ = 1.86 yjn yjn >/30

T h e 9 0% co n fid en ce interv al fo r/y g iv en as 3c ± 1 .6 4 5 y¡n w h ere /2 0 0 j

c

W h en n = 4 0 0 , the w id th is 2 '■9 6 7 m

8.31

= 2 ( i -9 6 ) = 3 -9 2 -

) - - 2 { -9 S ) - - L 9 6 -

W ith n = 40, x = 3 .7 a n d s = .5 and a = .0 1 , a 9 9 % c o n fid e n c e in te rv a l fo r ¡u is ap p ro x im ated by x ± 2 .5 8 -4 = = 3 .7 ± 2 .5 8 —¿ = = 3 .7 ± .2 0 4 or 3 .4 9 6 < lí < 3 .9 0 4 •Jn V 40 In rep e a te d sam pling, 9 9 % o f all in terv als c o n stru c te d in th is m a n n e r w ill e n d o s e / i . H enee, w e are fairly c e rta in th a t this p a rtic u la r in te rv a l c o n ta in s f j . (In o rd e r fo r this to be true, th e sam p le m u st be ran d o m ly selec te d .)

8.35

a

T h e p o in t estim ate o f p is p = — == . 1 3 6 , an d the a p p ro x im a te 9 5 % " n 500

c o n fid en ce in terv al for p is p ± 1.96

= . 136 ± 1.96 | - 3 6 (:8 6 4 ) = . 136 ± .030

500 or

. 106 < / ? < . ! 66 .

72

b In o rd e r to in c rea se the a c c u ra cy o f the co n fid e n c e in terv al, y o u m u st d e c re a se its w idth. Y ou can a c c o m p lish th is by (1 ) in creasin g the sam p le size n, o r (2 ) d ecrea sin g Za/2 by d e c reasin g th e c o n fid e n ce co efficien t.

8.3 9

a

W h en e stim a tin g th e d iffe re n c e /y, - / y 2 , the (1 - a ) 100% c o n fid e n c e ijiterv al is

(3c. - x , ) ± z a/2 — + — 1

",

. E stim a tin g o \ and o \ w ith s] and s 2 , the a p p ro x im ate

"2

9 5 % c o n fid en ce in te rv a l is l¡ 38 4 14 (1 2 .7 - 7 .4 ) ± 1 .9 6 J — + — = 5 .3 ± .6 9 0 b

o r 4.61 < / / , - f i 2 < 5.9 9 .

S in ce th e v a lu é //, - / y , = 0 is n o t in the c o n fid e n ce interv al, it is no t lik e ly that

/y, = / y , . Y ou sh o u ld c o n c lu d e th a t th ere is a d iffe re n ce in the tw o p o p u la tio n m eans. 8 .4 3

a

T h e p a ra m e te r to be estim a te d is / y , th e m ean sco re fo r th e p o stte st fo r all B A C C

classes. T h e 9 5 % c o n fid e n ce in terv al is a p p ro x im ately

I ± 1 . 9 6 - ^ = 1 8 .5 ± 1 .9 6 -? = £ L = 1 8 .5 ± .8 2 4 o r 1 7 .6 7 6 < /y < 19.324 V 365 b

T h e p a ra m e te r to be estim a te d is / y , the m ean sco re fo r the p o stte st fo r all

trad itio n al classe s. T h e 95% c o n fid e n c e interv al is ap p ro x im ate ly 5 6 96 3c ± 1 .9 6 4 = = 1 6 . 5 ± 1 . 9 6 - t = = 1 6 .5 ± .7 9 0 o r 15.710 < /y < 17.290 yfn V 29 8 c

N o w w e are in tereste d in th e d iffe re n c e b e tw een p o stte st m ean s, /y, - / y , , fo r

B A C C v ersu s trad itio n al classe s. T h e 9 5 % co n fid e n c e in terv al fo r /y, ~ /y 2 is ap p ro x im ately

( ,8 .5 - 1 6 .5 , ± l . 9 6 j M * + « « l v 7 v 365 298 2 .0 ± 1 .1 4 2 d

or

.858 < (/y, - / y 2 ) < 3.142

S ince the c o n fid en ce interv al in p a rt c h as tw o p o sitiv e e n d p o in ts, it d o e s not

co n tain the valu é /y, - / y 2 = 0 . H en ee, it is not lik ely th at the m ean s are e q u al. It ap p e a rs that there is a real d iffe re n c e in the m ean scores. 8 .4 7

R efer to E x ercise 8.18.

73

a

T h e 9 5 % c o n fid en c e in terv al fo r //, - / / 2 is ap p ro x im ate ly

(170 —160) ± 1.96. 1 0 ± 6 .6 6 7 b

- 1 5 ± 7 .0 4 0 c

or

3.333 < (//, - / / , ) < 16.667

T h e 9 9% co n fid en ce in terv al fo r

or

is ap p ro x im a te ly

- 2 2 .0 4 0 < (//, - / ¿ 2 ) < - 7 .9 6 0

N eith er o f the in terv als co n ta in the v alu é ( / v , - / / 2) = 0 . I f (//, —/ / 2 ) = 0 is

c o n tain ed in the co n fid e n c e in terv al, th en it is no t u n lik e ly th at H\ c o u ld eq ual / / 2 , im p ly in g no d ifferen c e in the a v erag e ro o m rates fo r the tw o h o tels. T h is w o u ld be o f in terest to th e ex p erim en ter. d S ince n eith er co n fid e n c e in terv al c o n ta in s th e valué //, = 0 , it is n o t likely that the m eans are eq u al. Y o u sh o u ld co n c lu d e th a t th ere is a d iffe re n c e in the a v erag e ro o m rates fo r the M a rrio tt an d W y n d h a m a n d a lso fo r the R ad isso n and the W y n d h a m chains.

8 .5 1

a

C alcú late A

x 337 x 374 = — = 1— = .42 a n d p 2 = — = -— = .5 8 . T h e ap p ro x im a te 90% n, 800 ' n 2 640

c o n fid en ce interval is

-.1 6 ± .0 4 3 b

or

- .2 0 3

117

T he tw o binom ial sam p les m u st be ra n d o m an d in d e p e n d e n t a n d the sa m p le sizes

m ust be large en o u g h th at the d istrib u tio n s o f p x a n d p 2 are ap p ro x im a te ly no rm al. A ssu m in g that fíle sam p les are ran d o m , th ese c o n d itio n s are m et in th is ex ercise.

74

8 .5 5

a

W ith p. = — — = .45 a n d p , = - ~ 2— = .5 1 . T h e ap p ro x im a te 9 9 % c o n fid e n c e 1 1001 1001

in terv al is (A -p ;) i2 .5 8 jM + M

.45 .5 5 ) .51 .49 ( .4 5 - .5 1 1 2 . 5 8 . — i------^ + — i---v y V 1001 1001 - .0 6 1 .0 5 8 b

or

- . 1 1 8 < ( p , - p 2) < -.0 0 2

S ince the in terv al in p art a c o n ta in s o n ly n eg a tiv e v alú es o f p , - p 2 , it is lik ely

that p | - p 2 < 0

P| < p 2 . T h is w o u ld indicate th at the p ro p o rtio n o f ad u lts w ho

claim to be fan s is h ig h e r in N o v e m b e r th an in M arch .

8 .5 9

r 120 x, 54 C a lcú late p, = — = ----- = .7 and P7 = — = ------ = .5 4 . T h e ap p ro x im ate 90% 1 n, 180 2 n2 100 c o n fid en c e in terv al is ( p , - p 2) ± l - 6 4 5 j ^ +

(.7 - .5 4 ) ± 1.645 v ' V 180 .1 6 1 .0 9 9

or

1 100

. 0 6 1 < ( p , - p 2) < . 2 5 9

In terv als c o n stru c te d in th is m a n n e r w ill e n d o s e the true v alu é o f p, - p 2 9 5 % o f the tim e in re p e a te d sam p lin g . H en ee, we are fairly certain th at th is p a rtic u la r in terv al e n d o s e s p, - p 2 . 8 .6 3

F o llo w the in stru ctio n s in th e M y P erso n al T ra in e r section . T h e a n sw e rs are sh o w n in the table below .

T y p e o f D a ta

B inom ial

O ne or Two S a m p le s O ne

M a r g in o f e rro r

pora

B ound, B

p~. 5

.05

S o lv e th is in e q u a lity

S a m p le size

n >385 1.96 ^ ^ < . 0 5 V n

Q u an titativ e

O ne 1 .9 6 -i-

(7 * 1 0

yjn

75

2

1 .9 6 4 1

«> 9 7

8.6 7

F o r the d iffe re n c e p x - / y , in the p o p u la tio n m ean s fo r tw o q u a n tita tiv e p o p u latio n s, the 9 5% u p p e r co n fid en ce b o u n d u ses z ^ = 1.645 a n d is c a lc u la te d as ( * , - J 2) + 1.545 \ n, 2 + 2 .0 0

8.71

or

+ ^ = ( 1 2 - 1 0 ) + 1 .6 4 5 J |- + ^ n2 V5 0 50 ( //,- //,) < 4

In th is ex ercise, the p a ra m e ter o f in te re st is p¡ - p 2 , nx= n 2 = n , and B = .05 . S in ce w e have no p rio r k n o w led g e ab o u t p x and p 2 . we assu m e th e la rg e st p o ssib le v ariatio n , w h ich o c cu rs if p x = p 2 = .5 . T hen z a/2 x (s td e rro r o f p x- p 2) < B ,

&

^ nx

\

r n2

£ 05 ^

y f ñ > —— — .05 8 .7 5

os

V =>

n > 1 0 8 5 .7 8

n

o r « , = « , = 1086 1 2

T h e stan d ard d ev iatio n is estim a te d as R / 4 = 1 0 4 /4 = 2 6 , and . 2.58.1— + — < 5 => n, ^

8 .7 6

n

a

> 2 .5 8 V Í3 5 2

2.58,1— + — V n n

3 5 9 9g 0 f ^ = ^ = 3 6 0

^

F o r the d iffe re n ce /y, - f i 2 in the p o p u la tio n m ean s th is y e a r a n d te n y e ars ag o , the

9 9% lo w er co n fid e n ce bound u ses z 01 = 2.33 and is c a lc u la te d as

K

s,2

^ nx

n2

_

252

28:

\ 400

400

( x - x , ) - 2 .3 3 í— + — =(73 —63) —2.33A/------+ ■ V1

1 0 - 4 .3 7

or

’

( jjx - / / , ) > 5.63

b S in ce the d iffe re n c e in the m ean s is p o sitiv e, y o u can c o n c lu d e th at th ere has been a d e crease in the a v erag e p er-c a p ita b e e f c o n su m p tio n o v e r th e last ten years. 8 .8 3

a

T h e po in t estím ate o f p is x = 29.1 an d the m arg in o f e rro r in e stim a tio n w ith

5 = 3.9 and n = 64 is 3.9 1.96(7, = 1 .9 6 -? = * 1 .9 6 -7 = = 1.96 = .9555 y fñ y fñ y ¡6 4 , b

T h e ap p ro x im ate 9 0 % co n fid en c e interv al is x ± 1 .6 4 5 -4 = = 2 9 .1 ± 1 .6 4 5 - ^ L = 29.1 ± .8 0 2 yfn V 64

76

o r 2 8 .2 9 8 < p < 2 9 .9 0 2

In te rv a ls c o n stru c te d in th is m a n n e r e n d o s e the true v alu é o f /y 9 0 % o f th e tim e in rep e a te d sa m p lin g . T h e re fo re , w e are fairly certain th at th is p a rtic u la r interv al en d o ses /y . c

T h e a p p ro x im a te 9 0 % lo w er c o n fid en c e b o u n d is 1 - 1 . 2 8 - ^ = 2 9 . 1 - 1 . 2 8 ^ = 2 8 .4 8 o r /y > 28 .4 8 -Jn V 64 _ W ith B = .5, o ~ 3 .9 , and 1 - a = .95 , w e m u st so lv e fo r n in the fo llo w in g

d

inequality : cr \ . 9 6 —j= < B ■Jn

=>

3.9 \ . 9 b —j= < .5 yjn

4 ñ > 15.288 => n > 2 3 3 .7 2 3 o r n > 2 3 4 8 .8 7

A ssu m in g m áx im u m v a ria tio n w ith p = .5 , solve 1.645J - ^ < .0 2 5

4~n >.

iW M =

3 2 ,

n > 1082.41

or

ai > 1083

.025 8.9 1

a

D efin e sa m p le #1 as the sa m p le o f 4 8 2 w om en a n d sam p le # 2 as the sam p le o f

3 5 6 m en. T h en p , = .5 and p : = .7 5 . b

T h e ap p ro x im a te 9 5 % c o n fid e n c e in terv al is

( P \ - M ± 1.96

n,

(.5 - .7 5 ) ± 1 .9 6 M + v ’ \ 482 -.2 5 ± .0 6 3 c

or

+

n2

356 - . 3 1 3 < ( p , - p 2) < - .1 8 7

S ince the v alu é p, - p 2 = 0 is n o t in the co n fid e n c e in te rv a l, it is u n lik ely that

p l = p 2 . Y o u sh o u ld no t co n c lu d e th a t th ere is a d iffe re n c e in th e p ro p o rtio n o f w o m en an d m en o n W all S treet w ho h av e ch ild ren . In fact, since all the p ro b ab le v alú es o f p, - p 2 a re n eg ativ e, the p ro p o rtio n o f m en o f W all S treet w h o have ch ild ren a p p e a rs to be la rg e r than th e p ro p o rtio n o f w om en. 8.9 5

A ssu m e th at o = 2.5 a n d the d e sire d b o u n d is .5. T hen 1 . 9 6 < B y¡n

1 .9 6 ^ p < .5 y¡n

=> n > 9 6 .0 4 o r n > 9 7

77

8 .9 9

a

I f you use p = .8 as a c o n se rv ativ e e stím a te fo r p , the m arg in o f e rro r is

ap p ro x im ately ± 1 . 9 6 . ® ^ = + .0 2 9 V 750 b

T o re d u c e the m argin o f e rro r in p a rt a to ± .0 1 , sol ve fo r « in th e e q u atio n

1 .9 6 .p í^ = ,0 1

V

8 .1 0 3

=>

«

V ^ = L 9 6 ( '4 > = 7 8 .4

=> « = 6 1 4 6 .5 6 o r « = 614 7

.01

It is assu m ed that p = .2 and that th e d e sired b o u n d is .01. H en ee, 1 .9 6 J —

< .01

=>

yjn >

í

= 4 2 .7 2

.01 « > 1 8 2 4 .7 6 o r « > 1 8 2 5 8 .1 07

a

T h e ap p ro x im ate 9 5 % c o n fid e n c e in terv al fo r p is í 529 I ± 1 . 9 6 - = = 2 .9 6 2 ± 1 .9 6 - = = = 2 .9 6 2 ± . 125 yjn V 69

or

2 .8 3 7 < p < 3.087 .

b In o rd e r to cu t the interv al in half, the sa m p le size m u st in cre ase b y 4. I f this is d o n e, the new h alf-w id th o f th e c o n fid e n c e in terv al is

l.9 6 -í= = - ( l . 9 6 - í \ . y[4n

2^

y fñ

H en ee, in th is case, the n ew sa m p le size is 4 ( 6 9 ) = 2 7 6 .

8.111

T h e ap p ro x im ate 98% co n fid e n ce interv al fo r p is T ± 2 .3 3 -4 = = 2 .7 0 5 ± 2 . 3 3 ^ 2 = 2 .7 0 5 ± .0 1 1 V« V 36 or

8 .1 1 5

2 .6 9 4 < p < 2 .7 1 6 .

F o r this ex ercise, B = .08 fo r the b in o m ial e stim a to r p , w h ere S E ( p ) = A — V « p = .3 , w e have 1.96A ^ ~ < B => 1 .9 6 ,p í í ! < .08 « r 1 .9 6 Í 2 ( \ 8 ) V « > ----------------- = > « > 9 . 8 .08

or « > 97. 78

or

« > 9 6 .0 4

. If

8 .1 1 7

U se the In ter p r e tin g C o n fíd en ce In te r v a ls a p p let. A n sw e rs w ill v ary , bu t the w id th s o f all th e in te rv a ls sh o u ld be the sam e. M o st o f the sim u la tio n s w ill sh o w b etw een 8 an d 10 in te rv als th a t w o rk co rrectly .

8 .121

U se the E x p lo r in g C o n fíd e n c e In te rv a ls applet. a-b M o v e the slid er on th e rig h t sid e o f th e a p p let to ch an g e the sam p le size. In creasin g th e sa m p le size re su lts in a sm a lle r sta n d a rd e rro r and in a n a rro w e r interval. c B y in c re a sin g th e sa m p le size n, y o u o b tain m o re in fo rm atio n and can o b ta in this m ore p re c ise e stím a te o f (J w ith o u t sacrificin g co n fíd en ce.

79

9: Large-Sample Tests of Hypotheses a

T h e c ritic a l v alu é th at se p a ra te s the reje c tio n and n o n rejectio n re g io n s fo r a rig h t-

tailed te st b a sed o n a z-sta tistic w ill be a v alu é o f z (c alled za ) su ch that / >( z > z a ) = t f = .01 . T h a t is, z 0I = 2.33 (see the fig u re b elo w ). T h e nuil h y p o th esis H 0 w ill be re je c te d if z > 2 .3 3 .

b

F o r a tw o -tailed te st w ith a = .05 , the critical v alu é fo r the reje c tio n reg ió n cu ts

o ff oc¡2 = .025 in the tw o tails o f th e z d istrib u tio n in F ig u re 9 .2 , so th at z 025 = 1.96 . T h e nuil h y p o th e sis H 0 w ill be re je c te d if z > 1.96 o r z < - 1 .9 6 (w h ich y o u can also w rite as

Id > 1. 9 6 ).

c S im ila r to p art a, w ith the re je c tio n reg ió n in th e lo w e r tail o f the z d istrib u tio n . T h e nuil h y p o th e sis H 0 w ill be re je cte d i f z < - 2 .3 3 . d

S im ila r to p art b , w ith a / 2 = .005 . T h e n uil h y p o th e sis H 0 w ill be re je c te d if

z > 2.5 8 o r z < - 2 .5 8 (w h ich y o u can also w rite as |z| > 2.58 ).

81

In this ex ercise, th e p a ra m e te r o f in te re st is fJ , the p o p u latio n m ean. T h e o b je c tiv e o f the e x p erim e n t is to sh o w th a t the m ean e x c e e d s 2.3. a

W e w ant to p ro v e the a ltern a tiv e h y p o th e sis th at /v is, in fact, g re a te r then 2.3.

H enee, the a lte rn a tiv e h y p o th e sis is H a :/y > 2 .3 and the nuil h y p o th e sis is H 0 : / / = 2 .3 . b

T h e best e stim a to r fo r f i is the sam p le a v erag e x , an d the test sta tistic is

Izñ . "

a /J ¡

w h ich rep rese n ts the d ista n c e (m e asu re d in un its o f sta n d a rd d e v ia tio n s) fro m x to the h y p o th esized m ean /y . H en ee, if th is v alu é is larg e in a b so lu te v alu é, o n e o f tw o co n clu sio n s m ay be d raw n . E ith e r a v ery u n lik ely e v e n t h as o c cu rred , o r the h y p o th esized m ean is in co rrect. R e fer to p a rt a . If a - .05, the critical v alu é o f z th at se p arates the re je c tio n and n o n -re jec tio n reg io n s w ill be a v alu é (d en o te d by z0 ) such that P ( z > z n ) = a = .05 T h at is, z 0 = 1.645 (see b elo w ). H en ee, H 0 w ill be re je c te d if z > 1 -645 .

a - 05

Reject H0

c T h e stan d a rd e rro r o f the m ean is fo u n d u sin g the sam p le sta n d ard d ev ia tio n 5 to ap p ro x im ate the p o p u latio n sta n d a rd d ev ia tio n c r : n r,

cr

5

.29

_

d

T o c o n d u c t the test, calc ú la te the v alu é o f the test sta tistic u sin g th e in fo rm atio n

c o n ta in e d in the sam p le. N o te that the v alu é o f the tru e sta n d a rd d e v ia tio n , o , is a p p ro x im a te d u sin g the sa m p le stan d ard d ev ia tio n 5 .

a¡4ñ

s/yfñ

.049

T h e o b se rv ed v alu é o f the test statistic, z = 2 .0 4 , falls in th e re je c tio n re g ió n a n d ti n uil h y p o th e sis is rejected . T h ere is su fficie n t e v id e n c e to in d íc ate th at p > 2 .3 . 9.11

a

In o rd e r to m ak e sure th at the av erag e w eig h t w as o n e p o u n d , y o u w o u ld test H0 :// = 1

Ha :p * 1

versus

b -c T h e test statistic is

z = x-Mo a *~Po a /y fñ

_

s/yfñ

1 .0 1 -1 = 33

.1 8 / >/35

w ith p -v a lu e = ^ ( |z | > -33) = 2 (.3 7 0 7 ) = .7414 . S in ce the /?-value is g re a te r th an .( the nuil h y p o th e sis sh o u ld not be reje cte d . T he m a n a g e r sh o u ld re p o rt th at th ere is in su ffic ien t e v id en c e to in d ícate th at the m ean is d iffe re n t fro m 1. 9 .1 5

a

T h e h y p o th e sis to be tested is H 0 :/y = 110

versus

H a : /y < 110

and the te st statistic is x-H o O \[ñ

= 1 0 7 -1 1 0 _ s/yfñ

1 3 /V i 00

w ith p -v a lu e = P ( z < - 2 .3 1 ) = .0 1 0 4 . T o d raw a c o n c lu sió n fro m th e p - v alu é, u se t g u id elin e s fo r statistical sig n ifican ce in S ectio n 9 .3 . S in ce th e p -v a lu e is b e tw e e n .( a n d .05, the test resu lts are sig n ifícan t at the 5% lev el, b u t n o t at the 1% level. b I f a = .05 , H0 can be re je c te d and y o u can c o n c lu d e th a t th e a v e ra g e score im p ro v e m e n t is less than claim ed . T h is w ould be th e m o st b e n e fic ia l w ay fo r the c o m p e tito r to State th ese c o n clu sio n s. c I f you w o rk ed fo r the P rinc eto n R e view , it w ould be m o re b en e ficia l to concludi th at th ere w as insufficient evid e n ce a t the 1% level to c o n c lu d e th a t the a v e ra g e scoi im p ro v e m e n t is less th an claim ed . 9 .1 9

T h e h y p o th e sis o f in terest is on e-tailed: H 0 :/A

= 0

v ersu s

Ha ://,- //2 < 0

T h e test statistic, calc u la te d u n d er the a ssu m p tio n th a t //, —Jl^ = 0 , is

83

w ith the unk n o w n cr2 an d a \ e stim a te d by s f an d s \ , resp ectiv ely . T h e stu d en t can use o n e o f tw o m e th o d s fo r d e c isió n m aking. p -v a lu e ap p ro a ch : C a lc ú la te p -v a lu e = P ( z < - 1 . 3 3 ) = .0918 . S in ce th is p -v a lu e is g reater th an .05, the nuil h y p o th e sis is n o t re je cte d . T h ere is in su fficien t e v id e n c e to in d icate that the m ean fo r p o p u la tio n 1 is sm alle r th an the m ean fo r p o p u la tio n 2. C ritica l v a lu é a p p ro a ch ; T h e re je c tio n reg ió n w ith a = .05 . is z < - 1 .6 4 5 . S in ce the o b se rv e d v alu é o f z d o e s no t fall in the reje c tio n reg ió n , H 0 is n o t reje c te d . T h ere is in su fficien t e v id e n c e to in d ic ate th at the m ean fo r p o p u la tio n 1 is sm a lle r th a n the m ean fo r p o p u latio n 2. a

T h e h y p o th e sis o f in te re st is tw o -tailed : H0 : p ,- p 2 = 0

v ersu s

H a :p , - p 2 * 0

T h e test statistic, ca lc u la te d u n d e r the assu m p tio n th at //, - p , = 0 , is

w ith p -v a lu e = P ( |z | > 2 .2 6 ) = 2 ( .0 1 19) = .0238 . S in ce the p -v a lu e is less th an .05, the nuil h y p o th e sis is re je cte d . T h ere is e v id en c e to in d ica te a d iffe re n ce in the m ean lead lev els fo r the tw o se c tio n s o f the city. b

F ro m S ectio n 8.6, the 9 5 % c o n fid e n c e interv al fo r Pj - p , is a p p ro x im ately

-1 .9 ± 1 .6 5 c

or

- 3 .5 5 < ( p , - p , ) < - . 2 5

Since th e v alu é p , - p 2 = 5 o r p 1 - p , = - 5 is n o t in th e co n fíd e n c e in terv al in part

b, it is not lik ely th at the d iffe re n c e w ill be m ore th an 5 p p m , an d hen ee the statistical sig n ifícan ce o f the d iffe re n c e is n o t o f p ractical im p o rta n c e to the en g in ee rs. a

T h e h y p o th e sis o f in te re st is tw o-tailed : H0 : p , - p 2 = 0

and the test sta tistic is

v ersu s

Ha :p ,- p 2 ^ 0

B -^ )-0

,9 4 - 2 .8 1-22

n, + n2

= _31g

2 .8 2

V 36 + 26

w ith p -v a lu e = P ( |z | > 3 .1 8 ) = 2 (.0 0 0 7 ) = .0 0 1 4 . S in ce th e p - \ a lu e is le ss th an .05, the nuil h y p o th e sis is rejected . T h ere is e v id en c e to in d ícate a d iffe re n c e in the m ean co n c e n tra tio n s fo r th ese tw o ty p e s o f sites, b

T h e 9 5 % co n fid e n c e interv al for/y, - / y , is ap p ro x im a tely

+~

(*i “ * 2 ) ± 1-96

n2

V "i

( . 9 4 - 2 . 8 ) ± 1 .9 6 J — + — v 1 V 36 26 - 1 . 8 6 + 1.15

or

—3.01 < (//, —/ / 2 ) < —.71

Since th e v alu é p ^ - p 2 = 0 d o es n o t fall in the in terv a l in p a rt b , it is n o t lik ely that /y, = p 2 . T h e re is ev id en c e to ind ícate th at the m ean s are d iffe re n t, c o n firm in g the c o n clu sió n in p art a. 9.2 9

a

T h e h y p o th esis o f in terest is tw o-tailed: h o : ^ i - ^

2

= 0

v e rsu s

H a

-Mi-Mi

560

and the te st statistic is _

(x, - x2) - 0 _ 9 8 .1 1 - 9 8 .3 9 _ js [ + s[ n,

n2

„ 22

I.72 | ,7 4 2 V 65

65

w ith p -v a lu e = P (\z\ > 2 .2 2 ) = 2 ( 1 - .9 8 6 8 ) = .0 2 6 4 . S in ce th e p - v a lu e is b etw ee n .01 an d .05, the nuil h y p o th e sis is rejected , a n d the re su lts are sig n ific an t. T h ere is ev id en c e to in d icate a d iffe re n c e in the m ean te m p e ra tu re s fo r m en v e rsu s w o m en . b S in ce the p -v a lu e = .0264, we can reject H0 a t the 5% level (p -v a lu e < .05), b u t n o t at the 1% lev el (p -v alu e > .01). U sin g the g u id elin e s fo r sig n ifica n ce g iv en in S ectio n 9.3 o f the text, w e d ecla re the resu lts statistically si g n ific a n t, b u t n o t hig hly significant. 9 .3 3

a

T h e tw o sets o f h y p o th esis b o th in v o lve a d ifferen t b in o m ia l p a ra m e te r p: H 0 : p = .6

versus

Ha : p

H 0 : p = .5

v ersu s

H a : p < .5 (p a rt b)

.6 (p a rt c)

x 35 F o r the se co n d test in p art a, x = 35 and n = 75 , so th at p = — = — = .4 6 6 7 , the n 75 te st statistic is b

85

z = PZPg = A 6 V -¿ = M o n

_ 5S

-5 ( '5 ) V 75

S ince no v alué o f a is sp ec ifie d in ad v an ce, w e calcú late p -v a lu e = P ( z < - .5 8 ) = .2 8 1 0 . S in ce th is p -v a lu e is g re ate r than .10, th e nuil h y p o th esis is not re je c te d . T h ere is in su fficie n t e v id e n c e to c o n tra d ict the claim . x 49 F o r the first test in p art a , x = 4 9 and n = 7 5 , so th at p = — = — = .6 5 3 3 , th e test n 75 statistic is

c

_ _ P - P o _ .6 5 3 3 -.6 _ n i M o

/.6 (.4 ) 75

w ith p -v alu e = P ( |z | > .9 4 ) = 2 (. 1736) = .3 4 7 2 . S in ce the p -v a lu e is g re a te r th an .10, the nuil h y p o th esis is not re je c te d . T h e re is in su fficien t e v id e n c e to co n tra d ic t the claim . ^

T h e h y p o th esis o f in terest is H 0 : p = .45

H a : p * .45

v ersu s

x 32 W ith p = — = — = .4 , the te st statistic is ai 80 z = P S P ^ = - 4 0 - .4 5 Polo

"

-

90

.45 (.55)

V

80

T h e rejectio n reg ió n is tw o -ta ile d a = .0 1 , o r |z¡ > 2 .5 8 a n d H 0 is no t re je c ted . T h ere is in su fficien t e v id en c e to d isp u te th e n e w sp a p e r’s claim . T h e h y p o th esis o f in terest is H 0 : p = .40 v ersu s

H a : p * .40

x 114 w ith p = — = ------ = .3 8 , the test sta tistic is ai 300 P -P o _ M o

-3 8 - . 4 0

_

n]

1.40 (.60) 300

T h e rejectio n reg ió n w k h a = .0 5 is | z |> 1.96 and the nuil h y p o th e sis is n o t re je c te d . (A ltern ativ ely , w e c o u ld c a lc ú la te p -v a lu e = 2 P ( z < - .7 1 ) = 2 (.2 3 8 9 ) = .4 7 7 8 . S ince this p -v a lu e is g reater than .05, the nuil h y p o th e sis is n o t reje cte d .) T h e re is in su fficien t e v id e n c e to in d ic ate th at the p ro p o rtio n o f h o u se h o ld s w ith at least one d o g is d ifferen t from that re p o rted by the H u m an e Society.

86

9 .4 5

a

T h e h y p o th e sis o f in te re st is: H 0 : P\ - p 2 = 0 v ersu s

Ha

: p, - p 2 < 0

+ n 2P i n, + n 2

C a lc ú la te p, = .3 6 , p , = .60 a n d p =

_ 18 + 3 0 = 4 3 . T h e test sta tistic is 5 0 + 50

then P '-P >

T-

-

-3 6 - - 60

l- f l+ jJ l V

V-4 8 (-5 2 ) ( 1/ 5 ° -+ 1/ 5 0 )

n2 J

Ui

T h e rejectio n reg ió n , w ith a = .05 , is z < - 1 .6 4 5 an d H 0 is re je c ted . T h e re is e v id en c e o f a d ifferen c e in th e p ro p o rtio n o f su rv iv o rs fo r th e tw o g ro u p s. b F ro m S ectio n 8.7, the a p p ro x im ate 9 5 % c o n fíd e n ce in te rv al is ( Á -P :)± 1 .9 6

M

+M n2

V ni ( .3 6 - .6 0 ) ± 1 .9 v 7 V -.2 4 ± .1 9 9 .4 7

or

+M 50

50

- .4 3 < ( p , - p , ) < - . 0 5

T h e h y p o th e sis o f in terest is H 0 : p, - p , = 0

v ersu s

H a : p, - p 2 ^ 0

C a lcú late p, = — = .214 , p 1 = — = .25 , an d p = + *2 = + ^ = .227 . 56 32 n, + n 2 5 6 + 32 T h e test statistic is then T=h z h

í —1

+

v n. i

—O

-2 1 4 ~ -2 5 yj.221 ( .7 7 3 ) (l/5 6 + 1/32)

™

n ,2 y

T h e rejectio n reg ió n , w ith or = .05 , is |z| > 1.96 an d H 0 is n o t re je c te d . T h e re is in su fficien t e v id e n c e to in d icate a d iffe re n c e in th e p ro p o rtio n o f re d M & M s fo r the p lain an d p e a n u t v arieties. T h e se resu lts m atch th e c o n c lu sio n s o f E x e rc ise 8.53. 9.51

T h e h y p o th e sis o f in terest is H 0 : p, - p 2 = 0

v ersu s

H a : p, - p , > 0

93 119 x +x 93 + 119 C a lcú late p, = — = .7 6 9 , p 0 = ------ = .5 9 8 , a n d p = — - = ---------------= .6 6 2 5 . 121 199 n , + n 2 121 + 199 T h e te st statistic is then z =

I w

1 1 —+— «2/

, -7 6 9 - 5 9 8 = 3 .1 4 ^/.6 625 (.3 3 7 5 ) (1/121 + 1/199)

87

w ith p -v a lu e = P { z > 3 .1 4 ) = 1 - .9 9 9 2 = .0008 . S in ce the p -v a lu e is less th an .01, the resu lts are re p o rte d as h ig h ly sig n ific a n t at the 1% lev el o f sig n ifica n ce . T h ere is ev id en ce to co n firm the re s e a rc h e r's co n clu sió n . 9.55

T h e p o w er o f th e test is 1 - / ? = ^ ( r e je c t H 0 w hen H 0 is f a l s e ) . A s p g e ts farth er from p 0 , the p o w e r o f the te st in creases.

9 .5 9

a -b S ince it is n e ce ssa ry to p ró v e that the a v erag e pH level is less th an 7 .5 , the h y p o th e sis to be tested is o n e-tailed : H n : p = 7.5 d

v ersu s

H a : p < 7.5

T h e test statistic is 1 Z _ ÍL x x - E = a /4 7 t

s/J ¡

-.2 .2/V 3Ó

a n d the reje c tio n reg ió n w ith a = .05 is z < - 1 .6 4 5 . T h e o b se rv ed valu é, z = - 5 .4 7 7 , falls in the re je c tio n reg ió n an d H0 is rejected . W e co n c lu d e th a t the a v erag e pH lev el is less th an 7 .5 . 9.63

L et p, be the p ro p o rtio n o f d e fe c tiv e s p ro d u ce d by m ach in e A and p , be the p ro p o rtio n o f d e fe c tiv e s p ro d u c e d by m ach in e B. T h e h y p o th esis to be tested is H(>: P\ ~ Pi - 0 v ersu s C a lcú late p, = 200

H a : p, - p 2 * 0

= .0 8 , p , = —— = .0 4 , and p = 200 K ni + n 2

— - .06 . T h e 2 0 0 + 200

test statistic is then , 1 1 N \p q — H----n. n 2

•Q 8~ 0 4 = 1.684 0 6 (.9 4 ) ( V 2 0 0 + 1 /2 0 0 )

T h e rejectio n reg ió n , w ith a = .05, is |z| > 1.96 and H 0 is n o t re jec ted . T h e re is in su fficien t e v id e n c e to in d ícate th at the m ach in es are p erfo rm in g d iffe re n tly in term s o f the p ercen tag e o f d e fe c tiv e s b ein g p ro d u ced . 9 .6 7

T h e h y p o th esis to be tested is Hn : p, - p : = 0

v ersu s

H a : p, - p : > 0

and the test statistic is ( I , - T 2) - 0 _

n,

n,

1 Q -8

V 40

88

40

_ /l

T h e re je c tio n reg ió n , w ith a = .05, is o n e -tailed o r z > 1 6 4 5 and the n uil h y p o th esis is rejected . T h ere is su fficien t e v id en c e to in d ícate a d iffe re n c e in the tw o m eans. H en ee, w e co n c lu d e th at d iet I has a g re a te r m ean w e ig h t lo ss th an d ie t II. 9.71

T h e h y p o th e sis to be tested is

a

H0 :/y ,- //2 = 0

v ersu s

Ha

>0

an d the te st statistic is

T h e reje c tio n reg ió n , w ith a = .05, is o n e -ta ile d o r z > 1.645 a n d th e n uil h y p o th e sis is re je c te d . T h e re is a d iffe re n ce in m ean yield fo r the tw o ty p e s o f spray. b

A n a p p ro x im ate 95% c o n fid e n ce in terv al for/y, -

13 + 5.88 9 .7 5

is

7.12 < (//, - f r ) < 18.88

or

T h e h y p o th e sis to be tested is H 0 : // = 5

versus

H a : // > 5

an d the te st statistic is __ x

Mo

a /J ñ

x

Mo _

s/^Jñ

7.2

5 _ ...

6 .2 /V 3 8

T h e rejec tio n reg ió n w ith a = .01 is z > 2 .3 3 . S in ce the o b se rv e d v a lu é , z = 2 .1 9 , d o e s n o t fall in the rejectio n reg ió n a n d H0 is not reje c te d . T h e d a ta d o n o t p ro v id e su fficien t e v id en c e to in d icate th at th e m ean p p m o f P C B s in th e p o p u latio n o f g am e b ird s e x c e e d s the F D A ’s rec o m m en d ed lim it o f 5 ppm . 9 .7 9

T h e h y p o th e sis to be tested is H 0 : P\ - Pi ~ 0

versus

Ha : p x- p2 * 0

= - — = .37 , and C alcú late p , = — = .4 0 , p ,2 = 5512 1 6 124 5 . _ x, + x 2 _ 6124(.4) 1) + + 5512(.37) 5 5 1 2 (.3 7 ) _ 11636 ^ ni + n 2 11636

■phe te st sta tistic is then

89

z =•

.4 0 - . 3 7

P \-P l

= 3.32

>/.3 8 6 ( .6 1 4 ) ( l/6 1 2 4 + l/5 5 1 2 )

T h e rejectio n reg ió n fo r a = .01 is | z \> 2 .5 8 and the nuil h y p o th esis is rejected . T h ere is su fficie n t e v id e n c e to ind ícate th at the p e rc e n ta g e o f stu d e n ts w ho are fluent in E n g lish d iffers fo r th ese tw o d istricts. 9 .8 3

a

T h e p a ra m e te r o f in te re st is / / , the a v e ra g e d aily w age o f w o rk ers in a giv en

industry. A sam p le o f n = 4 0 w o rk e rs has been d raw n fro m a p a rtic u la r co m p an y w ithin th is in d u stry a n d x , the sa m p le av erag e , has b een ca lc u late d . T h e o b je c tiv e is to d eterm in e w h e th e r th is c o m p a n y p ay s w ag es d ifferen t from the to ta l in d u stry . T h at is, assu m e that th is sam p le o f fo rty w o rk ers has been d ra w n fro m a h y p o th etica l p o p u latio n o f w o rk e rs. D o e s th is p o p u la tio n h av e as an a v erag e w age / / = 5 4 , o r is / / d ifferen t from 5 4 ? T h u s, th e h y p o th e sis to be tested is H 0 : ¡x = 5 4

v ersu s

H : // * 54 .

b -c T h e test statistic is

z~

x-M

5 1 .5 0 - 5 4

s/y/ñ

1 1 .8 8 /7 4 0

= -1 .3 3 1

an d the L a r g e - S a m p le T e s t o f a P o p u la tio n M e a n a p p let g iv es p -v a lu e = . 1832 . (U sing T ab le 3 w ill p ro d u c e a p - \a lu e o f . 1836.) d S in ce a = .01 is sm a lle r than the p -v a lu e , . 1832, H0 c a n n o t be re je c te d and we ca n n o t co n c lu d e th at th e c o m p a n y is p ay in g w ag es d ifferen t fro m the in d u stry averag e. e S ince n is g rea te r than 30, the C en tral L im it T h e o re m w ill g u a ra n te e the n o rm ality o f x re g a rd le ss o f w h e th e r the o rig in al p o p u la tio n w as n o rm al o r not.

90

10: Inference from Small Samples 10.1

R efer to T a b le 4 , A p p e n d ix I, in d ex in g d f a lo n g th e left o r rig h t m a rg in a n d ta ac ro ss th e to p . a tM = 2 .0 1 5 w ith 5 d f

10.5

b

t 02S = 2 .3 0 6 w ith 8 d f

c

t 02i » 1.96 w ith 3 0 d f

c

/ 10 = 1.330 w ith 18 d f

a

U sin g th e fo rm u las g iv e n in C h a p te r 2 , calc ú late X * (. = 70.5 an d X * 2 = 4 9 9 .2 7 .

T hen y .Z i.T O í.™ n 10

/ =

b

,

4 9 9 .2 7 - ^ --------------2— = ------------------1^— = .2 4 9 4 44 n -\ 9

and

5 = .4994

W ith d f = n - 1 = 9 , th e ap p ro p ria te v alu é o f t is / 01 = 2.821 (fro m T a b le 4 ) and

th e 9 9 % u p p e r o n e-sid ed co n fid e n c e b o u n d is c x + t 0i- f =

01

19 4 9 4 4 4 => 7.05 + 2 .8 2 1 J :

V io

=> 7.05 + .446

o r /y < 7 .4 9 6 . In terv als c o n stru c te d u sin g th is p ro ce d u re w ill e n d o s e /y 9 9 % o f the tim e in rep e a te d sam p lin g . H en ee, w e a re fairly ce rta in th a t th is p a rtic u la r in terv al en d o ses / y . c

T h e h y p o th e sis to be tested is H 0 : /y = 7.5

v ersu s

H a : /y < 7.5

a n d th e test statistic is 7 0 5 - 7 .5 , s/J ñ

1.2 4 9 4 4 4 10

T h e re je c tio n re g ió n w ith a = .01 an d « - 1 = 9 d eg rees o f fre ed o m is lo c a te d in the lo w e r tail o f th e /-d istrib u tio n and is fo u n d fro m T a b le 4 as / < - / 01 = - 2 .8 2 1 . S in ce th e o b se rv e d v alu é o f th e te st statistic falls in th e re je c tio n reg ió n , H 0 is re je c te d a n d w e c o n c lu d e th a t /y is less th a n 7.5. d

N o tic e th a t th e 9 9 % u p p e r o n e-sid ed co n fid e n c e b o u n d fo r /y d o es n o t in c lu d e the

v alu é /y = 7.5 . T h is w o u ld co n firm th e re su lts o f th e h y p o th e sis te st in p a rt c , in w h ic h w e c o n c lu d e d th a t /y is less th a n 7.5. 10.9

a

S im ila r to p rev io u s ex erc ise s. T h e h y p o th esis to b e te ste d is H 0 :/y = 100

v ersu s

91

H a :/ y < 1 0 0

Ijc,. _ 1797.095 C alcú late x = — L = 8 9 .8 5 4 7 5 n ~ 20

165,6 9 7 .7 0 8 1 n -1

(1797.095 )2 = 222.1150605

19

and

5 = 14.9035

T h e test statistic is

V2Ó T h e critical valué o f t w ith or = .0 1 a n d n - \ = 1 9 d e g re e s o f freed o m is r 01 = 2.539 and the re je c tio n reg ió n is t < - 2 .5 3 9 . T h e nuil h y p o th esis is reje c te d and w e co n c lu d e th a t /J is less th a n 100 D L. b T h e 9 5% u p p e r o n e -sid e d c o n fid e n c e b o u n d , b ased on n - 1 = 19 d e g ree s o f freedom , is x + í0 yjn

=> 8 9 .8 5 4 7 5 + 2 .5 3 9 14'9Q^ 1 2 5 11 => / / < 9 8 .3 1 6 V 20

T h is c o n firm s the re su lts o f p a rt a in w h ich w e c o n c lu d e d th at th e m ean is less than 100 D L. 10.13

a

T h e h y p o th e sis to be tested is H 0 : n = 25

v ersu s

H a : ¡u < 25

T h e test statistic is

T h e critical valu é o f t w ith a = .05 a n d n - 1 = 2 0 d e g re e s o f freed o m is t 05 = 1 .7 2 5 and the reje c tio n reg ió n is t < - 1 . 7 2 5 . S in ce the o b se rv ed v alu é d o es falls in the rejectio n reg ió n , H 0 is re jec ted , and w e co n c lu d e th at p re -tre a tm en t m ean is less than 25. b

T h e 9 5% c o n fid e n c e in terv al b ase d o n d f = 2 0 is 7.4 => 2 6 .6 ± 2 . 0 8 6 - = V 2l

=> 2 6 .6 ± 3 .3 7

o r 2 3 . 2 3 < 2 9 .9 7 . c 10.17

T h e p re -tre a tm en t m ean lo o k s c o n sid e ra b ly sm a lle r th an the o th e r tw o m eans.

R efer to E x erc ise 10.16. If w e use th e larg e sa m p le m eth o d o f C h a p te r 8, the large sam p le c o n fid en c e in terv al is

=> 246.96±12.98 92

o r 2 3 3 .9 8 < p < 2 5 9 .9 4 . T h e in terv als are fairly sim ilar, w h ich is w h y w e c h o o se to x — u. p - w ith a z d istrib u tio n w h en n > 3 0 . s/y jn

a p p ro x im a te th e sa m p lin g d istrib u tio n o f

10.19

a

_ (/,,

+ ( „ , - 1 ) k _ 9 ( 3 .4 ) + 3 ( 4 .9 ) n] + n 2 —2 + (n2 - \ ) s ;

b

ll( 1 8 ) + 2 0 ( 2 3 )

nt + n 2 —2 10.25

a

2 7?;

10 + 4 - 2

12 + 21

—

^

2

T h e h y p o th e sis to b e tested is H 0 :/ /, ~ n 2 = 0

v ersu s

Ha :p, - p 2 * 0

F ro m th e M i n i t a b p rin to u t, th e fo llo w in g in fo rm a tio n is a v ailab le: ^ = .896

s f = (.4 0 0 )2

n, = 1 4

x 2 = 1 .1 4 7

s \ = ( 6 7 9 )2

rc2 = 11

an d the test statistic is t=

(T, - x 2 ) - 0 U

= - 1 .1 6

«2 T h e rejectio n re g ió n is tw o -tailed , b ased on n, + n 2 - 2 = 23 d e g re e s o f freed o m . W ith a = .05 , fro m T a b le 4 , the reje c tio n reg ió n is \t\ > tm5 = 2 .0 6 9 a n d Hq is n o t re jec ted . T h e re is not e n o u g h e v id e n c e to in d ícate a d ifferen c e in the p o p u la tio n m eans. b It is not n e c e ssa ry to b o u n d the p -v a lu e u sin g T a b le 4 , sin c e th e e x a c t p -v a lu e is g iv en on th e p rin to u t as P -V a lu e = .260. c I f you ch e c k th e ratio o f the tw o v a rian ces u sin g the ru le o f th u m b g iv en in this sectio n you w ill find: \2 l a r g e r r _ (.6 7 9 )2 =

sm a lle r s 2

2.1

(.4 0 0 )2

w h ic h is less th an th ree. T h erefo re, it is rea so n a b le to assu m e th a t the tw o p o p u la tio n v a rian ces are equ al. 10.27

a

C h e c k th e ra tio o f the tw o v a rian ces u sin g the ru le o f th u m b g iv en in th is section: J a r g e r r _ _ 2/78 0 9 5 _ lf i2 2 sm aller 5

.17143

w h ich is g reater th an th ree. T h erefo re, it is n o t re a so n a b le to assu m e th at th e tw o p o p u la tio n v a rian c es are equal. b Y ou sh o u ld use the u n p o o led t test w ith S a tte rth w a ite ’s a p p ro x im atio n to the d e g re e s o f freed o m fo r testing H 0 :p, ~ n 2 = 0

v ersu s

93

H a : //, - ¡i2 * 0

T h e test statistic is 3 . 7 3 - 4 .8

t =

2 .7 8 0 9 5 i— + n, n

15

= -2 .4 1 2

.17143 +15

w ith r s¡ s — +— n, n 2

df =

( .1 8 5 3 9 7 + .0 1 1 4 2 8 7 )2 2

2

.0 0 2 4 5 5 1 3 7 + .0 0 0 0 0 9 3 3

U y - + l n2 ; n ,«2W ith d f ~ 15 , the p -v a lu e fo r this te st is b o u n d ed b e tw een .02 a n d .05 so th at H 0 can b e re je c te d at the 5% lev el o f sig n ifican ce. T h e re is e v id e n c e o f a d iffe re n c e in the m ean n u m b e r o f u n co n tam in ate d eg g p lan ts fo r th e tw o d isin fectan ts. a I f sw im m e r 2 is faster, h is(h er) a v erag e tim e sh o u ld be less th an th e a v erag e tim e fo r sw im m e r 1. T h erefo re , th e h y p o th esis o f in terest is H 0 : //, ~ / j 2 = 0

v ersu s

Ha

>0

a n d th e p re lim in a ry calc u la tio n s are as follow s: S w im m er 1

S w im m e r 2

I * , , = 5 9 6 .4 6

I jc2í = 5 9 6 .2 7

= 3 5 5 7 6 .6 9 7 6 n, = 10

1 4

= 3 5 5 5 4 .1 0 9 3

n2 = 10

T hen

, í •*]/

( W

+y , ( i ^ ) 2 1- x2¡ !h_______________ n2 n, + n2 - 2

3 5 5 7 6 .6 9 7 6 - ( 5 9 6 4 6 ) + 3 5 5 5 4 .1 0 9 3 - ,(5 9 6 '2 7 ) 10 10 5+ 5 -2

A l so,

x¡ = 1

5 9 .6 4 6 10

_

and

= .0 3 1 2 4 7 2 2

5 9 6 .2 7 _ = ------------ = 59.6 27

2

10

T h e te st statistic is t=

( x , - * 2) - 0 1. ( ' O —+— r v n\ n i )

5 9 .6 4 6 - 5 9 .6 2 7 .031247221

1

V io

= 0.24 10

F o r a o n e-ta ile d test w ith d f = «, + « , - 2 = 1 8 , the p -v a lu e c a n be b o u n d e d using T a b le 4 so th at p -v a lu e > .1 0 , and H 0 is not rejected . T h e re is in su ffic ie n t e v id e n c e to in d ícate th a t sw im m e r 2 ’s a v erag e tim e is still fa ste r than th e a v e ra g e tim e for sw im m e r 1. 10.35

a

T h e test statistic is t=

d -n d

.3 -0

= 2.372

Jy fñ w ith « - 1 = 9 d e g re e s o f freedom . T h e p -v alu e is then 1 P (\t\ > 2 .3 7 2 ) = 2 P ( t > 2 .3 7 2 ) so th at P ( t > 2 .3 7 2 ) = - p -v a lu e S in ce th e v alu é t = 2 .3 7 2 falls b e tw een tw o tab led e n trie s fo r d f = 9 ( f 025 = 2 .2 6 2 and f 01 = 2 .8 2 1 ), y o u c a n co n c lu d e that .01 < — p -v a lu e < .025 2 .02 < p -v a lu e < .05 S in ce th e p -v a lu e is less th an a = .05 , the nuil h y p o th esis is re je cte d an d w e co n c lu d e th a t th e re is a d iffe re n c e in the tw o p o p u la tio n m eans. b

A 9 5% c o n fid e n ce in terv al fo r p 1 - p , = p ^ is d ± t a a -^=

=>

3 ± 2.2 62 . —

025

=> ,3 ± .2 8 6

VIO

o r ‘.0 1 4 < ( p , - p 2) < .5 8 6 . c

U sin g sf¡ = .1 6 and B = . 1, the in eq u ality to b e so lv e d is a p p ro x im a te ly 1 .9 6 - ^ < .l ■Jn r 1 .9 6 V 4 6 f n > -------------- = 7.84

« > 6 1 .4 7

or

« = 62

.1

S in ce th is v alu é o f n is g rea te r than 30, the sam p le size, « = 62 p airs, w ill be valid. 10.37

a It is n e ce ssa ry to u se a p aire d -d iffe re n c e test, sin c e the tw o sa m p les are no t ra n d o m and in d ep en d e n t. T he h y p o th esis o f in terest is H 0 : P] - P 2 = 0

or

H0 :p d = 0

Ha : p ,- p 2 * 0

or

Ha : p ^ 0

T h e table o f d iffe re n ce s, a lo n g w ith the calc u latio n o f d and s] , is p re se n te d below . d¡

.1

.1

0

.2

-.1

I d, = .3

df

.01

.01

.00

.04

.01

Z d f = .07

95

d = ^ - = - = .0 6 n 5

JJ

and

_

.0 7 -

(■3): = .013

T h e test statistic is d -jU j .0 6 -0 , t=— = = 1 .1 7 7 sjy jn Im 3 w ith n - 1 = 4 d eg rees o f freed o m . T h e rejec tio n reg ió n w ith a = .05 is H > ¿ 025 = 2 .7 7 6 , and H 0 is n o t re je c te d . W e c an n o t co n c lu d e th at th e m ean s are differen t. b T h e p -v a lu e is P flíl > 1 .1 7 7 ) = 2 /> (f > 1 .1 7 7 ) > 2 ( . 1 0 ) = .2 0 A 95% c o n fid en c e in te rv a l fo r //, —/ / 2 = / i d is

c

.013

.06 + 2.776.

¿ ± '0 2 5 - f vn

.06 ± .1 4 2

o r - .0 8 2 < ( / / , - / / , ) < .2 0 2 . d In o rd e r to u se th e p a ire d -d iffe re n c e test, it is n e cessary th at th e n p aired o b se rv atio n s be ra n d o m ly se le cte d fro m n o rm ally d istrib u te d po p u latio n s. 10.41

a E ach su b ject w as p re se n te d w ith b o th sig n s in ra n d o m ord er. I f h is re a c tio n tim e in g en eral is h ig h , b o th resp o n se s w ill be high; if h is re actio n tim e in g en eral is low , b o th re sp o n se s w ill be low . T h e larg e v ariab ility fro m su b jec t to su b je ct w ill m ask the v ariab ility d u e to th e d iffe re n c e in sign ty p es. T h e p a ire d -d iffe re n c e d e sig n will elim in a te the s u b je c t to su b je c t variab ility . b T h e h y p o th esis o f in te re st is - / /2 = °

or

H (): / r d = 0

H a :/r ,- /r 2 * 0

h o:^ i

or

H a ://rf* 0

T h e tab le o f d ifferen c e s, a lo n g w ith the calc u la tio n o f d and s] , is p re se n te d below . D riv er d,

2 141

1 122

3 97

4 107

5 37

6 56

7 110

8 146

9 104

10 149

T o tals 1069

j =I 4 =1069=106.9 n

Z d? -.

10

(£ 4 )2

126,561 -

(1 0 6 9 )2 10

“

n - \1

and the te st statistic is J - ^ = 1 0 6 .9 - 0 s d /4~n

3 6 .9 4 5 8

Vio 96

= 1364.

and

= 3 6 .9458

S in ce t = 9 .1 5 0 w ith d f = n - 1 = 9 is g rea te r th an th e tab led valué t m , p -v a lu e < 2 (.0 0 5 ) = .01 fo r this tw o ta ile d te st and Ho is re je cte d . W e c a n n o t co n c lu d e th at the m ean s are differen t. c

T h e 9 5 % c o n fid e n c e interv al fo r /i, - / / , = fid is

d ± t m, - ^ =

' y fn

~~~

=> 106.9 ± 2 .2 6 2 --6-^ i 5-8 => 106.9 ± 2 6 .4 2 8

Vio

o r 8 0 .4 7 2 < ( f i x ~ n 2 ) < 1 3 3 .3 2 8 . 10.45

a

U se the M i n i t a b p rin to u t g iv en in th e tex t b elo w . T h e h y p o th e sis o f in te re st is H 0 'V a ~Mb = 0

Ha :^ - /y B>0

an d the te st sta tistic is d -n d r- ^ 7 X

1 .4 8 7 5 - 0 - T 4 9 Í 3 T - 2 -82 V8

T h e p -v a lu e sh o w n in the p rin to u t is p -v a lu e = .013 . S in ce the p -v a lu e is less than .05, H0 is re je cte d a t the 5% lev el o f sig n ifican ce. W e co n c lu d e th at a sse sso r A giv es h ig h er a sse ssm en ts th an a sse sso r B. b

A 9 5 % lo w e r o n e -sid e d c o n fid e n c e b o u n d fo r p , - / / 2 = n d is s d - 105 —j= V»

1 49134 => 1 .4 8 7 5 -1 .8 9 5 ' V8

=> 1 .4 8 7 5 - .9 9 9

o r (//, - / / 2 ) > .4885 . c In o rd e r to a p p ly the p a ire d -d iffe re n c e test, th e 8 p ro p e rtie s m u st be ran d o m ly and in d ep en d e n tly selected and th e a ssessm e n ts m u st be n o rm ally d istrib u ted . d Y es. I f the in d iv id u al asse ssm e n ts a re n o rm ally d istrib u ted , th en the m ean o f fo u r asse ssm e n ts w ill be n o rm a lly d istrib u te d . H enee, the d iffe re n c e x A —x w ill be n o rm ally d istrib u te d a n d the t test on th e d iffe re n c e s is v alid as in c. 10.49

F o r this ex ercise , s 2 = .3 2 1 4 and n = 15 . A 90% c o n fid e n c e in terv al fo r c r w ill be (n -l)s 2

Xa/2

2 ( m —1) < o~ <

X(i-a/2)

w h ere Xa/i rep rese n ts the valué o f x ~ su c h th at 5% o f th e a rea u n d e r the c u rv e (show n in the figure on the n ex t p a g e) lies to its rig h t. S im ilarly , X~ v alué su ch th a t an a rea .95 lies to its right.

97

x]\-a¡i)

he the

H enee, w e have lo cated o n e -h a lf o f a in each tail o f the d istrib u tio n . In d ex ing Xm 2 and x l s w ith n - 1 = 14 d eg re e s o f freed o m in T a b le 5 y ields

X l s = 2 3 .6 8 4 8

x l s = 6 .57 0 6 3

and

an d the c o n fíd en c e interv al is 14 (.3 2 1 4 ) , 1 4 Í.3 2 1 4 ) — -------- - < a < — ---------2 3 .6 8 4 8 6 .5 7 0 6 3 10.53

a

or

.1 9 0 tms = 3 .1 8 2 . S in ce the o b se rv e d v alu é o f th e test statistic d o e s not fall in the reje c tio n reg ió n , Ho is not rejected . T h e re is in su fficien t e v id en c e to sh o w that the m ean d iffe rs fro m 5 m g/cc. b T he m an u fac tu re r cla im s th a t th e ra n g e o f the p o ten ey m e a su rem en ts w ill e q u a l .2. S ince th is ran g e is g iv en to eq u al 6cr . w e k n o w th at o ~ .0333 . T h en H 0 : .0 0 1 1

T h e test statistic is y

,

(n -l)í2 3 Í.0 0 7 4 ) = 4 J - = - ± -------- = 20.18 a.0011

and the o n e -ta ile d re je c tio n reg ió n w ith a = .05 and n - 1 = 3 d e g re es o f freedo m is r

> * 2 5 = 7.81

H0 is rejected ; th ere is su ffic ie n t e v id e n c e to in d icate th a t the ran g e o f the poten cy w ill e x c e ed the m a n u fa c tu re r’s claim . 10.57

H 0 : a = 150

T h e h y p o th e sis o f in te re st is

H a : a < 150

C alcú late (n - 1 )s 2 = I * 2 -

=

= 6 6 2 ,2 3 2 .8

9 2 ,3 0 5 , 6 0 0 -

and th e test statistic is y 2 = — — íp—- = ^ > 2 ,2 3 2 .8 = 2 9 .4 3 3 . T h e o n e-tailed rejectio n 1.059) > 2 (. 10) = .20 Sin ce the p - \ a lu e is so larg e , H0 is n o t rejected . T h e re is no e v id en c e to in d icate that the v a rian ce s a re d ifferen t. 10.65

F o r e a c h o f the th ree te sts, the h y p o th e sis o f in terest is H 0 : of =

of

v ersu s

H a : of

* of

and the te st sta tistic s are F -2 S-s

3 .9 2 2

1.03 '

F = 4 = 4 ^ = 2.01 si 3 .4 9 2 '

and

F = ¿ 5 ,2

=

^ = 14.29 4 .4 7 2

T h e critical v alú es o f F fo r v ario u s v a lú e s o f a are giv en on the fo llo w in g page, u sin g d f = 9 an d d f 2 = 9 .

99

a

.10 2 .4 4

.05 3.18

.025

.01

4.03

5.35

.005 6.54

H enee, fo r the first tw o tests, p -v a lu e > 2 (. 10) = .20 w hile fo r the last test, p -v a lu e < 2 (.0 0 5 ) = .01 T h e re is n o e v id en c e to in d íca te th a t th e v a ria n c e s are d iffe re n t fo r the first tw o tests, but H0 is re je c te d fo r the th ird v ariab le. T h e tw o -sam p le r-test w ith a p o o le d estím ate o f o 1 ca n n o t be used fo r the th ird variab le. 10.69

P aired o b serv atio n s are u sed to e stím a te th e d ifferen c e b etw ee n tw o p o p u latio n m ean s in p referen ce to an e stim a tio n b a sed on in d ep en d en t ra n d o m sa m p les se lected from the tw o p o p u latio n s b ec a u se o f the in c re a sed in fo rm atio n c a u se d by b lo c k in g the ob serv atio n s. W e e x p e c t b lo c k in g to crea te a larg e re d u c tio n in the stan d ard dev iatio n , i f d ifferen c es d o e x is t a m o n g th e blocks. P aired o b se rv a tio n s are no t a lw a y s p refe ra b le. T h e d e g ree s o f freed o m th at are av ailab le fo r e stim a tin g o 2 are less fo r p a ired than fo r u n p aired o b serv a tio n s. I f there w ere no d iffe re n c e b e tw een the b lo ck s, the p a ire d e x p e rim e n t w o u ld th en be less beneficial.

10.73

Since it is n ecessa ry to d e te rm in e w h eth er the in jected rats d rin k m o re w ater than n o n in jected rates, the h y p o th e sis to be tested is H 0 : p = 2 2 .0

H a : p > 22.0

an d the te st statistic is f=

3 1 .0 - 2 2 .0 = 5 9 8 5 s/'yjn

62

U sin g th e critical va lu é a p p ro a c h , th e re je c tio n reg ió n w ith a = .05 and n - 1 = 16 d eg rees o f freed o m is lo c ate d in the u p p e r tail o f the r-d istrib u tio n a n d is fo u n d from T a b le 4 as t > r 05 = 1.746 . S in ce th e o b se rv e d v alu é o f the te st statistic falls in the reje c tio n reg ió n , H 0 is re je c te d an d w e c o n c lu d e th at the in je c te d rats do d rin k m ore w ater th an th e n o n in je c te d rats. T h e 9 0 % co n fid e n ce in terv al is I ± 7 ;, ~ yfn

=> 3 1 . 0 ± 1 . 7 4 6 4 ¿ V Í7

3 1 .0 ± 2 .6 2 5

o r 28.375 < p < 33.6 2 5 . 10.76

T h e stu d en t m ay use the ro u n d e d v alú es fo r x an d s g iv en in the d isp la y , o r he m ay w ish to c a lc ú late x a n d s a n d u se th e m o re e x a c t c a lc u la tio n s fo r the co n fid en ce in tervals. T h e ca lc u la tio n s are sh o w n o n the n ex t page.

100

_ I jc¿ 1845 x = L = --------= 184.5 n 10

a

3 4 4 ,5 6 7 - í^ s 2 = ---------- *— n -1

= ---------------------------------= 4 6 2 .7 2 2 2 9

5 = 21.511 an d the 9 5 % co n fid e n ce in terv al is

5 x ± tm -= yjn o r 169.1 < /u < 199.9 .

v- 2 (X -v .)2 Z x ,2 - V ^ 52 = — 5 — = n- 1

I = ^ Í L = — = 7 3 .0 n 10

-

b

_____________ 21.511 => 1.845 ± 2 .2 6 2 — ^ = — => 184.5 ± 1 5 .4 -s/l 0

(7 3 0 )‘ 5 3 5 1 4 -> ' —— = 2 4 .8 8 8 9 9

s = 4 .9 8 9 and the 9 5 % co n fid e n c e interv al is j x ± t m. - 7= Jü

4 989 => 7 3 . 0 ± 2 . 2 6 2 - = VTÓ

=> 7 3 .0 ± 3 .5 7

o r 6 9 .4 3 < /J < 7 6 .5 7 . c

_ I x 25.42 x = i- = --------- = 2 .5 4 2 n 10 S x j _ ( £ ü ) _

6 5 .8 3 9 8 _ ( 2 5 '4 2 )

^ ------= .1 3 5 7 9 5 5 6 n- 1

9

5 = .3685 and the 9 5 % co n fid e n ce interv al is

x ± tms Vn o r 2.28 < f i < 2 .8 0 .

=> 2 . 5 4 ± 2 . 2 6 2 = > VlO

2 .5 4 ± .2 6

d N o. T h e re la tio n sh ip b e tw een the co n fid en ce in te rv als is n o t the sam e as the re la tio n sh ip b e tw een the o rig in al m easu rem en ts. 10.79

U se the c o m p u tin g fo rm u las o r y o u r scien tific c a lc u la to r to calcú late _ Ix ,. 322.1 v = -----= ----------- = 24.777 n 13 „ 2 ( I ± )2 ( 3 2 2 .1)2 Ijc ,2 - V 8 114.59 — }— 5 2 = ---------------'-1— = —— = 11.1619 n- 1 12 5 = 3 .3 4 0 9 a n d the 9 5 % co n fid e n c e interv al is

101

t. "u n o 2 4 .7 7 7 ± 2 . 1 7 9 ^ = ^ vi 3

c I ± t m5 —= yjn

2 4 .7 7 7 ± 2 .0 1 9

o r 2 2 .5 7 8 < m < 2 6 .7 9 6 . 83

a T h e ra n g e o f the first sam p le is 4 7 w hile the ra n g e o f the se co n d sam p le is only 16. T h e re is p ro b ab ly a d iffe re n c e in th e varian ces. b T h e h y p o th esis o f in terest is H (l: a y = o \ 1 7 7 ,2 9 4 -

H a : a ,2 * a \

v ersu s

(8 3 8 )2

C a lc ú la te s i =

1 9 2 ,3 9 4 - M = 5 7 7 .6 6 6 7

*,2 =

6

= 29.6

and the te st statistic is F = ii_ = 5TL666T_ _ si

29.6

T h e critic a l v alú es w ith d f x = 3 and d f 2 = 5 a re sh o w n b e lo w fro m T a b le 6. a

.10 3.62

Fa

.05 5.41

.025

.01 12.06

7.76

.005 16.53

H enee, p -v a lu e = 2 P ( F > 19 .5 1 6 ) < 2 (.0 0 5 ) = .01 S in ce th e p -v a lu e is sm alle r th an .01, H 0 is re jec ted at the 1% level o f sig n ifican ce. T h e re is a d ifferen ce in variab ility . c S in ce th e S tu d e n t’s t test re q u ire s the assu m p tio n o f eq u al v arian ce, it w o u ld be in a p p ro p ria te in this in stan ce. Y o u sh o u ld use the u n p o o le d t test w ith S a tte rth w a ite ’s a p p ro x im a tio n to the d eg ree s o f freedom . 87

A p a ire d -d iffe re n c e test is u sed , sin ce the tw o sa m p les are n o t ran d o m an d in d ep en d e n t. T h e h y p o th e sis o f in terest is H0 :A ~

H a :/il - l i 2 > 0

=0

a n d the tab le o f d ifferen ces, a lo n g w ith the calc u la tio n o f d an d s] , is p resen te d b elow . P air 4

d-

=M n

=^ = 4

5.5

1 -1

2 5

3 11

Zd>-

4 7

( 1 4 )2

n- 1

sd = 5 a n d the test statistic is

102

T o tal s 22

196-

(22)2 = 25

and

I=

d - n (i 5 .5 -0 y-£= = = 2.2 sJyJn A

T h e o n e -ta ile d p -v a lu e w ith d f = 3 can be b o u n d ed b e tw e e n .05 and .10. S in ce thi v alu é is g re a te r th an .10. H 0 is n o t rejected . T h e resu lts are n o t sig n ific a n t; th ere is in su ffic ie n t e v id e n c e to in d icate that lack o f sch o o l ex p e rie n c e h as a d e p re ssin g efl on IQ scores. 10.91

T h e o b ject is to d eterm in e w h eth e r o r n o t there is a d iffe re n c e b etw ee n the m ean resp o n se s fo r the tw o d iffe re n t stim uli to w h ich the p e o p le h av e been su b je cte d . T sa m p le s a re in d ep en d e n tly a n d ran d o m ly selected , and the a ssu m p tio n s n e cessary 1 the t test o f S ectio n 10.4 are m et. T h e h y p o th esis to be tested is H a : //, - f ¿ 2 * 0

H o ; A i - / /2 = 0

an d the p re lim in a ry c a lc u la tio n s are as follow s: 15 T, = — = 1.875 a n d

21 x 2 = — = 2.625

(1 5 )2 3 3 --— —

(2 1 )2 61 - - — = .69 6 4 3

and

s2 =

= .83 9 2 9

S in ce th e ra tio o f the v a rian ces is less than 3, you can u se th e p o o led t test. T h e p o o le d e stim a to r o f t m¡ = 2 .1 4 5 , a n d 1 is n o t rejected . T h e re is in su fficien t e v id e n c e to in d icate th at th ere is a d iffe re n c e i m eans. 10.97

It is p o ssib le to test the n uil h y p o th e sis H 0 : tJ,‘ = cr2 a g a in st a n y one o f th ree a lte rn ativ e h y p o th eses: (1 ) H a i o f * o \

(2 )

H a : o \

a T h e first a ltern ativ e w o u ld be p referred by the m a n a g e r o f th e d airy. H e d o e s n k n o w an y th in g a b o u t the v ariab ility o f the tw o m ach in es an d w ould w ish to d e tec t d e p a rtu re s fro m e q u a lity o f th e type o \ < o \ o r o \ > o \ . T h ese a lte rn a tiv e s are im p lie d in (1).

103

b T h e salesm an for c o m p a n y A w o u ld p re fer th at the ex p e rim e n te r se le ct the seco n d altern ativ e. R ejectio n o f the nuil h y p o th e sis w ould im ply th at h is m a c h in e had sm a lle r v ariab ility . M o reo v e r, even if the nuil h y p o th e sis w ere n o t rejected , th ere w o u ld be n o e v id en c e to in d icate th at the v ariab ility o f the c o m p an y A m a ch in e w as g re a te r th an the v ariab ility o f the c o m p an y B m achine. c T h e salesm an for c o m p a n y B w o u ld p refe r the th ird altern ativ e fo r a sim ilar reaso n .

10.101 A p aire d -d iffe re n c e a n aly sis is used. T o test H (): //, —/ / 2 = 0 v ersu s H a : /v, w h ere

> 0,

is th e m ean reactio n tim e a fte r injectio n and //, is th e m ean re a c tio n tim e

b efo re in jectio n , c a lc ú la te the d iffe re n c e s ( x 2 - x , ) : 6 , 1 ,6 , 1

( W T h en

d=

^ n

= — = 3.5 4

=— d

*— n- 1

74_ M l = ------------ 4— = g.33 3

and

s d = 2 .8 8 6 7 5 and the test sta tistic is t=

d - n d 3 .5 -0 — = -------------- = 2 425 s J J i 2 .8 8 6 7 5 '

F o r a o n e -ta ile d test w ith d f = 3 , the reje c tio n reg ió n w ith a = .05 is t > tM = 2 .3 5 3 , an d H 0 is rejected . W e co n c lu d e th at the d ru g sig n ifican tly in creases w ith reactio n tim e.

10.105 T h e u n d erly in g p o p u latio n s are ratin g s and can o nly tak e on the fin ite n u m b e r o f v alú es, 1, 2, ..., 9, 10. N e ith e r p o p u la tio n has a n o rm al d istrib u tio n , b u t b o th are d iscrete . F u rth er, the sa m p les are n o t in d ep en d e n t, sin ce the sam e p erso n is ask ed to ran k e ach c a r d esig n . H en ee, tw o o f the assu m p tio n s re q u ire d fo r the S tu d e n t’s t test have b een violated. 10.109

T h e M i n i t á b p rin to u t b elo w sh o w s th e su m m ary statistics fo r the tw o sam p les:

Descriptive Statistics: Method 1, Method 2 Variable Method 1 Method 2

N 5 5

Mean SE 137.00 147.20

Mean 4.55 3.29

StDev 10.17 7.36

S in ce th e ratio o f the tw o sa m p le v a rian ces is less th an 3, y o u can use the p o o le d t test to c o m p are the tw o jn e th o d s o f m easu rem en t, u sin g th e re m a in d e r o f the M i n it a b prin to u t on the n ext page:

104

Tw o-Sam ple T-Test and Cl: Method 1, Method 2 Difference = mu (Method 1) - mu (Method 2) Estímate for difference: -10.2000 95% CI for difference: (-23.1506, 2.7506) T-Test of difference = 0 (vs not =): T-Value = -1.82 Both use Pooled StDev = 8.8798 —

P-Value = 0.107

DF

T h e test sta tistic is t = - 1 .8 2 w ith p -v a lu e = . 107 and th e resu lts a re n o t sig n ific a n t T h e re is in su fficie n t e v id e n c e to d e cla re a d iffe re n c e in th e tw o p o p u la tio n m ean s. 1 0 .1 1 3

T h e h y p o th e sis to be tested is H 0 : n = 280

v ersus

H a : a > 280

T h e te st statistic is

Vio T h e critical valué o f t w ith a = .01 a n d n - 1 = 9 d e g re e s o f free d o m is í 0l = 2.821 the re je c tio n reg ió n is t > 2 .8 2 1 . S in ce th e o b se rv e d v alu é fa lls in the rejec tio n reg ió n , H 0 is re je cte d . T h e re is su fficien t e v id e n c e to in d ica te th at the a v e ra g e n u m b e r o f c a lo rie s is g re ate r than ad v ertised . 1 0 .1 1 9

U se the In ter p r e tin g C o n fid e n c e In te rv a ls ap p let. A n sw e rs w ill vary fro m studt to stu d en t. T h e w id th s o f th e te n in terv als w ill no t be the sam e, sin c e the v alu é o f c h an g e s w ith e a c h new sam p le. T h e stu d en t sh o u ld fin d th a t a p p ro x im a te ly 9 5 % i the in terv als in the first ap p le t co n tain / / , w h ile ro u g h ly 9 9 % o f the in te rv a ls in tf se co n d a p p le t co n ta in / / .

1 0 .1 2 3

U se the T w o S a m p le t T est: In d ep en d e n t S a m p les ap p le t. T h e h y p o th e sis to b( tested c o n c e rn s the d ifferen c es b e tw een m ean rec o v e ry ra te s fo r th e tw o su rg ical p ro ce d u re s. L e t //, be the p o p u la tio n m ean fo r P ro c e d u re I and f i 2 be th e p o p u la ti m ean fo r P ro c e d u re II. T h e h y p o th e sis to be tested is H 0 : A, - A2 = 0 v ersu s H a : A, - A2 ^ 0 S in ce the ratio o f th e v a rian ces is less than 3, y o u c a n use the p o o le d t test. E n te r a p p ro p ria te statistics into the a p p le t an d y o u w ill fin d th a t te st sta tistic is

w ith a tw o -ta ile d p -v a lu e o f .0030. S in ce the p -v a lu e is v e ry sm all, H 0 can be reje fo r any v alu é o f a g rea te r than .003 and the resu lts a re ju d g e d h ig h ly sig n ifican t. T h e re is su ffic ie n t e v id e n c e to ind icate a d ifferen c e in the m e a n rec o v e ry ra te s fo r tw o p ro ced u res.

105

11: The Analysis of Variance 1 1 .1

In c o m p a rin g 6 p o p u latio n s, th ere are k - 1 d e g re es o ff re e d o m fo r tre atm e n ts and n = 6 (1 0 ) = 6 0 . T h e A N O V A tab le is sh o w n below .

11.4

S ource T reatm en ts E rro r

df 5 54

T o tal

59

S im ila r to E x ercise 11.1. W ith n = 4 ( 6 ) = 2 4 and k = 4 , th e so u rc es o f v aria tio n a sso ciate d d f are sh o w n below . S ource T reatm en ts E rro r T o tal

11.7

df 3 20 23

T h e fo llo w in g p re lim in a ry calc u latio n s a re n ecessary: 7¡ = 1 4 a

(3 8 ):

CM =

7; = 1 9

7¡ = 5

G = 38

= 103.142857

14 T o tal SS = S x l - C M = 32 + 2 2 + • • • + 22 + 12 - C M = 130 -1 0 3 .1 4 2 8 5 7 = 2 6 .8í SST = Z - — CM = n,

- + — + - — C M = 1 1 7 .6 5 -1 0 3 .1 4 2 8 5 7 = 14.5071 5 5 4

, SST 14.5071 an d M S T = -------= ------------- = 7 .2 5 3 6 k - 1 2 c B y su b tra c tio n , S S E = T o ta l S S - S S T = 2 6 .8 5 7 1 -1 4 .5 0 7 1 = 1 2 .3 5 0 0 an d the d e g ree s o f freed o m , by su b tractio n , a re 1 3 - 2 = 1 1 . T h en M SE = SSE = 123500 d

e

11 11 T h e in fo rm a tio n o b tain ed in p arts a -c is C onsolidated in an A N O V A table. Source SS MS df T reatm en ts 2 14.5071 7 .2 5 3 6 E rro r 12.3500 11 1.1227 T o tal 13 26.8571 T h e h y p o th e sis to be tested is H 0 : //, =

= fiy

v ersu s

H a : a t least o n e p a ir o f m ean s are d iffe re n t

107

f

T h e rejectio n reg ió n fo r the te st sta tistic F = ---------= —;— — = 6 .4 6 is b a se d o n an M SE 1.1227

F -d istrib u tio n w ith 2 a n d 11 d e g re e s o f freed o m . T h e critical v alú es o f F for b o u n d in g th e p -v a lu e fo r th is o n e-ta ile d test a re show n below . a .10 .05 .025 .01 .005 2 .8 6

3.98

5 .2 6

7.21

8.91

S ince the observ ed valu é F = 6 .4 6 is b e tw e e n F 01 a n d F 025 , .01 < p -v a lu e < .025 and H 0 is rejected at the 5% level o f sig n ific a n c e . T h ere is a d iffe re n c e am o n g the m eans. 11.10

a

T h e follow ing p re lim in a ry ca lc u la tio n s are n ecessary: 7; = 3 8 0

CM =

T2 = 199

F, = 261

G = 84 0

( I jÜ 2 (8 4 0 )2 — = --------- = 6 4 ,1 4 5 .4 5 4 5 n 11

T o tal SS = I x¡ - C M = 6 5 ,2 8 6 - C M = 1140.5455 S S T = Z —— C M = ^ 5 _ + — n, 5 3

+ — — C M = 6 4 1 .8 7 8 8 3 3

C a lcú late M S = S S / d f an d c o n so líd a te the in fo rm atio n in an A N O V A table. S ource T reatm en ts E rro r T otal b

df 2 8 10

SS 6 4 1 .8 7 8 8 4 9 8 .6 6 6 7 1140.5455

MS 3 2 0 .9 3 9 62.3 3 3

T h e h y p o th esis to be tested is

H 0 : //, = f i 2 = p ,

v ersu s

H a : at lea st o n e p air o f m ean s are d ifferen t

and the F test to d etec t a d iffe re n c e in m ean stu d en t resp o n se is F = M 5 I = 5, 5 . M SE T h e rejectio n reg ió n w ith a = .05 a n d 2 and 8 d f i s F > 4 .4 6 and H 0 is rejected . T h e re is a sig n ifican t d iffe re n c e in m ean resp o n se d u e to the th ree d iffe re n t m ethods. .13

a W e w ould be re a so n a b ly c o n fid e n t th a t the d a ta satisfied the n o rm ality a ssu m p tio n b ecau se each m easu rem e n t re p re se n ts th e a v erag e o f 10 co n tin u o u s m e asu rem en ts. T h e C en tral L im it T h e o re m a ssu re s u s th a t this m ean w ill be ap p ro x im a te ly n o rm ally d istributed. b W e have a co m p letely ra n d o m iz e d d e sig n w ith fo u r treatm en ts, e ac h c o n tain in g 6 m easu rem en ts. T h e an aly sis o f v a rian c e ta b le is giv en in the M i n it a b prin to u t. T he F test is f

= M S T = ^ 5 8 0 = 5 7 38 M SE

.115

108

w ith p -v a lu e = .0 0 0 (in the c o lu m n m ark ed “ P ” ). S in ce the p -v a lu e is v ery sm all i th an .0 1 ), H 0 is re je c te d . T h ere is a sig n ifícan t d ifferen c e in the m ean le a f len g th am o n g the fo u r lo c a tio n s w ith P < .01 o r ev en P < .001. c

T h e h y p o th e sis to be tested is H 0 : p , = p 4 v ersu s

sta tistic is

H a : p , * p 4 a n d the test

~~~ x ,-x á

t =

6 .0 1 6 7 - 3 .6 5

1

1

1

1

n,

n.

6

6

M SE

= 12.09

T h e p -v a lu e w ith d f = 2 0 is 2 P ( t > 12.09) is b o u n d ed (u sin g T a b le 4) as p -v a lu e < 2 ( .0 0 5 ) = .01 and the n uil h y p o th e sis is reject. W e co n c lu d e th at th ere is a d iffe re n c e b e tw e en t m eans. d

T h e 9 9 % co n fid e n c e interv al fo r p , - p 4 is (x¡- x

4) ± r m

1 1 ^M SE — + —

(6 .0 1 6 7 - 3 .6 5 ) ± 2 . 8 4 5 ^ . 1 1 5 ^ + ^ ) 2 .3 6 7 ± .5 5 7

or

1.810 < p , - p 4 < 2.924

e W h en co n d u c tin g the t te sts, rem e m b e r th at the stated c o n fid e n c e c o e ffic ie n ts b a se d on ra n d o m sam p lin g . If y o u lo o k ed at the d a ta and o n ly co m p a re d th e large an d sm allest sam p le m eans, the ra n d o m n ess assu m p tio n w o u ld b e d istu rb ed . 11.17

a b

T h e d e sig n is a c o m p lete ly ran d o m iz ed d esig n (fo u r in d e p e n d e n t sam p le s). T h e fo llo w in g p re lim in a ry c a lc u la tio n s are n ecessary: 7; = 1 2 1 1

CM =

( I * ,) 2

(4 6 8 6 )2

20

T o tal S S = SST =

T2 = 1074

Z——

r 3 =1158

T4 = 1 2 4 3

G = 4686

= 1 ,0 9 7 ,9 2 9 .8

- C M = 1 ,1 0 1 ,8 6 2 - C M = 393 2 .2 CM =

n;

1 2 1 12 5

10742 +5

1 1582

12432

5

5

- C M = 3 2 7 2 .2

C a lc ú la te M S = S S / d f and c o n so líd ate the in fo rm a tio n in an A N O V A table. S ource T re a tm en ts E rro r T o tal

SS 3 2 72.2 660 3932.2

df 3 16 19

MS 1090.7333 41 .2 5

T h e h y p o th esis to be tested is H 0 :p , = p 2 = p 3 = p 4

versus

109

H a : at least o n e p a ir o f m ean s are d iffe r

and th e F test to d ete c t a d iffe re n c e in a v erag e p ric es is M ST F = ---------= 2oa6 .4 4 . M SE T h e rejectio n reg ió n w ith a = .05 and 3 and 16 d f is ap p ro x im a te ly F > 3 .2 4 a n d Ho is rejected . T h ere is en o u g h e v id e n c e to in d icate a d iffe re n c e in the a v e ra g e p ric e s fo r th e fo u r States.

11.21

a

c o = q 0, ( 4 , 1 2 ) - ^ = 4 . 2 0 - ^ = 1.878s V5 V5

b

c o = q m ( 6, 1 2 ) - ^ = 6 . 1 0 - ^ = 2 .1 5 6 7 s

V8 11 .25

V8

T h e d esig n is co m p le te ly ran d o m iz ed w ith 3 tre a tm en ts a n d 5 rep lic a tio n s p er treatm en t. T h e M i n i t a b p rin to u t belo w sh o w s the a n a ly sis o f v a rian ce fo r this ex p erim en t.

One-way ANOVA: mg/dl ve rsu s Lab Source Lab Error Total

DF 2 12 14

S = 5. 933

Level 1 2 3

N 5 5 5

SS 42.6 422 .5 465.0

MS 21.3 35 .2

F 0.60

R-Sq = 9.15%

Mean 108.86 105.04 105.60

StDev 7.47 6 .01 3 .70

P 0.562

R-Sq(adj) Individual 95% CIs For Mean Based on Pooled StDev . . +------ ........+ ------- --- + ----- ---- + ----(---------) (-------

(------

100.0

---) 104.0

108.0

112.0

Pooled StDev = 5 . 9 3 Tukey 95% Simultaneous Confidence Intervals All Pairwise Comparisons among Levels of Lab Individual confidence level = 97.94% Lab = 1 subtracted from: Lab Lower Center Upper 2 -13.824 -3.820 6.184 3 -13.264 -3.260 6.744 14.0 Lab = 2 subtracted from: Lab Lower Center Upper 3 -9.444 0.560 10.564

-7.0

0.0

7.0

+ --------- + ----------+ ----------+ ----(...... -..... -*............... ) -14.0

110

-7.0

0.0

7.0

a

T h e an aly sis o f v aria n c e F te st fo r H 0 : //, = / / 2 = //? is F = .60 w ith

p -v a lu e = .562 . T h e resu lts are n o t sig n ific a n t and H 0 is n o t re je cte d . T h e re is in su fficien t e v id e n c e to in d ícate a d iffe re n c e in th e tre a tm en t m eans. b S in ce the tre atm e n t m ean s are no t sig n ific a n tly d ifferen t, th ere is no need to use T u k e y ’s test to search fo r the p a irw ise d iffe re n c e s. NotiCe th at all th re e in terv als g e n erated b y M i n i t a b co n tain zero , in d icatin g th a t the p a irs c a n n o t be ju d g e d differen t. 11.29

R efer to E x e rc ise 11.28. T h e g iv en su m s o f sq u ares are in se rte d a n d m issin g en trie s fou n d by su b tractio n . T h e m ean sq u ares a re fo u n d as M S = S S / d f . Source T re atm e n ts B locks E rro r T o tal

11.33

SS 11.4 17.1

df 2 5 10

14.2 4 2 .7

17

MS 5.7 0 3.42 1.42

F 4.01 2.41

U se M i n it a b to o b ta in an A N O V A p rin to u t, o r u se the fo llo w in g calc u la tio n s: (I* y ) CM = 4 n

(1 1 3 )“ ^ = 1 0 6 4 .08333 12

T o ta l SS = X x 2 - C M = 6 2 + 1 0 2 +••■ + 142 - C M = 1 2 1 3 -C M = 1 4 8 .9 1 6 6 7 T2 SST = I - C - C M 3

222 + 3 4 2+ 2 7 2 + 3 0 2 = ~U - C M = 2 5 .5 8 3 3 3 3

S S B = I í - - C M = 33~ + 2 5 2 + 5 5 ~ - C M = 120.66667 and 4 4 S S E = T o tal SS - S S T - S S B = 2 .6667 C a lc ú la te M S = S S / d f and c o n so líd a te th e in fo rm a tio n in an A N O V A table. S o u rce T reatm en ts B locks E rro r T o tal a

df 3 2 6 11

SS 2 5 .5 8 3 3 120.6667 2.6 6 6 7 148.9167

MS 8.5 2 7 8 6 0 .3 3 3 3 0 .4 4 4 4

F 19.19 135.75

T o test the d ifferen c e a m o n g trea tm e n t m ean s, th e te st sta tistic is P _ M S T _ ^ 5 2 8 _ ]9 19 M SE

.4444

and the rejectio n reg ió n w ith a = .05 and 3 a n d 6 d f is F > 4 .7 6 . T h e re is a sig n ifican t d iffe re n c e a m o n g th e tre atm e n t m eans. b T o te st the d iffe re n c e a m o n g b lo c k m ean s, th e test statistic is F = M SB = 6 0 3333 M SE

.4444 111

an d Ihe reje c tio n regió n w ith a = .05 and 2 and 6 d f sig n ifican t d iffe re n c e am o n g the blo ck m eans. c

\s

F > 5 .1 4 . T h e re is a

W ith k = 4 , d f = 6 , n, = 3 , ÍA V Ñ ÍSE 1.4444 a ) = /M S É

m = q= q m (

3

,

=

434

T h e ran k ed m ean s a re show n below . 3 2 .6 1 2 5 3 4 .1 8 7 5

= 1 -885

3 4.8875

E stim a to rs A a n d B sh o w a sig n ific an t d iffe re n c e in av erag e costs. 11.45

a -b T h e re a re 4 x 5 = 2 0 trea tm e n ts an d 4 x 5 x 3 = 6 0 total o b serv atio n s. c In a factorial ex p erim e n t, v ariatio n d u e to th e in teractio n A x B is iso lated from SSE . T h e so u rces o f v ariatio n a n d a sso ciated d e g re e s o f freed o m are g iv e n on the next page.

113

1.49

Source

df

A B A xB E rro r

3 4 12 40

T o tal

59

a B ased on the fact that the m ean resp o n se fo r the tw o lev els o f facto r B b e h av es very d iffe re n tly d e p e n d ín g o n the lev el o f facto r A u n d er in v e stig a tio n , th e re is a stro n g in te ractio n p resen t b e tw een facto rs A a n d B. b T h e test statistic fo r in te ra ctio n is F = M S ( A B ) /M S E = 3 7 .8 5 w ith p -v a lu e = .0 0 0 fro m the M i n i t a b p rin to u t. T h e re is ev id e n c e o f a sig n ifican t in teractio n . T h a t is, the effect o f fa c to r A d e p e n d s u p o n the lvel o f fa c to r B at w hich A is m easu red . c In ligh t o f th is ty p e o f in teractio n , th e m ain effe c t m ean s (av e ra g ed o v e r th e lev els o f th e o th e r factor) d iffe r o n ly slig h tly . H enee, a test o f th e m ain -effe ct term s p ro d u c e s a n o n -sig n ifican t result. d N o. A sig n ifican t in terac tio n in d icates th at the effect o f o n e facto r d e p e n d s u p o n the lev el o f th e other. E ach facto r-lev el co m b in a tio n sh o u ld be in v estig a te d in d iv id u ally . e A n sw e rs w ill vary.

.5 3

a

T h e d esig n is a 2 x 4 facto rial ex p e rim e n t w ith r = 5 rep lic atio n s. T h e re are tw o

facto rs, G e n d e r and S ch o o l, o n e at tw o lev els and one at fo u r levels. b T h e a n aly sis o f v arian c e tab le can be fo u n d u sin g a C om puter p rin to u t o r the fo llo w in g calcu latio n s: S ch o o ls G ender 1_____ 2______ 3______ 4 T o tal M ale F em ale T o tal

2919 3082 6001

3257 3 629 6886

3330 3344 6674

2461 2410 4871

11967 12465 24432

?44322 C M = ----------- = 14923065.6 40 T o tal SS = 15281392 - C M = 3 5 8 3 2 6 .4

20 iC =

6 0 0 12 + 6 8 8 6 2 + 6 6 7 4 2 + 4 8 7 12

10

- C M = 2 4 6 7 2 5 .8

- S S G - S S ( S c ) - C M = 10574.9

114

S ou rce G Se G xSc E rro r T o tal c

df 1

SS 6200.1

MS 6 2 0 0 .1 0 0

F 2 .0 9

3 3

2 4 6 7 2 5 .8 10574.9

8 2 2 4 1 .9 3 3 3 5 2 4 .9 6 7

2 7 .7 5 1.19

32

9 4 8 2 5 .6 3 5 8 3 2 6 .4

2 9 6 3 .3 0 0

39

T h e test statistic is F = M S ( G S c ) /M S E = 1.19 and th e re je c tio n reg ió n is

F > 2 .9 2 (w ith a = .05 ). A ltern ately , y o u c a n b o u n d the p -v a lu e > . 1 0 . H e n e e ,?

is n o t re jec ted . T h e re is in su ffícien t ev id en c e to in d ícate in te ra c tio n b e tw e e n gende a n d schools. d Y o u can see in th e in teractio n p lo t th at th ere is a sm all d iffe re n c e b etw ee n the a v e ra g e sco res fo r m ale an d fem ale stu d en ts at sc h o o ls 1 an d 2, b u t n o d iffe re n c e te

e

T h e test statistic fo r testin g g e n d er is F = 2 .0 9 w ith F os = 4 .1 7 (o r p -v a lu e > .11

T h e te st statistic fo r sch o o ls is F = 2 7 .7 5 w ith F 05 = 2 .9 2 (o r p -v a lu e < .005 ). T h e is a sig n ifica n t e ffe c t d u e to schools. U sin g T u k e y ’s m eth o d o f p a ire d co m p a riso n s w ith a = .0 1 , c a lc ú la te „ = í„ , ( 4 , 3 2 ) ^ 4 , o J f L yjn, V 10 T h e ra n k e d m ean s are sh o w n below . 487.1 600.1 *4

82.63

6 6 7 .4

6 8 8 .6

*3

x2

115

11.59

T h e o b jectiv e is to d e te rm in e w h eth e r o r not m ean reactio n tim e d iffe rs fo r the five stim uli. T h e fo u r p e o p le u sed in the e x p e rim e n t act as b lo ck s, in an a tte m p t to isolate the v ariation from p erso n to p erso n . A ra n d o m iz e d blo ck d e sig n is used, and the an aly sis o f varian ce tab le is g iv en in the p rin to u t. a T h e F statistic to d e te c t a d iffe re n c e d u e to stim u li is = ^

f

I

= 2 7 .7 8

M SE w ith p -v a lu e = .0 0 0 . T h e re is a sig n ific a n t d iffe re n c e in the e ffe c t o f the five stim uli. b

T h e treatm en t m ean s c a n be fu rth e r c o m p a red u sin g T u k e y ’s test w ith , c i M > /M S É „ C1 ¡.00708 in n ú) = q os( 5 ,1 2 ) — — = 4 . 5 l J — — = . '9 0 V "r V 4

T h e ran k ed m eans are sh o w n below . E .525

c

A .7

B .8

D 1.025

C 1.05

T h e F te st for blo ck s p ro d u c e s F = 6 .5 9 w ith p -v a lu e = .007 . T h e blo ck

d ifferen ces are sig n ifican t; b lo c k in g has been effectiv e. 1 1 .6 3

T h is is sim ilar to p rev io u s ex erc ises. T h e c o m p le te A N O V A tab le is sh o w n below . Source SS MS F df A 1.14 1 1.14 6.51 B 2 2.58 1.29 7 .3 7 2 A xB 0 .4 9 0 .2 4 5 1.40 E rror 24 4 .2 0 0 .1 7 5 T otal 29 8.41 a

T h e test statistic is F = M S ( A B ) /M S E = 1.40 an d the rejec tio n reg ió n is

F > 3 .4 0 . T h ere is in su ffic ie n t e v id e n c e to in d icate an in teractio n . b

U sing T a b le 6 w ith d f = 2 and d f 2 = 24 , the fo llo w in g critical v alú es are

obtained. a

.10 2.54

.05 3 .4 0

.025 4 .3 2

.01 5.61

.005 6.66

T h e o b serv ed valué o f F is less th an F 1(), so th at p -v a lu e > . 10 . c

T h e test statistic fo r testin g fa c to r A is F = 6.51 w ith F ()5 = 4 .2 6 . T h e re is

e v id en c e that fac to r A affee ts the resp o n se. d

T h e test statistic fo r fa c to r B is F = 7.37 w ith Fm = 3 .4 0 . F a c to r B a lso affeets

th e response. 1 1.67

a T h e d esign is a ra n d o m ize d b lo ck d esig n , w ith w eek s re p rese n tin g b lo c k s and sto res as treatm ents. b T h e M i n it a b C om puter p rin to u t is sh o w n on the next page.

116

Two-way ANO VA: Total ve rsu s Week, Store DF Source Week 3 Store 4 Error 12 Total 19 S = 4.799

c

F SS MS P 571.71 8 .27 190.570 0.003 684.64 171.159 7 .43 0 .003 276.38 23.032 1532.73 R-Sq = 81.97% R-Sq(adj) = 71.45%

T h e F test fo r tre a tm e n ts is F = 7.43 w ith p -v a lu e = .003 . T h e p -v a lu e is sm all

en o u g h to a llo w re je c tio n o f H 0. T h ere is a sig n ifican t d iffe re n c e in the a v e ra g e w eek ly to tals fo r th e five su p erm ark ets.

d W ith k = 5 , d f = \2, n = 4 , te ,^ V M S E , (2 3 .0 3 2 a>= qtí5 (5 ,1 2 ) - j = = 4 .5 i y — - — = 10.82 T h e ra n k ed m ean s are sh o w n below . 1 5 4 2 4 0 .2 3 2 4 9 .1 9 2 5 2 .1 8

11.71

3 2 5 4 .8 7

2 256 .9 9

a T h e d e sig n is a c o m p lete ly ran d o m iz ed d esig n w ith th ree sam p le s, e ac h h a v in g a d iffe re n t n u m b e r o f m easu rem en ts. b U se the c o m p u tin g fo rm u las in S ectio n 11.5 o r th e M i n it a b p rin to u t below .

One-way ANOVA: Iron ve rsu s Site Source DF Site 2 Error 21 Total 23 S = 0.7221

SS MS 132.277 66.139 10.950 0.521 143.227 R-Sq = 92.36%

F 126.85

P 0.000

R-Sq(adj)

= 91.63%

T h e F test fo r tre atm e n ts has a test statistic F = 126.85 w ith p -v a lu e = .000. T h e nuil h y p o th e sis is re je c te d and w e co n clu d e th at th ere is a sig n ific an t d iffe re n c e in the a v erag e p e rc e n ta g e o f iro n o x id e at th e th ree sites. c T h e d ia g n o stic p lo ts a re sh o w n on the next p ag e. T h ere ap p ea rs to be n o v io latio n o f the n o rm ality assu m p tio n s; the v a rian ces m ay be u n eq u al, ju d g in g by th e d iffe rin g b a r w id th s ab o v e and b elo w th e c e n te r line.

117

Normal Probability P lo to f the Residual! ( r e s p o n s e is Iron)

118

12: Linear Regression and Correlation 12.1

T h e line co rresp o n d in g to the e q u a tio n y = 2 x +1 can be g caphed by lo catin g the y v alú es c o rre sp o n d in g to x = 0 , 1, and 2. W hen x = 0, y = 2 (0 ) + 1 = 1 W hen x = 1, >• = 2(1) -1-1=3 W hen x = 2 , y = 2 (2 ) + 1 = 5 T h e g rap h is sh o w n below .

N o te th at the e q u atio n is in th e form y = a + J3x. T h u s, the slo p e o f the lin e is /? = 2 and th e y -in te rc e p t is a = 1. 12.5

A d eterm in istic m ath e m a tic a l m o d el is a m odel in w h ich the v alu é o f a resp o n se y is e x a ctly p re d ic te d from v alú es o f the v aria b le s th a t a ffec t the resp o n se. O n the o th e r hand, a p ro b ab ilistic m ath e m atica l m odel is one th a t c o n ta in s ra n d o m e le m e n ts w ith sp ecific p ro b a b ility d istrib u tio n s. T h e valué o f the re sp o n se y in th is m o d el is not e x actly d eterm in ed .

12.9

a

C alcú late = 8 5 0 .8

2 > , = 3 .7 5 5

= 1 0 1 ,4 9 5 .7 8

£

y f = 1 .9 4 1 4 6 7

T h en S *> =

_ ( £ * , ) ( £ » ) = 88 8 0 0 0 3 333

119

I * , ? , = 4 4 3 .7 7 2 7 n=9

s ** = Z x ? ~ ^

X¡^ = 2 1 ,0 6 6 .8 2 n 2

b-c =

88-800033 = ft0 ()4 2 S_„

21 0 6 6 .8 2

and a = y - b x = 0 .4 1 7 2 2 -0 .0 0 4 2 1 5 1 6 (9 4 .5 3 3 3 ) = 0 .1 8 7 T h e least sq u ares lin e is y = a + b x = 0 .1 8 7 + 0 .0 0 4 2 * . T h e g ra p h o f the least sq u ares line and th e n ine d a ta p o in ts are sh o w n below .

d

W h en x = 1 0 0 , the v alu é fo r y can be p re d ic te d u sin g the least sq u a re s line as y = 0 .1 8 7 + 0 .0 0 4 2 (1 0 0 ) = 0 .4 4 C alcúlate SSR = ( ^ S„

= m 0 3 3 3 3 ) l = 3743064 2 1 ,0 6 6 .8 2

and (5 l2 - ± - 2 - L = .3 7 4 7 9 7 6 - .3 7 4 3 0 6 4 1 7 = .0 0 0 4 9 1 1 8 $xx T h e A N O V A table w ith 1 d f fo r re g re ssio n and n - 2 d f fo r e rro r is sh o w n on the next

S S E = T o tal SS - S S R = S

page. R em em b er th at the m ean sq u ares are calc u la ted as M S = S S / d f .

120

S o u rce R eg ressio n E rro r T o tal 12.13

di 1 7

SS .3 7 4 3 0 6 4

MS

.0 0 0 4 9 1 2

.0000010

8

.3 7 4 7 9 7 6

.3 7 4 3 0 6

a

T h e sca tte rp lo t g e n e ra ted b y M i n it a b is sh o w n below . T he a ssu m p tio n o f

b

C a lcú late I-*

=1192

I *

£ x 2 = 9 6 ,9 9 0

=725

£ y 2 = 3 6 ,4 6 1

= 5 9 ,3 2 9 n = 15

T hen 5

= Y xy

-Í5^1í5iil = 1710.6667 ,2

Sa = y xf

= 2 2 6 5 .7 3 3 3 n 2

1419.3333

--------= .75502 b = -2 - = S„ 2 2 6 5 .7 3 3 3 and a = y - b x = 4 8 .3 3 3 3 -(0 .7 5 5 0 2 )(7 9 .4 6 6 7 ) = -1 1 .6 6 5 (u sin g full accu racy ). T h e least sq u ares lin e is y = a + fox = - 1 1 .6 6 5 + 0.755*. c

W h en x = 85, the v alu é fo r y can be p re d ic te d u sin g the least sq u a re s line as 121

y = a + b x = - \ \ .665 + .755( 85) = 5 2 .5 1 . 12.17

T h e h y p o th esis to be tested is

a

H 0 : (1 = 0

v ersu s

H a : /? * 0

and the te st statistic is a S tu d e n t's t, c a lc u la te d as b -/3 0

_

Vm se J s 2

1 .2 - 0

V 0-533/10

T h e critical valué o f t is b ased o n n - 2 = 3 d e g re es o f freed o m a n d th e reje c tio n reg ió n for a = 0 .0 5 is |?| > t m = 3 .1 8 2 . S in ce the o b se rv e d v alu é o f t falls in the rejectio n reg ió n , w e reject H0 and co n c lu d e th a t /? * 0 . T h a t is, x is u sefu l in the p re d ictio n o f y. b F rom the A N O V A tab le in E x e rc ise 12.6, calc ú la te f = M S R = J ± 4 _ = 2 7 0 () M SE

0.5 3 3 3

w hich is the sq u are o f the t statistic from p art a: r = ( 5 .2 0 )2 = 2 7 .0 . T h e critical v alué o f t from p a rt a w as t m5 = 3 .1 8 2 , w h ile the c ritica l v alu é o f F

c

fro m p a rt b w ith d f = 1 and d f 2 = 3 is F 05 = 10.13 . N o tice th at the rela tio n sh ip b etw een the tw o critical v alú es is F = 10.13 = (3 .1 8 2 )2 = t 2 12.21

a T h e d ep en d en t v a riab le (to be p re d ic te d ) is >’ = co st an d the in d e p en d e n t v a riab le is x = distance. b P relim in a ry calc u latio n s: =5052

^ x j , = 7 ,5 6 9 ,9 9 9

Y , x f = 3 7 ,7 6 3 ,3 1 4

= 1 ,6 9 5 ,9 3 4

n = 18

x iy¡ ~ ^

) ( ^ ^ } = ^ 5 2 7 ^2 4 5 .6 6 7

: = 2 1 ,5 3 0

T hen

\2

= 1 2 ,0 1 1 ,0 4 1 .7 8 n S /, = - 5 1 = 0 ^ 2 7 1 5 3 a = y - b x = 2 8 0 .6 6 6 7 -0 .1 2 7 1 5 3 (1 1 9 6 .1 1 1 1 ) = 128.57699 a n d the least sq u ares lin e is y = a + b x = 128.57699 + 0 .1 2 7 1 5 3 * .

122

c

T h e p lo t is sh o w n b elo w . T h e line a p p ea rs to fit w ell th ro u g h the 18 d a ta poii Fitted Line Plot y = 128.6 +0.1272 x s R-Sq R-Sq(adJ)

d

C a lc ú la te T o tal SS = S y y = ' £ y ? - ^ i L n

72.3755 69.9% 68.0%

= 1 ,6 9 5 ,9 3 4 -

= 2 7 8 ,0 0 6 . 18

T hen

(5 „ ) SSE = 5 ”

S

(1 ,5 2 7 ,2 4 5 .6 6 7 )" = 2 7 8 , 0 0 6 - — ------------------- ¿ - = 8 3 ,8 1 1 .4 1 0 5 5 1 2 ,0 1 1 ,0 4 1 .7 8

SSF 83 811 41055 and M S E = -------- = — ------ 1---------- = 5 2 3 8 .2 1 3 . T h e h y p o th esis to be tested is n- 2 16 H 0 : [3 = 0

v ersu s

H a : /? * 0

and the te st statistic is t_

b -P Q ^M S E /S „

O 0 .1197151 2 7 1 5 3—- 00

= 6.09

V5 2 3 8 -213/1 2 ,0 1 1 ,0 4 1 .7 8

T h e critical valué o f t is b ased on « - 2 = 16 d eg re e s o f fre e d o m an d th e reje c tio n reg ió n fo r a = 0 .0 5 is |/| > t m = 2 .1 2 0 , and H 0 is rejected . T h e re is e v id e n c e at tf 5 % level to in d ícate th at x and y a re lin early related . T h a t is, the re g re ssio n m odel y = a + f i x + e is usefu l in p red ic tin g c o st y. 12.25

a T h e sca tte rp lo t g e n erated b y M i n it a b is sh o w n on the n ex t p ag e. T h e a ssu m p tio n o f lin earity is reaso n ab le.

123

Scatter plot o f Final Exam v s Posttest 100

• • •

90

• E

• •

80

• •

• •

• •

•

5 i

• ” 60

50

•

•

•

4070

75

80

85

90

95

100

Posttest

b U sin g th e M i n i t a b p rin to u t, th e e q u atio n o f the re g re ssio n line is y = - 2 6 .8 2 + 1 .2 6 1 7 * . c

T h e h y p o th esis to be tested is Hn :p = 0

v ersu s

H a : /? * 0

and th e te st statistic is a S tu d e n t’s t, ca lc u la te d as

t = - r b ~ P Q = = 1.49

Vm se f s „ w ith p -v a lu e = .0 0 0 . S in ce the p -v a lu e is le ss th an a = .0 1 , w e re je c t H 0 and co n clu d e that /? * 0 . T h a t is, final ex am sco res an d p o sttest sc o re s are linearly related. d

T h e 9 9% co n fid en ce interv al fo r th e slope /? is b ± t a/2yl M S E / S a => 1.2617 ± 2 .8 7 8 (0 .1 2 5 3 ) => 1.2617 + 0 .4 8 4 9

o r 0 .7 7 6 8 < /? < 1.7466 . 12 .2 9

U se a p lo t o f re sid u als v ersu s fits. T h e p lo t sh o u ld a p p e a r as a ra n d o m sc a tte r o f points, free o f an y pattern s.

12 .3 3

a

I f you lo o k carefu lly , there a p p e ars to be a slig h t cu rv e to the fiv e points.

b T h e fit o f the reg re ssio n lin e, m ea su red as r 2 = 0 .9 5 9 in d ic ate s th a t 9 5 .9 % o f the o v erall v ariation ca n b e e x p la in e d b y th e stra ig h t lin e m odel. c W hen w e lo o k at the re sid u a ls th ere is a stro n g c u rv ilin e a r p attern th at has not b een ex p lain ed by the straig h t lin e m odel. T h e re la tio n sh ip b e tw een tim e in m o n th s an d n u m b er o f b e o k s ap p ears to be cu rv ilin ear. 1 2 .3 9

a A lth o u g h very slig h t, the stu d en t m ig h t n o tice a slig h t c u rv a tu re to the d ata points. b T h e fit o f the lin e ar m odel is v ery g o o d , a ssu m in g th at this is in d e e d the c o rre c t m odel for this d ata set.

124

c T h e n o rm al p ro b ab ility p lo t fo llo w s the c o rre c t p attern fo r the a ssu m p tio n o f n o rm ality . H o w ev e r, th e re sid u a ls sh o w th e pattern o f a q u a d ra tic cu rv e, in dicatin¿ th at a q u a d ra tic ra th c r th an a lin ea r m odel m ay have been the c o rre c t m o d el fo r this data. 12.42

a

a

U se a C om puter p ro g ra m o r the h and c a lc u la tio n s sh o w n below .

£ > ,= 1 1 6 =2818

¿*= 1480

¿ > ,* = 3 6 ,1 3 3

£ y 2 = 4 6 7 ,6 0 0

n=5

T h en

n ■ S „ = 2 > ,2 - —

n

^ - = 126.8 2

S

= V v2 - ^ — ^ 2 - = 2 9 ,5 2 0 n

1797 b = - 2 - = _ _ = 14.17192 S„ 126.8 a = y - b x = 2 9 6 -1 4 .1 7 1 9 2 ( 2 3 .2 ) = - 3 2 .7 8 9 and the le a st sq u ares lin e is y = - 3 2 .7 8 9 + 14.172*. b

T h e p ro p o rtio n o f the total v ariatio n e x p la in e d by reg ressio n is S~ 179 7 2 r2= — 3 - = — ---------- = 0.8 6 2 7 S X XS m (1 2 6 .8 X 2 9 ,5 2 0 ) yv

c T h e d ia g n o stic p lo ts, g e n erate d by M i n it a b are sh o w n b elo w . T h e p lo ts d o m sh o w any stro n g v io latio n o f a ssu m p tio n s.

R e s id u a ls V e r s u s the F itte d V a lú e s (response is Total Yante) •

5040 3020s

io -

•

•

-10-20 -30

• 150

200

250 Htted Valué

125

300

350

1 2.43

R efer to E x ercise 12.42 and calc ú late S S E = S . - ^(SJ L L . = 29520-

1797'

= 4 0 5 3 .0 5 2 0 5

1268

a

SSE

4 0 5 3 .0 5 2 0 5

= 1351.01735. ti —2 3 T h e p o in t estim a to r fo r E ( y ) w hen x = 21 is

an d M S E =

y = - 3 2 .7 8 9 + 1 4 .1 7 2 (2 1 ) = 2 6 4 .8 2 3 a n d the 95% co n fid en c e interv al is 2A $ ± f 0 2 5 jM S E

2 6 4 .8 2 3

± 3 .1 8 2 . /l 351 .0 1 7 3 5

1

( 2 1 - 2 3 .2 )

5+

126.8

2 6 4 .8 2 3 ± 57.0 7 9 or b

2 0 7 .7 4 4 < E ( y ) < 3 2 1 .9 0 2 . T h e p o in t e stim a to r fo r y w h en x = 21 is still y = - 3 2 .7 8 9 + 14.1 7 2 (2 1 ) = 2 6 4 .8 2 3

an d the 9 5 % p red ictio n in terv al is M SE

,

1

(tp -x )

n

S„

1+ - + —

2A

---------

1 (21 - 2 3 . 2 ) 2 ^ 2 6 4 .8 2 3 ± 3 .1 8 2 ^ 1 3 5 1 .0 1 7 3 5 1 + - + 5 126.8

or

2 6 4.82 3 ± 1 3 0 .1 4 3 134.680 < > '< 3 9 4 .9 6 6 .

126

c T h is w o u ld n o t be ad v isa b le , sin c e y o u are try in g to e stím a te o u tsid e the ra n g e o f e x p e rim e n ta ro n . 12.47

a

R e fe r to fig u re below . T h e sam p le c o rre la tio n c o e ffic ie n t w ill be p o sitiv e. Scatterplot of y vs x •

4.0 -

3.5-

•

>. 3 .0 -

2 .5 -

2 .0 -

• -

• 2

-

1

0

1

2

X

b

C alcú late

S

„ =

^ - ^ - 1

0

- y

= 10

= « - í f =4 = 0 .9 4 8 7 and r 2 = (0 .9 4 8 7 )2 = 0 .9 0 0 0 . A p p ro x im a te ly

T h en r = V io

9 0 % o f th e total su m o f sq u ares o f d e v ia tio n s w as re d u c ed by u sin g the lea st sq u ares e q u a tio n in ste a d o f y as a p re d ic to r o f y. 12.51

W h en th e p re -te st sco re x is h ig h , th e p o st-te st sco re y sh o u ld also be h ig h . T h ere sh o u ld be a p o sitiv e co rrelatio n .

127

C alcú late . 70 , 0 0 6 - ^ Z p = 4 6 8 .4 2 8 5 7

s *x

n

^

Syy = Z X2 - ^ Sv

T, T h en r =

- =

= 6 5 , 9 9 3 - - ^ - = 5 1 7 .4 2 8 5 7 7

y‘ ^ = 7 4 , 5 8 5 = 733.42857

n

7

4 6 8 .4 2 8 5 7 n _ .... = 0 .7 6 0 . ^ 5 1 7 .4 2 8 5 7 (7 3 3 .4 2 8 5 7 )

T h e te st o f h y p o th esis is H0 : p = 0

v ersu s

Ha :p > 0

and th e test statistic is ,_ rV ^ 2 _

í

r

VI - r 2

i

0 .7 6 0 ,/5

„

— «Z.O I ^

y l\- ( 0 .7 6 0 ) 2

T h e reje c tio n reg ió n fo r a = 0.05 is t > tM = 2 .0 1 5 a n d H 0 is re jec ted . T h e re is su fficien t e v id en c e to in d ícate p o sitiv e co rrelatio n . 1 2.55

a S in ce n eith er o f the tw o v a ria b les, am o u n t o f so d iu m o r n u m b e r o f c a lo rie s, is c o n tro lle d , th e jn e th o d s o f c o rrelatio n ra th e r th an lin e a r re g re ssio n a n a ly sis sh o u ld be used. b U se a C om puter p ro g ra m , y o u r scien tific c a lc u la to r o r th e c o m p u tin g fo rm u las g iv en in the text to calc ú late th e c o rre la tio n c o e ffic ie n t r. T h e M i n i t a b p rin to u t for th is d a ta set is sh o w n o n the n ex t page.

128

Correlations: Sodium, Calories Pearson correlation of Sodium and Calories = 0.981 P-Value = 0.003

T h e re is e v id e n c e o f a h ig h ly sig n ifíc a n t c o rre la tio n , sin ce th e p -v a lu e is so sm al T h e co rrela tio n is po sitiv e. 12.61

A n sw e rs w ill vary. T h e M i n i t a b o u tp u t for this lin ea r re g re ssio n p ro b le m is sho b elo w .

Regression A nalysis: y ve rsu s x The regression equation is y = 21.9 + 15.0 x Predictor Constant x

Coef SE Coef 21.867 3.502 14.9667 0.9530

S = 3.69098

R-Sq = 96.1%

T 6.24 15.70

P 0.000 0.000

R-Sq(adj)

= 95.7%

Analysis of Variance Source Regression Residual Error Total

DF 1 10 11

SS 3360.0 136.2 3496.2

MS 3360.0 13.6

F 246.64

P 0.000

Correlations: x, y Pearson correlation of x and y = 0.980 P-Value = 0.000

a

T h e c o rre la tio n c o e ffic ie n t is r = 0.980.

b

T h e c o e fficie n t o f d e te rm in a tio n is r 2 = 0.961 (o r 9 6.1% ).

c

T h e least sq u ares lin e is y = 2 1.867 + 1 4:9 6 6 7 x .

d W e w ish to e stím a te th e m ean p e rc e n ta g e o f kill fo r an a p p lic a tio n o f 4 po i o f n em a tic id e p e r acre. S in ce the p e rc e n t kill y is actu a lly a b in o m ial percen tag e v arian ce o f y w ill c h a n g e d e p e n d in g o n th e v alu é o f p , th e p ro p o rtio n o f n em atoi k illed fo r a p a rtic u la r a p p lic a tio n rate. T h e resid u al p lo t v ersu s th e fítted v alú es sh o w s th is p h e n o m e n o n a s a “ fo o tb a ll-sh ap e d ” p attern . T h e n orm al p ro b a h ility also sh o w s so m e d e v ia tio n from n o rm ality in the tails o f th e plo t. A tran sfo rm a t m ay be n e e d e d to assu re that the re g ressio n assu m p tio n s are satisfied. 12.6 5

a U se a C om puter p ro g ram , y o u r scien tific c a lc u la to r o r the c o m p u tin g fo rm i g iv en in the te x t to c a lc ú la te th e c o rre latio n c o e ffic ie n t r. = | , 2 3 3 . 9 8 7 - 5028i( f 5 6 ) = 3 7 ,3 2 3

=Y x? - n= ^

2,21'2,i 78 - 12= 1 t M5 = 2 .3 0 6 an d w e re je c t H 0. T h e re is su fficien t e v id en c e to in d ícate th at x a n d >' are lin e a rly related .

d

T h e 95% co n fid e n c e in terv al fo r th e slo p e (3 is b ± ía/27 M S E ¡ s 2 => - .7 5 8 ± 2 .8 9 6 ^ .5 8 2 2 5 /5 0 0 => - .7 5 8 ± .099

o r - .8 5 7 < e

< - .6 5 9 .

W h en x = 14, the e stím a te o f e x p e c te d fresh n e ss E (y) is

y = 2 0 .4 7 - .7 5 8 ( 1 4 ) = 9.858 and th e 9 5 % c o n fid en c e in terv al is i , o ,

9 .8 5 8 ± .5 6 2 o r 9 .2 9 6 < E { y ) < 10.420. f

C alcú late _ _ S S R _ _ 2 8 7 2 8 2 _ Q gg4

T o tal SS 2 9 1.94 T h e total v ariatio n has b een red u c e d by 9 8 .4 % % b y u sin g th e lin e a r m o d el. 1 2 .76-77 U se th e H o w a L in e W o r k s ap p let. T h e line y = 0 .5 x + 3 h as a slo p e o f 0 .5 and a yin tercep t o f 3, w hile th e lin e y = - 0 .5 x + 3 h as a slo p e o f - 0 .5 and a y -in te rc e p t o f 3. T h e seco n d line slo p es d o w n w a rd at th e sam e rate as th e first lin e slo p e s u p w ard . T hey b o th cro ss th e y ax is at th e sam e point. 1 2 .80

a U se a C om puter, y o u r scien tific c a lc u la to r o r th e c o m p u tin g fo rm u la s to find the c o rrelatio n b etw een x and y. T h e M i n it a b c o rre la tio n p rin to u t b e lo w sh o w s r = 0.231 w ith p -v a lu e = 0 .5 4 9 w h ich is n o t sig n ific a n t at the 5% lev el o f sig n ific a n c e . Y ou c a n n o t co n c lu d e th at th ere is a sig n ifican t p o sitiv e c o rre la tio n b e tw een m e d ia n rate and sco re fo r “b u d g e t” hotels.

Correlations: Median Rate, S core Pearson correlation of Median Rate and Score = 0.231 P-Value = 0.549

b -c U se th e C o r r e la tio n a n d th e S c a tte r p lo t a p p let. T h e re is a ra n d o m ser o f p o in ts, w ith no o u tliers. T h e stu d e n t’s p lo t sh o u ld lo o k sim ilar to the M i n i t a b sh o w n below .

Scatterplot of Score vs Median Rate~ •

80-

•

75-

• •

v 70• 65-

• •

• 60• 40

42

44

46 48 Median Rate

133

50

52

54

56

13: Múltiple Regression Analysis 13.1

a

W h en

= 2, £ (> •) = 3 + * , - 2 ( 2 ) =

-1 .

jc2 = 1,E ( y ) = 3 + * , - 2 ( 1 ) = x, + 1 .

W hen

= 0 ,E ( y ) = 3 + x l - 2 ( 0 ) = x l + 3 .

W hen

T h e se th ree straig h t lines are g ra p h ed below .

x1

13.5

b

N o tice th at the lines are p a ra lle l (th ey h av e the sam e slope).

a

T h e m odel is q u ad ratic.

b

S in ce R~ = .8 1 5 , th e sum o f sq u ares o f d e v ia tio n s is re d u c e d by 81.5% u sin g the

q u ad ratic m o d el ra th e r th an y to p re d ic t y. c

T h e h y p o th e sis to be te ste d is H 0 : /?, = p : = 0

H a: at le a st one /?( d iffe rs fro m zero

and the test statistic is F = ^ = 37.37 M SE w h ich has an F d is trib u tio n w ith df¡ = k = 2 and d f 2 = n - k - \ = 2 0 - 2 - 1 = 1 7 . T he p -v alu e g iv en in the p rin to u t is P = .0 0 0 and H 0 is rejected . T h e re is e v id e n c e th at the m odel c o n trib u te s in fo rm a tio n fo r the p red ic tio n o f y. 13.9

a

R ate o f in crease is m easu re d by the slope o f a line ta n g e n t to the cu rv e; this lin e is

g iv en by an e q u a tio n o b tain ed as d y / d x , the d e riv a tiv e o f y w ith re sp e ct to x. In p articu lar,

135

^ = ^ ( A ,+ / ? ,* + A * 2) = f l+ 2 /¡ ,* dx dx w hich has slo p e 2/3-,. If /?, is n eg ativ e, then the rate o f in crease is d ecreasin g . H enee, the h y p o th e sis o f in terest is H n :/? 2 = 0 , b

H a: £ < 0

T he individ u al r-test is / = -8 .1 1 as in E x ercise 13.8b. H o w ev er, the te st is o ne-

tailed, w hich m ean s th at the p -v a lu e is h a lf o f the a m o u n t g iv en in the p rin to u t. T h at is, p -v a lu e = -^ (.0 0 0 ) = .0 0 0 . H en ee, H0 is ag ain re jec ted . T h ere is e v id en c e to in d ícate a d e c rea sin g rate o f in crease.

13.15

a T he M i n it a b p rin to u t fittin g the m odel to the d a ta is sh o w n on the n ex t page. T he least sq u ares lin e is y = - 8 .1 7 7 + 0.292a:, + 4.434a:2

Regression Analysis: y v e rsu s x1, x2 The regression equation is y = - 8.18 + 0.292 xl + 4.43 x2 Predictor Coef SE Coef Constant -8.177 4.206 xl 0.2921 0.1357 x2 4.4343 0.8002 S = 3.30335 R-Sq = 82.3%

T P -1.94 0.093 2.15 0.068 5.54 0.001 R-Sq(adj) = 77.2%

Analysis of Variance Source DF SS Regression 2 355.22 Residual Error 7 76.38 Total 9 431.60

MS 177.61 10.91

Source xl x2

DF 1 1

F 16.28

P 0.002

Seq SS 20.16 335.05

b T h e F test fo r the o v erall u tility o f the m odel is F = 16.28 w ith P = .002 . T he resu lts are h ig h ly sig n ific a n t; the m o d el c o n trib u te s sig n ific a n t in fo rm atio n fo r the p red ictio n o f y. c T o test the effe c t o f ad v e rtisin g ex p e n d itu re , the h y p o th e sis o f in terest is '

H 0 : /?, = 0,

H a: & * 0

an d the test statistic is t = 5 .5 4 w ith p -v alu e = .0 0 1 . S in ce a = .0 1 , H0 is rejected . W e co n clu d e th at a d v e rtisin g e x p e n d itu re c o n trib u te s sig n ifica n t in fo rm a tio n fo r the p red ictio n o f y, g iv en th at cap ita l in v estm e n t is alre ad y in the m odel. d F ro m the M i n i t a b p rin to u t, R - S q = 8 2 . 3%. w h ich m ean s th at 8 2 .3 % o f the total v ariatio n can be e x p la in e d by the q u a d ratic m odel. T h e m odel is very effectiv e. 13.19

a T h e v ariable x 2 m u st be the q u a n tita tiv e v ariab le, since it ap p e a rs as a q u a d ra tic term in the m odel. Q u a litativ e v ariab les a p p e a r o n ly w ith e x p o n e n t 1, alth o u g h they

136

m ay a p p e a r as the co efficie n t o f a n o th e r q u a n titativ e v a ria b le w ith e x p o n e n t 2 or greater. b

W h en x, = 0 , y = 12.6 + 3.9x2 w h ile w h en x, = 1, y = 12.6 + . 5 4 ( l ) - 1 . 2 x 2 + 3 .9 x 2

_

= 1 3 .1 4 - 1 .2 x2 + 3 .9 x2 c

T h e g ra p h b elo w sh o w s the tw o p aráb o las.

x2

13 .2 3

T h e b asic resp o n se e q u atio n fo r a sp ecifíc ty p e o f b o n d in g c o m p o u n d w o u ld be E ( y ) = A , + P \ X\ + P 2X\ S in ce th e q u a lita tiv e v a riab le “b o n d in g c o m p o u n d ” is a t tw o lev els, o n e d u m m y v a riab le is n e ed e d to in co rp ó rate this v ariab le in to th e m o d el. D e fin e th e d u m m y v a riab le x2 as follo w s: x 2 = 1 if b o n d in g c o m p o u n d 2 = 0 o th erw ise T h e ex p a n d e d m odel is n o w w ritten as £ (> 0 = A ) + $ x i + M

13.25

2 + P i * 2 + P*x \x i + M 2*2

a

F ro m the p rin to u t, the p red ic tio n e q u atio n is y = 8 .5 8 5 + 3 .8 2 0 8 x - 0 .2 1 6 6 3 x 2 .

b

R 2 is lab eled “R -sq ” o r R 2 = .9 4 4 . H en ee 9 4 .4 % o f the total v ariatio n is

acco u n te d fo r by u sin g x a n d x 2 in th e m odel. c T h e h y p o th esis o f in te re st is H 0 : /?, = p 2 = 0

H a: at least o n e

d iffers fro m zero

and th e te st statistic is F = 3 3 .4 4 w ith p -v a lu e = .003 . H en ee, H 0 is re je c te d , and we co n c lu d e th at the m o d el co n trib u tes sig n ifican t in fo rm a tio n fo r the p re d ic tio n o f y. d T h e h y p o th e sis o f in te re st is H0 :A = 0

H a:/? 2 * 0

137

an d the test statistic is t = - 4 .9 3 w ith p -v a lu e = .008 . H en ee, H0 is reje c te d , a n d we co n clu d e that the q u ad ra tic m o d e l p ro v id e s a b e tte r fit to the d ata than a sim p le lin e a r m odel. e T h e p attern o f the d ia g n o stic p lo ts d o e s not in d ícate any o b v io u s v io latio n o f the reg ressio n assu m p tio n s. 1 3.29

a

T he m odel is

>’ = fio +M *i+ PiX2 + M 2 +

+ A * í*2 + £

and the M in itab p rin to u t is sh o w n below .

Regression A nalysis: y ve rsu s x1, x2, x ls q , x1x2, x1sqx2 The regression equation is y = 4.5 + 6 .39 xl - 50.9 x2 + 0.132 xlsq ■ Predictor Coef SE Coef Constant 4 .51 42 .24 xl 6.394 5 .777 x2 -50.85 56 .21 xlsq 0.1318 0.1687 xlx2 17.064 7 .101 xlsqx2 -0.5025 0.1992 S = 71.6891 R-Sq = 76.8% Analysis of Variance Source DF SS Regression 5 664164 Residual Error 39 200434 Total 44 864598

T P 0.11 0.916 1.11 0 .275 -0.90 0.371 0.78 0.439 2.40 0.021 -2.52 0.016 R-Sq(adj) = '

MS 132833 5139

F 25.85

P

0 .0 0 0

b T h e fitted p red ic tio n m o d el u se s the co e ffic ie n ts giv en in the c o lu m n m ark ed “C o e f ’ in the printout: y = 4.51 -t- 6.394jc, - 5 0 .8 5 x , + 17.064x,;t2 + .131 8.x,2 - . 5 0 2 5 x f x 2 T h e F te st fo r the m o d e l’s u tility is F = 25.85 w ith P = .0 0 0 and R~ = .768 . T he m odel fits q u ite w ell. c

If the d o lp h in is fem ale, x 2 = 0 and the p red ic tio n e q u atio n b eco m e s y = 4.51 + 6.394.V, + .1 3 1 8jc2

d

If the d o lp h in is m ale, x 2 = 1 and the p red ic tio n e q u a tio n b eco m e s y = - 4 6 .3 4 + 23.458.x, - .3707a:2

e

T h e h y p o th esis o f in te re st is H „ :/? 4 = 0

H a :/? 4 * 0

an d the test statistic is t = .78 w ith p -v a lu e = .4 3 9 . H0 is no t reje cte d a n d w e co n c lu d e that the q u ad ratic term is not im p o rta n t in p re d ic tin g m e rcu ry c o n ce n tra tio n fo r fem ale dolp h in s. 13.31

a - b T h e dató-is p lo tte d on the n ex t page. It ap p e a rs to be a c u rv ilin e a r rela tio n sh ip , w hich co u ld be d e sc rib e d u sin g the q u a d ra tic m odel y = /?0 + ¡3xx + f i 2x 2 + £ .

138

X -----------------

—

c

T h e M i n it a b p rin to u t is sh o w n below .

R egressio n A nalysis: y v e rsu s x, x_sq The regression equation is y = 4114749 - 4113 x + 1.03 x_sq Predictor Coef SE Coef Constant 4114749 343582 x -4113.4 343.2 x_sq 1.02804 0.08568 S = 0.523521 R-Sq = 99.7% Analysis of Variance Source DF SS Regression 2 297.16 Residual Error 3 0.82 Total 5 297.98

d

T P 11.98 0.001 -11.99 0.001 12.00 0.001 R-Sq(adj) = 99.5%

MS 148.58 0.27

F 542.11

P 0.000

T h e h y p o th e sis o f in terest is H0 : f l = A = 0

and the te st statistic is F = 542.11 w ith p -v a lu e = .0 0 0 . H 0 is re je c te d a n d we co n c lu d e th a t th e m odel p ro v id e s valu ab le in fo rm a tio n fo r th e p re d ic tio n o f y. e

R 2 = .9 9 7 . H enee, 9 9 .7 % o f the total v ariatio n is a c c o u n te d fo r b y u sin g x a n d x 2

in the m odel. f T h e resid u al p lo ts a re show n b elo w . T h e re is no rea so n to d o u b t the v a lid ity o f ____ the re g re ssio n a ssu m p tio ns. R e s id u a ls V e r s u s t he Fitted V a lú e s (ra p a rs e • y)

0M -------------------------------------------

| aoo •0.25

0

5

10 ntted Valué

139

15

20

1 3.35

a R 2 = .999 . H enee, 9 9 .9 % o f the total v ariatio n is a c c o u n te d fo r b y u sin g x a n d x 2 in th e m o d el. b T h e h y p o th esis o f in terest is H0 :/? ,= A = 0 a n d the test statistic is F = 1676.61 w ith p -v a lu e = .0 0 0 . H 0 is re je c te d a n d we co n clu d e th at the m o d el p ro v id es v alu ab le in fo rm a tio n fo r th e p re d ic tio n o f y. c T h e h y p o th esis o f in terest is H0 :f l= 0 a n d the test statistic is t = - 2 .6 5 w ith p -v a lu e = .045 . H 0 is re je c te d and w e c o n c lu d e th a t th e lin e a r re g re ssio n co e ffic ie n t is sig n ific a n t w hen x 1 is in the m odel. d T h e h y p o th esis o f in terest is H (, : A = 0 and the te st statistic is t = 15.14 w ith p -v a lu e = .0 0 0 . H0 is reje cte d a n d w e co n c lu d e that th e q u a d ra tic re g re ssio n c o e ffic ie n t is sig n ific a n t w hen x is in th e m odel. e W h en th e q u a d ratic term is re m o v e d fro m th e m o d el, th e v alu é o f R 2 d e c re a se s by 9 9 .9 - 9 3 . 0 = 6 .9 % . T h is is the ad d itio n al co n trib u tio n o f th e q u a d ra tic term w h en it is in clu d ed in the m odel. f T h e c le a r p a tte m o f a c u rv e in the resid u al p lo t in d icates th a t the q u a d ratic term should be in clu d ed in th e m odel.

140

14: Analysis of Categorical Data 14.3

F o r a test o f sp ec ifie d ce ll p ro b ab ilities, the d e g re e s o f fre e d o m are k - 1 . U se T a b le 5, A p p e n d ix I: a

d f = 6; x l s = 12.59; re je c t H0 if X 2 > 12.59

b

d f = 9; X m = 2 1 .6 6 6 ; re je c t H 0 if X 2 > 2 1 .666

c d f = 13; x%a = 2 9.814; re je c t H 0 if X 2 > 2 9 .8 1 9 4 d

14.7

d f = 2; Xas = 5-99; re je c t H 0 if X 2 > 5.99

O ne th o u san d c a rs w ere each c la ssifie d acco rd in g to th e lañ e w h ich th e y o ccu p ie d (one th ro u g h fo u r). I f n o lañ e is p referred o v er an o th er, th e p ro b a b ility th a t a car will be d riv en in lañe i, i = 1 ,2 ,3 ,4 is Va. T h e nuil h y p o th e sis is then H n : P x = P 2 = Pi = P< = j a n d the test statistic is

w ith £ , = n p i = 1 0 0 0 (1 /4 ) = 2 5 0 fo r i = 1 ,2 ,3 ,4 . A tab le o f o b se rv e d an d e x p ected cell c o u n ts fo llo w s: L añe o¡

1 294

2 276

3 238

4 192

E,

250

250

250

250

T hen v2 _ (2 9 4 - 2 5 0 )2 i (2 7 6 -2 5 0 )2 _ (2 3 8 -2 5 0 )2 _ (1 9 2 -2 5 0 )2 250

250

250

25 0

_ 6 H 0 _ 24 4g 250 T h e re je c tio n reg ió n w ith k - 1 = 3 d f is X 2 > x l s = 7 .8 1 . S in ce th e o b se rv e d v alué o f X 2 fa lls in the reje c tio n reg ió n , w e reject H 0. T h ere is a d iffe re n c e in p refe ren c e fo r the fo u r lañes. 14.11

S im ila r to p rev io u s ex ercises. T h e h y p o th esis to b e tested is H 0 : P\ = Pi = " ‘ = P n ~ v ersu s

H a : at least one p¡ is d ifferen t from th e oth ers

w ith E. = nPi = 4 0 0 ( 1 /1 2 ) = 3 3 .3 3 3 . T h e te st statistic is

141

, ( 3 8 - 3 3 .3 3 ) ( 3 5 - 3 3 .3 3 ) X 2 = - ---------------- ¿ - + - + ^-------------- — = 13.58 3 3 .3 3 33.33 T h e u p p e r tailed reje c tio n reg ió n is w ith a = .05 a n d k - 1 = 11 d f is X 2 > Xas - 19.675 . T h e nuil h y p o th e sis is n o t re je c te d and w e c a n n o t c o n c lu d e that the p ro p o rtio n o f c ases v aries from m o n th to m onth. 14.15

It is n ecessary to d ete rm in e w h eth er ad m issio n rates d iffe r fro m th e p rev io u sly re p o rted rates. A tab le o f o b se rv e d and ex p ected cell co u n ts follow s: U n co n d itio n al T rial 329 43 300 25 E¡ T h e nuil h y p o th e sis to be tested is O,

H 0 : P, = -60; p 2 = .05;

R efu sed 128 175

T o ta ls 500 500

= .35

a g a in st th e altern ativ e th at at least o n e o f th ese p ro b a b ilitie s is in co rrect. T h e test statistic is x2 = (3 2 9 -3 0 0 )- + ( « z ^ 300

l + ( 1 2 L - 17 S y

25

175

T h e n u m b er o f d e g re es o f freed o m is k - 1 = 2 an d the rejec tio n reg ió n 2 2 X > X a s = ^ -9 9 . T h e nuil h y p o th e sis is rejected , and w e c o n c lu d e th a t th e re has b een a d e p artu re fro m p rev io u s ad m issio n rates. N o tic e th a t the p e rc e n ta g e o f u n c o n d itio n al a d m issio n s has risen slig h tly , the n u m b e r o f c o n d itio n a l a d m issio n s has in creased , and the p e rc en tag e refu sed ad m issio n has d e cre a se d at the e x p e n se o f the fist tw o catego ries. 14.17

R e fe r to S ectio n 14.4 o f the text. F o r a 3 x 5 co n tin g en cy ta b le w ith r = 3 a n d c = 5 , th ere are ( r - l ) ( c - l ) = ( 2 ) ( 4 ) = 8 d e g re es o f freed o m .

14.21

a T h e h y p o th e sis o f in d e p en d e n c e b etw een a tta c h m e n t p a tte rn a n d ch ild c are tim e is tested u sin g the ch i-sq u a re statistic. T h e co n tin g e n cy tab le, in c lu d in g c o lu m n and ro w to tals and th e estim a te d e x p e c ted cell c o u n ts, follow s.________ A tta c h m e n t Secure A nx io u s T o ta l

C h ild C a re M o d érate 24 35 (24.09) (30.97) 11 10 (10.91) (14.03) 111 51 L ow

H igh 5 (8 .9 5 )

T o ta l 64

8 (4 .0 5 ) 297

29 459

T h e te st statistic is v 2 _ ( 2 4 - 2 4 . 0 9 ) 2 | ( 3 5 - 3 0 . 9 7 )2 <

24.09

30.97 142

(8 - 4 .0 5 ) + -------------— = 7 .2 6 7 4.05

an d the reje ctio n reg ió n is X 2 > X m = $-99 w 'th 2 d f .

is re je c ted . T h e re is

e v id e n c e o f a d e p e n d en c e b etw ee n a tta c h m e n t p attern an d c h ild c a re tim e, b

T h e v alu é X 2 = 7 .2 6 7 is b e tw e e n x l ¡ and Xms so l^ at

< p -v a lu e < .05 .

re su lts are sig n ifican t. 14.25

a T he h y p o th e sis o f in d e p e n d e n c e b e tw een sa la ry a n d n u m b e r o f w o rk d ay s at he is tested using the c h i-sq u a re statistic. T h e co n tin g e n c y ta b le , in clu d in g c o lu m n a ro w to tals a n d th e estim a te d e x p e c te d cell c o u n ts, g en e ra te d by M i n i t a b follow s.

Chi-Square Test: L e ss than one, At least one, not all, All at home Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts

Less than one 38 36.27 0.083

At least one, not all 16 21.08 1.224

All at home 14 10.65 1.051

2

54 49.07 0.496

26 28.52 0 .223

12 14 .41 0.404

92

3

35 35.20 0 .001

22 20 .46 0.116

9 10 .34 0 .174

66

4

33 39.47 1. 060

29 22 .94 1.601

12 11.59 0.014

74

Total

160

93

47

300

1

Total 68

Chi-Sq = 6.447, DF = 6, P-Value = 0.375

T h e test statistic is v 2 _ (3 8 - 3 6 .2 7 ) 2 | (1 6 - 2 1 .0 8 )2 ( 36.27

| ( 1 2 - 1 1 .5 9 )2 _

2 1 .0 8

^

11.59

a n d the rejectio n reg ió n w ith a = .05 and d f = 3 (2 ) = 6 is X

2 > Xas

= 12.59 and

nuil h y p o th esis is no t re je cte d . T h e re is in su ffic ien t e v id e n c e to in d ícate th at salai d ep en d en t on the n u m b e r o f w o rk d ay s sp en t a t hom e. b

T h e o b serv e d v alu é o f th e test statistic, X

2=

6 .4 4 7 , is less th a n X a o = 10.64¿

th at the p -v a lu e is m ore th a n .10. T h is w ould co n firm th e n o n -reje c tio n o f th e nul h y p o th e sis fro m p art a. 14.29

B e cau se a set n u m b e r o f A m e ric a n s in e a ch su b -p o p u la tio n w ere e a c h fíx ed at 20' w e have a c o n tin g e n c y tab le w ith fix ed ro w s. T h e tab le, w ith e stim a te d e x p ected co u n ts ap p earin g in p a ren th e se s, is sh o w n on th e n e x t page.

143

W h ite -A m e rica n A frican -A m erican H isp an ic-A m erican A sian -A m erican T otal

Y es 40 (62) 56 (62)

No 160 (1 3 8 ) 144 (1 3 8 )

T otal 200

68 (62) 84 (62)

132 (1 3 8 ) 116 (1 3 8 ) 552

200

248

200

200 800

T h e test statistic is v2 _ ( 4 0 - 6 2 )2 t (5 6 - 6 2 ) 2 | 62

, (1 1 6 -1 3 8 )2

62

21

138

and the reje c tio n reg ió n w ith 3 d f '\s X 2 > 1 1 .3 4 4 9 . H 0 is re je cte d and w e co n c lu d e that th e in cid en ce o f p aren tal su p p o rt is d e p en d en t on the su b -p o p u la tio n o f A m erican s. 1 4.33

T h e n u m b er o f o b serv a tio n s p e r co lu m n w ere se lected p rio r to the ex p e rim e n t. T he test p ro c e d u re is identical to th at u sed for an r x c c o n tin g e n c y tab le. T he c o n tin g e n c y ta b le, in clu d in g c o lu m n and ro w to tals and the estim a te d e x p e c te d cell c o u n ts, follow s. F am ily M em b ers 1 2 3 4 o r m ore T otal T h e test statistic is . X-

A p artm en t

T yp e D úplex

8 (9.6 7) 16

20 (9.67) 8

(11) 10 (1 1 .3 3 ) 6

(11) 10 (1 1 .3 3 ) 2

(11) 14 (1 1 .3 3 ) 16

(8) 40

(8) 40

(8) 40

S ingle R esid en ce 1 (9 .6 7 )

T o ta l

9

33

(8 —9 .6 7 )2 ( 2 0 - 9 . 6 7 )2 ------------’- + ±----------------- + 9 .6 7 9.67

29

34 24 120

(1 6 -8 )2 +± — = 3 6 .4 9 9 8

using C om puter accu racy . W ith ( r - l ) ( c - l ) = 6 d f a n d a = .0 1 , the re je c tio n reg ió n is X 2 > 16.8449 . T h e nuil h y p o th e sis is rejected . T h ere is su fficie n t e v id e n c e to in d icate th a t fam ily size is d ep e n d en t on type o f fam ily resid e n c e . It ap p e a rs th at as the fam ily size in creases, it is m ore likely th at p eo p le w ill liv e in sin g le re sid e n c e s. 1 4.35

I f the h o u se k e e p e r actu a lly has no p referen ce, he o r she h as an e q u a l ch a n c e o f p ick in g any o f the five flo o r p o lish es. H enee, the nuil h y p o th e sis to be tested is

144

H,

P\ = Pi = Py = P\ = Ps = ^

T h e v alú es o f O, are the a c tu a l co u n ts o b se rv e d in the e x p e rim e n t, and E. = / i p , = 1 0 0 (1 /5 ) = 2 0 .

-

P o lish

T hen

X =

o¡

A 27

E¡

20

B 17 20

C 15 20

D 22 20

E 19 20

(2 7 -2 0 )'

(1 7 -2 0 )"

(1 9 -2 0 )2

20

20

20

= 4 .4 0

T h e p -v a lu e w ith d f = k - 1 = 4 is g re ate r th an .1 0 an d Ho is n o t rejected . W e can n o t co n clu d e th at there is a d iffe re n c e in the p re feren c e fo r th e fiv e flo o r p o lish es. E ven if this h y p o th e sis h a d been re je c te d , the c o n c lu sió n w o u ld be th at at least o n e o f the v alué o f the p , w as sig n ific a n tly d iffe re n t fro m 1/6. H o w e v er, th is d o e s n o t im p ly that p¡ is n ecessarily g re a ter th an 1/6. H en ee, w e c o u ld n o t c o n c lu d e th at p o lish A is su perior. If th e o b jectiv e o f the e x p e rim e n t is to sh o w th at p o lish A is su p e rio r, a b etter p ro ced u re w ould be to test an h y p o th e sis as fo llo w s:

H 0 '■P\ = 1 /6

H a : p, > 1 /6

F ro m a sam p le o f n = 100 h o u se w iv e s, x = 27 are fo u n d to p re fe r p o lish A. ca n be p erfo rm ed o n the sin g le b in o m ial p a ra m e ter p¡. 14.39

A s-te st

a T o test fo r h o m o g e n e ity o f the five b in o m ial p o p u la tio n s, w e use c h i-sq u are statistic and the 5 x 2 c o n tin g e n c y ta b le show n b e lo w T h e nuil h y p o th e sis is th at v o ter ch o ice and c h u rc h a tte n d a n c e are in d ep en d en t, w ith p be the p ro p o rtio n o f v o ters w ho in ten d to v o te fo r G .W . B u sh in the 2 0 0 4 ele c tio n fo r a p a rtic u la r c h u rc h atten d an ce g roup. T h e c o n tin g e n cy ta b le g e n erated by M i n i t a b is sh o w n belo w

Chi-Square Test: G.W. Bush, Democrat Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts G.W. Bush Democrat Total 89 53 142 i 73 .64 68.36 3 .205 3 .453 2

87 80.38 0.546

68 74 .62 0.588

155

3

93 92.30 0.005

85 85.70 0.006

178

4

114 128.60 1.658

134 119.40 1.786

248

145

5

22 30.08 2.169

36 27.92 2.336

58

Total

405

376

781

Chi-Sq = 15.752,

DF = 4, P-Value = 0.003

T h e o b serv ed v alu é o f the te st statistic is X 2 = 15.752 w ith p -v a lu e = .003 a n d the nuil h y p o th e sis is re je c te d at the 5% level o f sig n ifícan ce. T h ere is su ffic ie n t e v id en c e to in d ic a te th a t th e p ro p o rtio n o f ad u lts w h o in ten d to v o te fo r G .W . B u sh in the 2 0 0 4 elec tio n is d e p en d e n t on c h u rc h atten d an ce. 14.43

T h e flo w er fall in to o n e o f fo u r cla ssific a tio n s, w ith th e o re tic al ratio 9 :3 :3 :1 . C o n v e rtin g th ese ra tio s to p ro b ab ilities, Pl = 9 / 1 6 = .5625

p 2 = 3 /1 6 = . 1875

p , = 3 /1 6 = .1875

p 4 = 1 /1 6 = .0625

W e w ill te st th e nuil h y p o th e sis th at the p ro b a b ilitie s are as ab o v e a g a in st the a ltern ativ e th at th ey differ. T h e tab le o f o b se rv e d and e x p e c ted cell c o u n ts fo llo w s: AB Ab aB aa o¡ E, T h e te st statistic is

95 90

30 30

28 30

7 10

v2 _ (9 5 -9 0 )2 ^ 3 0 -3 0 )* t (2 8 -3 0 )2 , (7 -1 0 )2 _ ^ ^ ] 90

30

30

10

T h e n u m b er o f d e g re e s o f freed o m is k - 1 = 3 a n d the reje c tio n re g ió n w ith a = .01 is X 2 > x \ \ = 1 1-3449 . S in ce the o b se rv ed v alué o f X 2 d o e s n o t fall in th e reje ctio n re g ió n , w e d o not reject H 0. W e d o n o t have en o u g h in fo rm atio n to c o n tra d ic t the th eo retical m odel fo r the c la ssific atio n o f flo w e r c o lo r and sh ap e. 14.48

In o rd e r to p erfo rm a c h i-sq u a re “g o o d n ess o f fit” test on th e g iv e n d a ta , it is n ecessary that th e v a lú es O, and E¡ are k n o w n fo r e a c h o f th e fív e c e lls. T h e O, (the n u m b e r o f m easu re m en ts fallin g in th e i-th cell) are g iv en . H o w e v e r, £ . = n p ( m ust be calc u lated . R em e m b e r th at p, is the p ro b ab ility th at a m e a su re m e n t falls in th e j'-th cell. T h e h y p o th e sis to be tested is H0 : the ex p e rim en t is b in o m ial

v ersu s

H a : the e x p e rim e n t is n o t b in o m ia l

L e t x be the n u m b e r o f su c c e sse s and p be the p ro b a b ility o f su ccess on a sin g le trial. T h en , a ssu m in g the nuil h y p o th e sis to be true, ^

= p ( x = 0 ) = Coy

(1 - P ) 4

p 2 = / > ( * = 2 ) = c 24p 2 ( l - p ) 2 p 4 = p ( x = 4 ) = C ; p 4( l - p ) °

146

p l = p ( x = l ) = c t4p ' ( l - p

)3

p 3 = p ( x = 3 ) = C * p 3( l - p ) ‘

H en ee, o n c e an e stím ate fo r p is o b tain ed , the e x p e c ted ce ll freq u en c ie s can be calc u lated u sin g the ab o v e p ro b a b ilitie s. N o te th a t e a c h o f th e 100 e x p erim en ts co n sists o f fo u r triáis a n d h en ee the c o m p lete e x p e rim e n t in v o lv es a total o f 4 0 0 triáis. T h e best e stim a to r o f p is p = x / n (as in C h a p te r 9 ). T hen, „ _ x _ n u m b e r o f su cc esses _ 0(1 l) + l ( l 7 ) + 2 ( 4 2 ) + 3(12)-t-'4‘(9 ) n

n u m b e r o f triáis

400

1 2

T h e ex p erim en t (c o n sistin g o f fo u r triáis) w as re p e a te d 100 tim es. T h e re are a to tal o f 4 0 0 triáis in w h ich th e re su lt “n o su c c e sse s in fo u r triá is” w as o b se rv e d 11 tim es, the resu lt “one su ccess in fo u r triá is” w as o b serv e d 17 tim es, a n d so on. T h en Po = c * ( 1 / 2 ) 0 ( l / 2 ) 4 = 1/16

p , = c ; (1 /2 )' (1 /2 )3 = 4 /1 6

p 2 = < T Í ( l / 2 ) '( l / 2 ) ’ = 6 /1 6

p , = C , ( l / 2 ) ? (1 /2 )' = 4 /1 6

P , = C 44 ( l / 2 ) 4 (1 /2 )° = 1 /1 6 T h e o b serv e d and e x p e c ted ce ll fre q u en c ie s are sh o w n in th e fo llo w in g table. X 0 2 1 3 4 11 17 42 21 9 o, 6.25 2 5 .0 0 3 7 .5 0 2 5 .0 0 6.25 E, and th e statistic is X =

(1 1 - 6 .2 5 )2

(1 7 - 2 5 .0 0 ) 2

( 9 - 6 .2 5 )

= 8.56 6.25 2 5 .0 0 6.25 In o rd e r to b o u n d the p -v a lu e o r set up a reje c tio n reg ió n , it is n e cessary to d e term in e the ap p ro p ria te d e g re e s o f freed o m asso c iate d w ith th e te st statistic. T w o d eg re e s o f freed o m are lo st because: 1

T h e cell p ro b a b ilitie s are restricte d by th e fact th a t Z p , = 1.

2

T h e b in o m ia l p a ra m e te r p is u n k n o w n and m u st b e e stim a te d b efo re c alc u latin g the ex p ected cell co u n ts. T h e n u m b er o f d e g re e s o f fre e d o m is e q u al to k - \ - \ = k - 2 = 3 . W ith d f = 3 , the p -v a lu e fo r X 2 = 8 .5 6 is b e tw een .025 and .05 and the nuil h y p o th e sis can be re jec te d at th e 5 % level o f sig n ifican ce. W e co n clu d e th at the e x p erim en t in q u e stio n d o es n o t fu lfill th e req u irem en ts fo r a b in o m ial ex p erim en t.

14.51

T h e nuil h y p o th e sis to be tested is

u H0 : p,

= p 2 = p 3 = -1

and the test statistic is x 2= I w ith E j = np¡ = 2 0 0 (1 /3 ) = 6 6 .6 7 fo r / = 1 ,2 ,3 . A ta b le o f o b se rv e d and e x p ected cell c o u n ts follow s:

147

E n tran c e

1

o¡ E¡

83 6 6 .6 7

2 61 6 6 .6 7

3 56 6 6 .6 7

T hen v 2 _ ( 8 4 - 6 6 . 6 7 ) 2 | ( 6 1 - 6 6 . 6 7 )2 | ( 5 6 - 6 6 .Ó 7 ) 2 _ 6 ^ 6 6 .6 7

6 6 .6 7

6 6 .6 7

W ith d f = k - 1 = 2 , th e p - v a lu e is b e tw een .025 and .05 án d w e can re je c t H 0 at the 5% level o f sig n ific a n c e . T h ere is a d iffe re n c e in p re feren ce fo r the th ree d o o rs. A 9 5% c o n fid e n c e in terv al fo r p \ is giv en as

n

± Z o 2 5 \r^ n

025V

=> —

± l - 9 6 j 4 1 5 ^'5 8 5 ^ => .4 1 5 ± .0 6 8

200

V 200

o r .347 < p, < .4 8 3 . 14.55

S in ce the c a rd s fo r each o f the th ree h o lid ay s w ill be e ith er “h u m o ro u s” o r “ not h u m o ro u s” , the tab le actu a lly co n sists o f tw o ro w s and th ree c o lu m n s, an d is sh o w n w ith e stim a te d e x p ected and o b serv ed cell c o u n ts in the M i n i t a b p rin to u t b elo w .

Chi-Square Test: Fathers, Mothers, Valentines Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Fathers Mothers Valentines Total 100 125 120 345 i 115.00 115.00 115.00 1. 957 0.870 0 .217 2 400 380 375 1155 385.00 385.00 385.00 0 .584 0 .260 0 .065 Total 500 500 500 1500 Chi-Sq = 3.953, DF = 2, P-Value = 0. 139

T h e test statistic fo r the e q u ality o f the th ree p o p u latio n p ro p o rtio n s is X 2 = 3.953 w ith p -v alu e = . 139 and H 0 is n o t rejected . T h ere is in su ffic ie n t e v id e n c e to in d ícate a d iffe re n c e in the p ro p o rtio n o f h u m o ro u s c a rd s fo r the th ree h o lid ay s. 1 4 .5 9

a T h e 2 x 3 co n tin g e n cy tab le is a n a ly z ed as in p re v io u s e x e rc ises. T h e M i n it a b p rin to u t b e lo w sh o w s the o b serv e d a n d estim a ted e x p ected ce ll c o u n ts, the test statistic and its a sso c iate d p -v a lu e .

Chi-Square Test: 3 or fewer, 4 or 5, 6 or more Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts 3 or fewer 4 or 5 6 or more Total 1 49 43 34 126 37.89 45.47 42 .63 T . 254 0.003 2 .895 2 31 47 62 140 42.11 47.37 50.53 2 .929 0.003 2 .605 Total 80 90 96 266 Chi-Sq = 11.690, DF = 2, P-Value = 0.003

148

T h e resu lts are hig h ly sig n ifican t ( p -v a lu e = .003 ) a n d w e c o n c lu d e th at th ere is a d iffe re n c e in the su sc e p tib ility to c o ld s d e p e n d in g on the n u m b e r o f re la tio n sh ip s you have. b T h e p ro p o rtio n o f p eo p le w ith c o ld s is c a lc u la te d c o n d itio n a lly fo r each o f the T h ree o r few e r

F o u r o r fiv e

S ix o r m o re

C oid

i » = .61 80

« = .4 8 90

— = .35 96

N o coid

— = .39 80

— = .52 90

— = .65 96

T otal 1.00 100 1.00 A s the re se a rc h e r su sp ects, th e su sce p tib ility to a c o id seem s to d e c rea se as the n u m b e r o f re la tio n sh ip s in creases! 14.65

U se the first G o o d n ess-o f-F it a p p let. E n te r the o b se rv e d v alú es in to the th ree c e lls in the first row . and the e x p e c te d cell co u n ts w ill au to m a tic a lly a p p e a r in the seco n d row . T h e v alué o f X : = 18.5 w ith p -v a lu e = .0001 p ro v id e su fficien t e v id e n c e to re je c t H0 and co n c lu d e th at c u sto m e rs have a p referen ce fo r o n e o f the th ree b ran d s (in this case, B ran d II). O reen

R ed

« 1 1 5

..............

Uoo o

■

T o« *l

B iu e

120

65

300

100.0

100 0

300

i

O b s e iv é d Fr« Q u «n e l*s

ChiSq(2) = 18.5, p-value = 14.69

1

OE-4

T h e d a ta is a n aly z ed as a 2 x 3 c o n tin g en cy tab le w ith e stim a te d e x p e c ted cell co u n ts sh o w n in p aren th ese s. U se th e C h i-S q u a re T est o f In d e p e n d e n c e ap p let. Y o u r

P arty R ep u b lican D em o crat T otal

1 114 (1 2 0 .8 6 ) 87 (8 0 .1 4 ) 201

O p in ió n 2 53 (4 8 .1 0 ) 27 (3 1 .8 9 ) 80

149

3 17 (1 5 .0 3 ) 8 (9 .9 7 ) 25

T o ta l 184 122 306

T h e test statistic is y 2 _ ( 1 1 4 - 1 2 0 .8 6 ) 2 t ( 5 3 - 4 8 . 1 0 ) 2 [ 120.86

4 8 .1 0

( (B - 9 .9 7 )2 = 2 9.97

W ith d f = 2 , the p -v a lu e is g re a te r th an .10 (the ap p let rep o rts p -v a lu e = .2 3 7 8 ) and H 0 is not rejecte d . T h e re is no e v id en c e th at p arty a ffiliatio n h as any e ffe c t on opinión.

15: Nonparametric Statistics 15.1

a

If d istrib u tio n 1 is sh ifted to th e rig h t o f d istrib u tio n 2 , th e ra n k su m fo r sa m p le 1

(T¡) w ill te n d to be large. T h e te st sta tistic w ill be T ¡ , the ra n k sum fo r sa m p le 1 if the o b se rv atio n s had b een ra n k ed fro m larg e to sm all. T h e n uil h y p o th e sis w ill be re je c te d if 7j’ is u n u su ally sm all.

15.5

b

F ro m T a b le 7 a w ith n, = 6 , n 2 = 8 and a = .05 , H 0 w ill be re je c te d i f 7'* < 3 1 .

c

F ro m T a b le 7c w ith n, = 6 , n2 = 8 a n d a = .0 1 , H 0 w ill be re je c te d if 7”, < 2 1 .

I f H a is tru e and p o p u la tio n 1 lies to th e rig h t o f p o p u la tio n 2 , th en 7 , w ill b e larg e and 7* w ill be sm all. H en ee, the te st sta tistic w ill be 7* and th e la rg e sa m p le a p p ro x im a tio n can be used. C alcú late 7* = n, (rtj + n2 + 1) - 7 , = 1 2 (2 7 ) —193 = 131 _ n , ( t t i + n 2 + l ) _ 1 2 (2 6 + 1) Mt

2

^

2

= nin 2( n ] + n 2 + 1) = 1 2 (1 4 ) (2 7 ) = 3 ?g r

12

12

T h e te st statistic is

Z—

7 ,- A r _ 1 3 1 -1 6 2 _ /

Gr

v.jy

V 378

T h e reje ctio n reg ió n w ith a = .05 is z < - 1 .6 4 5 a n d Hq is n o t reje c te d . T h e re is in su ffic ien t e v id e n c e to in d ic a te a d iffe re n c e in th e tw o p o p u latio n d istrib u tio n s. 15.9

S im ila r to p rev io u s ex ercises. T h e d ata, w ith c o rre sp o n d in g ran k s, are sh o w n in the fo llo w in g table.

___________ ______________ H e a r in g (2) D e a f (1) 2 .7 5 ( 1 5 ) 0 .8 9 ( 1 ) 2 .1 4 ( 1 1 ) 1.43 (7) 1.06 (4) 3 .2 3 (1 8 ) 2 .0 7 (1 0 ) 2 .4 9 ( 1 4 ) 2 .1 8 (1 2 ) 3 .1 6 (1 7 )

1.01 (3) 0 .9 4 (2 )

2 .9 3 ( 1 6 ) 2 .2 0 ( 1 3 )

2.01 (9) 1 .1 2 (5 .5 )

1.79 (8) 1 .1 2 (5 .5 )

7, = 1 2 6 C a lc ú la te

151

T{ = 126 T¡ = « , ( « , + n 2 + l ) - r , = 9 (1 9 ) —126 = 4 5 T h e test statistic is T = m in (7 J,7 ’1‘ ) = 45 W ith n, = n 2 = 9 , th e tw o -ta ile d rejec tio n reg ió n w ith ar = .05 is fo u n d in T a b le 7 b to be 7j’ < 6 2 . T h e o b serv ed v alu é, T = 4 5 , falls in th e re je c tio n reg ió n and H 0 is rejected . W e co n clu d e th at th é d e a f c h ild re n d o d iffe r fro m the h e a rin g c h ild re n in e y e-m o v em en t rate. 15.13

a

If a p a ire d d ifferen c e e x p erim en t h as been u sed a n d th e sign test is o n e -ta ile d

( H a : p > .5 ) , then th e e x p e rim e n te r w o u ld lik e to sh o w th at o n e p o p u la tio n o f m e asu rem en ts lies ab o v e the o th e r p o p u la tio n . A n e x a c t p ra c tic a l sta te m e n t o f the a ltern ativ e h y p o th e sis w o u ld d e p en d on the e x p e rim en tal situ atio n . b It is n ecessary th at a (the p ro b a b ility o f re je c tin g th e nuil h y p o th e sis w h en it is tru e) take v a lú e s less than a = .15 . A ssu m in g th e n uil h y p o th e sis to be tru e, th e tw o p o p u la tio n s a re id en tical an d co n se q u e n tly , p = P { A ex ceed s B fo r a g iv en p a ir o f o b se rv a tio n s) is 1/2. T h e b in o m ial p ro b a b ility w as d iscu sse d in C h a p te r 5. In p a rtic u la r, it w as n o ted th a t th e d is trib u tio n o f the ra n d o m v ariab le * is sy m m etrical a b o u t the m ean n p w h en p = 1/2 . F o r ex am p le, w ith n = 25, P ( x = 0 ) = P ( x = 2 5 ) . S im ila rly , P { x = 1) = P ( x = 2 4 ) a n d so on. H en ee, the lo w e r tailed p ro b a b ilitie s ta b u la ted in T a b le 1, A p p e n d ix I w ill b e id en tical to th eir u p p er tailed eq u iv ale n t p ro b a b ilitie s. T h e v alú es o f a a v a ila b le fo r th is u p p e r tailed test an d th e c o rre sp o n d in g reje ctio n re g io n s are sh o w n below .

15.17

R ejec tio n R eg ió n

a

*>20

.002

*>19

.007

*>18

.022

*>17

.054

*>16

.115

a I f assesso rs A and B are eq u al in th e ir p ro p e rty a ss essm e n ts, th e n p, the p ro b a b ility th a t A ’s assessm e n t e x c e e d s B ’s asse ssm e n t fo r a g iv en p ro p e rty , sh o u ld eq u al 1/2. I f o n e o f the a sse sso rs ten d s to be m o re co n se rv a tiv e th an th e o th er, then eith e r p > 1/2 o r p < 1/2 . H enee, w e c a n te st th e e q u iv a le n c e o f th e tw o a sse sso rs by testin g th e h y p o th esis H n : p = l/2

v ersu s

Ha :pí¿l/2

u sin g the test statistic *, th e n u m b er o f tim es th at a sse sso r A e x c e e d s a sse sso r B fo r a p a rtic u la r p ro p e rty assessm en t. T o find a tw o -ta ile d reje c tio n reg ió n w ith a c ió s e to .05, u se T ab le 1 w ith n = 8 and p = .5 . F o r th e reje c tio n reg ió n {* = 0, * = 8) the

152

valu é o f a is .004 + .004 = .008 , w h ile fo r th e rejec tio n reg ió n { x = 0 ,1 ,7 ,8 } the v alu é o f a is .0 3 5 + .035 = .0 7 0 w h ich is c lo se r to .05. H en ee, u sin g th e reje c tio n reg ió n {jc < 1 or * > 7 } , th e nuil h y p o th esis is not rejected , sin ce x = n u m b e r o f p ro p e rtie s fo r w h ich A ex ce e d s B = 6 . T h e p -v a lu e fo r this tw o -ta ile d te st is p -v a lu e = 2 P ( x > 6 ) = 2(1 - . 8 5 5 ) = .290 S in ce th e p -v a lu e is g rea te r th a n .10, th e re su lts are n o t sig n ifica n t; H 0 is n o t re je c te d (as w ith th e critical v alu é a p p ro ach ). b T h e t sta tistic u se d in E x e rc ise 10.45 allo w s the ex p e rim e n te r to re je c t Ho, w h ile th e sig n te st fails to re je c t H 0. T h is is b ecau se th e sig n test u sed less in fo rm atio n and m ak es fe w e r a ssu m p tio n s th a n d o e s th e t test. I f all n o rm a lity a ssu m p tio n s a re m et, the t test is th e m o re p o w e rfu l test and can re je c t w hen the sign te st c an n o t. 15.21

a

Ho'. p o p u la tio n d istrib u tio n s 1 a n d 2 are identical H a: the d istrib u tio n s d iffe r in lo catio n b S in ce T a b le 8, A p p en d ix I g iv es critical v alú es fo r rejec tio n in th e lo w e r ta il o f the

d istrib u tio n , w e use th e sm a lle r o f T + and T~ as th e test statistic. c F ro m T a b le 8 w ith n = 3 0 , a = .05 an d a tw o -ta ile d test, th e re je c tio n re g ió n is T < 13 7 . d

S in ce T + = 2 4 9 , w e c a n calc ú late n ( n + 1) . 3 0 (3 1 ) T~ = — --T * = — - 2 4 9 = 216 . 2

2

T h e te st statistic is th e sm a lle r o f T + a n d T~ or T = 2 1 6 and H 0 is n o t re jec ted . T h ere is no e v id e n c e o f a d iffe re n c e b e tw een the tw o d istrib u tio n s. 15.25

a

T h e h y p o th e sis to be tested is H 0: p o p u la tio n d istrib u tio n s 1 and 2 are identical H a: th e d istrib u tio n s d iffe r in lo catio n

an d the te st statistic is T, the ra n k sum o f the p o sitiv e (o r n eg ativ e) d ifferen ces. T he ran k s are o b tain ed by o rd e rin g the d ifferen ces acco rd in g to th eir a b so lu te valué. D efin e d¡ to be the d iffe re n c e b e tw een a p a ir in p o p u la tio n s 1 a n d 2 (i.e., x u - x 2¡). T h e d iffe re n c e s, alo n g w ith th e ir ran k s (acc o rd in g to a b so lu te m ag n itu d e), are sh o w n in the fo llo w in g table._______ .7 .3 -.1 .5 .2 .5 .1 di 4 5.5 7 1.5 5.5 3 R a n k | d¡ \ 1.5 T h e ra n k sum fo r p o sitiv e d iffe re n c e s is T + = 26.5 and the ra n k su m fo r n eg ativ e d iffe re n c e s is T~ = 1 .5 w ith n = 7 . C o n sid e r the sm a lle r ran k su m an d d e term in e the a p p ro p ria te lo w e r p o rtio n o f the tw o -ta ile d reje c tio n reg ió n . In d ex in g n = l and a = .05 in T a b le 8, the reje ctio n reg ió n is T < 2 and Hq is rejected . T h e re is a d iffe re n c e in th e tw o p o p u latio n lo catio n s.

153

b T h e resu lts d o n o t ag ree w ith th o se o b tain ed in E x e rc ise 15.16. W e are able to re je c t H0 w ith the m o re p o w erfu l W ilc o x o n test. 15.29

T h e p aire d d ata are given in th e ex ercise. T h e d iffe re n ce s, a lo n g w ith th e ir ran k s

a

el R ank

i

2

3.5

7.5

-1 3.5

1 3.5

3 10

1 3.5

-1 3.5

3 10

-2 7.5

3 10

1

0

2.5

-

K l L et p = P ( A e x c e ed s B fo r a giv en in te rse c tio n ) an d x = n u m b e r o f in te rse c tio n s at w h ich A ex c ee d s B . T h e h y p o th e sis to be tested is H 0 : p = 1/2

v ersu s

H a : p * 1/2

u sin g the sign te st w ith x as th e test statistic. C ritica l v a lu é a p p ro a ch : V ario u s tw o tailed reje c tio n reg io n s are tried in o rd e r to find a reg ió n w ith a ~ .05 . T h e se are sh o w n in th e fo llo w in g table. R ejectio n R eg ió n jc < 1;jc > 10

a .012

x < 2 ,x > 9

.066

* < 3 ; jc > 8

.226

W e ch o o se to reject H o i f x < 2 o r ; t > 9 w ith a = .066 . S in ce x = 8 , H q is not rejected . T h ere is in su fficien t e v id en ce to in d íca te a d iffe re n c e b etw een th e tw o m ethods. p -v a lu e a p p ro a ch : F o r the o b serv e d v alu é x = 8 , c a lc ú late the tw o -ta ile d p -v a lu e : p -v a lu e = 2 P ( x > 8) = 2(1 - . 8 8 7 ) = .226 S ince th e p -v a lu e is g reater th a n .10, H 0 is n o t rejected . b T o use the W ilco x o n sig n ed ran k test, w e use th e ran k s o f th e a b so lu te d iffe re n c e s show n in th e ta b le abo v e. T h e n T + = 5 1 .5 a n d T~ = 1 4 .5 w ith n = 1 1 . In d ex in g « = 11 and a = .05 in T a b le 8, th e lo w er p o rtio n o f th e tw o -ta ile d re je c tio n re g ió n is T < 11 and H0 is no t re je c te d , as in p a rt a. 15.31

a S ince th e e x p e rim e n t h as b een d e sig n e d as a p a ire d e x p e rim e n t, th e re a re th ree tests av ailab le fo r testin g the d iffe re n c e s in the d istrib u tio n s w ith an d w ith o u t im ag ery - (1) th e p a ire d d iffe re n c e t test; (2 ) the sign te st o r (3 ) th e W ilc o x o n sig n e d ra n k test. In o rd e r to use the p a ire d d iffe re n c e t test, th e sc o re s m u st be a p p ro x im a tely n o rm al; since th e n u m b e r o f w ords rec a lle d h as a b in o m ial d istrib u tio n w ith n = 25 and u n k n o w n recall p ro b a h ility , th is d istrib u tio n m ay n o t b e a p p ro x im a te ly norm al, b U sin g the sig n test, the h y p o th e sis to be tested is H 0 : p = 1/2

v ersu s

H a : p > 1/2

F o r the o b se rv e d v alué x = 0 w e c a lc ú la te the tw o -tailed p -v alu e: -

p -v a lu e = 2 P ( x < 0 ) = 2 (.0 0 0 ) = .000

T h e resu lts are h ig h ly sig n ifican t; Ho is re je c te d a n d w e c o n c lu d e th ere is a d ifferen c e in the recall sco re s w ith and w ith o u t im ag ery . U sin g the W ilc o x o n sig n e d -r a n k test, th e d iffe re n c e s w ill all be p o sitiv e ( x = 0 for the sign test), so th at and

154

. n ( n + 1) 2 0 (2 1 ) T + = — -------- = — -— - = 2 1 0 2

and

T ~ = 2 1 0 -2 1 0 = 0

2

In d ex in g n = 2 0 an d a = .01 in T a b le 8, the lo w er p o rtio n o f the tw o -ta iled reje c tio n reg ió n is T < 37 a n d H0 is rejected . 15.35

T h e d a ta w ith c o rre sp o n d in g ra n k s in p a re n th e se s are sh o w n b e lo w. A ge 2 0 - 39

10-19 2 9 (21) 33 (2 9 .5 ) 2 6 ( 1 2 .5 ) 27 39 35 33

(15) (40) (36) (2 9 .5 )

2 9 (21) 36 (3 7 .5 ) 2 2 (5 .5 )

40-59

6 0 -6 9

2 4 (8) 2 7 (1 5 ) 33 (2 9 .5 ) 31 (24)

3 7 (3 9 ) 2 5 ( 1 0 .5 ) 22 (5 .5 ) 33 (2 9 .5 )

28 (18) 29 (21) 34 (34)

21 (3) 2 8 (1 8 ) 2 4 (8) 34 (34)

2 8 (1 8 ) 2 6 ( 1 2 .5 )

36 (37.5) 21(3) 20(1)

30 (23) 34 (34) 2 7 (1 5 )

2 5 ( 1 0 .5 ) 2 4 (8 ) 33 (2 9 .5 )

21(3) 32 (2 5 .5 )

33 (2 9 .5 )

32 (2 5 .5 )

7j = 247.5

T2 = 168

T3 = 2 1 6 .5

7; = 1 8 8

n, = 1 0

n 2 = 10

n3 = 1 0

n 4 = 10

T h e test sta tistic , b ased on the ra n k sum s, is H =

12

3 ( „ + l)

n ( n + 1)

/i,

12

(2 4 7 .5 )2

(1 6 8 )2

(2 1 6 .5 )2

(1 8 8 )2

4 0 (4 1 )

10

10

10

10

- 3 ( 4 1 ) = 2.63

T h e reje c tio n reg ió n w ith a = .01 and k - 1 = 3 d f is based o n the c h i-sq u a re d istrib u tio n , o r H >

= 11.35 . T h e nuil h y p o th e sis is n o t re je c ted . T h e re is no

e v id e n c e o f a d iffe re n c e in location. b

S in ce th e o b se rv e d valué H = 2.63 is le ss th an

%2l0 =

6.25 , the p -v a lu e is g re a ter

than .10. c -d F ro m E x e rc ise 11.60, F = .87 w ith 3 and 3 6 df. A gain, th e p -v a lu e is g reater than .10 and the re su lts are the sam e. 15.39

T h e ra n k s o f the d a ta are show n on the next page.

155

T reatm ent B lock

1

2

3

4

1 2

4 4

3 4

1 1.5 1

4

2 1.5 3 2

3 3 2 3 2.5 3 2 3

7 ; = 2 1 .5

4

5 6 7 8

a

1 1 1

4

4

1

2.5 2 3 2

7; = 3 2

T2 = 8.5

7;=18

4

T h e test statistic is Fr = ,12 , y T - - ? , b { k + \) r bk(lc + 1 ) ^ ' V ’

= 8( ^ 5 )[(32)2 + (8*5)2 +182 + (21-5)2] - 3(8)(5) = 2119 an d th e reje ctio n reg ió n is Fr > ^

05 = 7 .8 1 . 2

H e n ee, Ho is re je c te d and w e co n c lu d e

that th e re is a d iffe re n ce am o n g the fo u r treatm en ts. b

T h e o b se rv e d valué, Fr = 2 1 . 1 9 , ex c e e d s x 2m , p -v a lu e < .005 .

c -e T h e a n aly sis o f v arian ce is p e rfo rm e d as in C h a p te r 11. T h e A N O V A ta b le is show n below . S ou rce T reatm en ts B lo ck s E rro r T o tal

SS 198.34375 2 2 0 .4 6 8 7 5 18.40625 4 3 7 .4 0 6 2 5

df 3 7 21 31

MS 6 6 .1 1 4 5 8 3 3 1 .4 9 5 5 3 6 0 .8 7 6 4 8 8

F 75.43

T h e an aly sis o f v arian ce F test fo r tre a tm e n ts is F = 7 5 .4 3 a n d the a p p ro x im a te p value w ith 3 and 21 d f is p -v a lu e < .005 . T h e resu lt is id en tical to th e p aram etric result. 15.43

T ab le 9, A p p e n d ix I giv es critical v a lú es r0 su ch th a t P ( r s > r 0 ) = a . H en e e , fo r an u p p er-ta ile d test, th e critical v alu é fo r re je c tio n can be re a d d ire ctly fro m the table. a

15.47

rs > . 4 2 5

a; > .6 0 1

b

a T h e tw o v a ria b le s (ratin g and d ista n ce ) a re ra n k ed fro m lo w to hig h , and the resu lts are show n in the fo llo w in g table. V o ter

X

v

V o ter

X

1 2

7.5 4 3 12

7 8 9

6

3 4

3 7 12

5

10

1 8

6

7.5

11

10 11 12

156

y

4

2

11 1 5 9

9 5.5

2

5.5

10

C alcú late

I j c ¡y, = 4 4 2 .5

I x f = 6 4 9 .5

X y f = 6 4 9 .5

n = \2

Ix , =78

X y, = 78

T hen 78: S a = 6 4 9 . 5 ---------= 142.5

S n, = 4 2 2 . 12

12

S yy = 6 4 9 . 5 12

and

b T h e h y p o th esis o f in tere st is H 0: no co rrela tio n v ersu s H a: n eg a tiv e c o rrelatio n . C o n su ltin g T a b le 9 fo r a = .05 , the critical valué o f rs, d en o te d by r 0 is -.4 9 7 . S in ce the valu é o f the test statistic is less th an the c ritical v alu é, th e nuil h y p o th e sis is rejected . T h e re is e v id e n c e o f a sig n ifican t neg ativ e c o rre la tio n b e tw een ratin g and d istan ce. 15.51

R e fe r to E x ercise 15.50. T o te st fo r p o sitiv e c o rre latio n w ith a = .05 , in d ex .05 in T a b le 9 an d the re je c tio n reg ió n is rs > .600 . W e re je c t the nuil h y p o th e sis o f no asso ciatio n a n d co n c lu d e th a t a p o sitiv e c o rrelatio n e x ists b e tw een th e te a c h e r’s ra n k s an d the ran k s o f th e IQ s.

15.55

a

D efin e p = P (r e s p o n s e fo r stim u lu s 1 ex ce e d s th at fo r stim u lu s 2 ) and x = n u m b e r

o f tim es the re sp o n se fo r stim u lu s 1 ex c ee d s th at fo r stim u lu s 2. T h e h y p o th e sis to be tested is H 0 : p = 1/2

v ersu s

Ha : p

1/2

u sin g the sign te st w ith x as th e te st statistic. N o tice th at fo r th is e x e rc ise n = 9 , and th e o b se rv e d v a lu é o f th e test sta tistic is x = 2 . V a rio u s tw o tailed reje c tio n reg io n s are tried in o rd e r to fin d a reg ió n w ith a ~ .05 . T h ese are sh o w n in the fo llo w in g table. a_ .004

R ejec tio n R eg ió n x = 0 ,x = 9 1;jc > 8

.040

x < 2 -,x > 7

.180

jc <

W e ch o o se to re je c t Hq if x < 1 o r x > 8 w ith a = .0 4 0 . S in ce x = 2 , H 0 is not re je c te d . T h e re is in su ffic ie n t e v id e n c e to in d ícate a d iffe re n ce b e tw een th e tw o stim uli. b T h e e x p e rim e n t h as b een d e sig n e d in a p aire d m an n er, and the p a ire d d ifferen c e te st is u sed . T h e d iffe re n c e s are sh o w n below . d,

- .9

- 1 .1

1.5

- 2 .6

-1 .8

-2 .9

T h e h y p o th e sis to be tested is H0

-//2 =0

Ha ://,-//2 * 0

C alcú late

157

- 2 .5

2.5

-1 .4

- = I < = -9 2 = _1 d = ^ L = — = -\m i n '9 , > „ '2 Zd?5: =

——

= 3 7 - 1 4 - 9 .4 0 4 = 3 4 6 7

n- 1 and the test statistic is

T h e reje c tio n reg ió n w ith a = .05 and 8 d f is |/| > 2 .3 0 6 an d H 0 is n o t re je c te d . 15.59

T h e d ata, w ith c o rre sp o n d ing ran k s, are sh o w n in th e fo llo w in g table. A (1) 6.1 (1) 9 .2 ( 1 7 ) 8 .7 (1 2 ) 8 .9 (1 3 .5 ) 7.6 (5) 7.1 (3) 9 .5 ( 1 8 ) 8.3 (9.5) 9 .0 (1.5)

B (2) 9.1 (16) 8.2 (8) 8 .6 ( 1 1 ) 6 .9 (2) 7.5 (4) 7.9 (7) 8.3 (9.5) 7.8 (6) 8 .9 (1 3 .5 )

7¡ = 9 4 T h e d ifferen ce in the b rig h tn e ss lev els u sin g the tw o p ro c e s se s can be te ste d u sin g the n o n p aram etric W ilc o x o n ra n k su m test, o r the p ara m e tric tw o -sa m p le t test. 1 T o test the nuil h y p o th e sis th at the tw o p o p u latio n d is trib u tio n s a re id en tical, calc ú late 7¡ = 1 + 1 7 h------h 1.5 = 9 4 7 T = * , ( n , + « 2 + l ) - 7 ¡ = 9 ( 1 8 + l ) - 9 4 = 77 T h e test statistic is T = m in (7 ¡,r,* ) = 77 W ith n x = n 2 = 9 , the tw o -tailed reje c tio n reg ió n w ith a = .05 is fo u n d in T a b le 7 b to be Tx < 6 2 . T he o b serv ed valué, T = 7 7 , d o e s no t fall in the re je c tio n reg ió n and H 0 is n o t rejected . W e c an n o t co n c lu d e th at th e d istrib u tio n s o f b rig h tn e ss m e a su re m e n ts is d iffe re n t fo r the tw o p ro cesses. 2 T o test the n uil h y p o th esis th at the tw o p o p u la tio n m ean s are id en tical, c a lc ú late _

I * ,;

74.4

_ 1*2, 73.2 jc, = ------ - --------- = 8 .1333

= 8.2667

n, 158

9

( 7 4 .4 )2 2 _ (w, -1 ) J ,2 + ( n 2 - l ) s 2 _

6 2 5 .0 6 -

+ 5 9 9 .2 2 -

(7 3 .2 ) 2 - —

= .8675

16

n\ + n 2 - 2

an d the test statistic is t=

8 .2 7 - 8 .1 3

*1 ~*2 1 1 — + -in, n2

= .304

J . 8675

T h e rejectio n re g ió n w ith a = .05 and 16 d e g ree s o f freed o m is |f| > 1.746 a n d H 0 is not rejected . T h e re is in su fficien t e v id en c e to in d icate a d iffe re n c e in th e av erag e b rig h tn ess m e a su rem en ts fo r the tw o p ro cesse s.N o tic e th at th e n o n p a ra m e tric and param etric tests re ac h the sam e c o n clu sio n s. 15.61

S in ce th is is a p a ire d ex p erim e n t, y o u can c h o o se e ith e r the sign test, the W ilco x o n sig n ed ran k test, o r the p ara m e tric p aire d t test. S in ce the te n d e rize rs have been sco red o n a scale o f 1 to 10, the p aram etric test is n o t ap p licab le. S ta rt b y u sin g the ea sie st o f th e tw o n o n p a ra m e tric tests - the sign test. D efin e p = ^ ( te n d e r iz e r A ex c ee d s B fo r a g iv en cut ) and x = n u m b er o f tim es th at A ex c e e d s B. T h e h y p o th e sis to be tested is H 0 : p = l/ 2

H a : p * 1/2

v ersu s

u sin g the sig n test w ith x as the test statistic. N o tice th at fo r this e x e rc ise n = 8 (th ere are tw o ties), an d the o b se rv e d valué o f the te st statistic is x = 2 . p -v a lu e a p p r o a c h : F o r th e o b se rv e d valué x = 2 , calc ú late p -v a lu e = 2 P ( x < 2 ) = 2 (.1 4 5 ) = .290 S in ce the p -v a lu e is g re a te r th an .10, H0 is n o t rejected . T h ere is in su ffic ie n t e v id e n c e to in d icate a d iffe re n c e b e tw een the tw o ten d erizers. I f you use the W ilc o x o n sig n ed ran k test, y o u will fin d T* = 7 and T~ = 29 w h ich w ill not a llo w reje c tio n o f H 0 at th e 5 % lev el o f sig n ifican ce. T h e re su lts are th e sam e. 15.65

T h e h y p o th esis to be tested is H 0: p o p u la tio n d istrib u tio n s 1 an d 2 are identical H a: th e d istrib u tio n s d iffe r in lo catio n an d the test statistic is T, the ra n k sum o f the p o sitiv e (o r n eg ativ e) d ifferen ces. T h e ran k s are o b tain e d by o rd e rin g the d iffe re n c e s a c c o rd in g to th eir a b so lu te valué. D efin e d¡ to be the d ifferen c e b e tw een a p a ir in p o p u la tio n s 1 an d 2 (i.e., x u - x 2i ). T h e d iffe re n c e s, a lo n g w ith th e ir ran k s (acc o rd in g to ab so lu te m ag n itu d e), a re show n di R a n k | d¡ | di R a n k | d¡ \

-3 1 14.5

-3 1 14.5

-1 1

7

12.5

7

-6 4.5

-9 10.5

159

-1 1 12.5 -2 2

-8 9

-9 10.5

-7 7

7 7

-1

-6

-3

1

4.5

3

T h e rank sum fo r positive d ifferen c es is T * = 14 a n d the ra n k su m fo r neg ativ e d iffe re n c e s is T~ = 106 w ith n = 15 . C o n sid e r th e sm a lle r ran k sum a n d d e term in e th e ap p ro p ria te lo w er p o rtio n o f the tw o -ta ile d reje c tio n reg ió n . In d e x in g n = 15 and a = .05 in T a b le 8, the rejectio n reg ió n is T < 25 an d H 0 is re jec ted . W e co n clu d e that there is a d ifferen c e b etw een m ath an d art scores. 15.69

a - b S in ce the e x p e rim e n t is a c o m p le tely ra n d o m iz e d d esig n , th e K ruskal W a llis H test is used. T he co m b in e d ran k s are sh o w n below . P la n t A B

R anks 9 12

1 19

7 14

C 13 9 D 20 17 9 16 T h e test statistic, b a se d on the ra n k sum s, is

6 18

11 3

5 4 2

15

34 63 33 80

12

H =

/ t ( n + l) 12

Z ^ ~ 3 ( « + l) (3 4 )2 , ( 6 3 ) 2 , (3 3 )2 _ (8 0 ) - 3 ( 2 1 ) = 9.08

20 ( 21)

W ith d f = k - 1 = 3 , the o b se rv e d v alu é H = 9 .0 8 is b e tw e e n

%m5 a n d

X 05 so l^ at

.025 < p -\a lu e < .05 . T h e n uil h y p o th e sis is reje c te d an d w e c o n c lu d e th at th ere is a d iffe re n c e am o n g the fo u r plants. c F ro m E x ercise 11.66, F = 5 .2 0 , and H 0 is re je cte d . T h e re su lts are th e sam e. 15.73

T h e d ata are alread y in ran k form . T h e “ su b sta n tial e x p e rie n c e ” sa m p le is d e sig n a te d as sam p le 1, an d n, = 5 , n 2 = 7 . C alcú late 7, = 1 9 T¡ = « ,( /! , + n 2 + \ ) - T { = 5 (1 3 ) —19 = 4 6 T h e test statistic is T = m in (7 ^ ,7 'l*) = 19 W ith n, = n 2 = 1 2 , th e o n e -ta ile d reje c tio n reg ió n w ith a = .05 is fo u n d in T a b le 7 a to be Tx < 2 1 . T h e o b se rv e d valu é, 7 = 1 9 , falls in th e re je c tio n reg ió n and H 0 is re je c te d . T h ere is su fficien t e v id e n c e to in d icate th at the rev ie w b o a rd c o n sid e rs e x p e rie n c e a p rim e facto r in the selectio n o f th e b est can d id ates.

15.77

T h e d a ta w ith c o rre sp o n d in g ran k s in p a re n th e se s are sh o w n on the n ex t page.

160

T rain in g P erio d s (h o u rs) .5

1.0

8 (9.5) 14(14) 9(11.5)

9(11.5) 7(7)

4(1.5) 6(5)

4(1.5) 7(7)

5 (3.5 )

7(7) 8 (9 .5 )

5 (3.5)

7, = 4 8

T2 = 2 2

T, = 2 3

7, = 1 2

«,=4

n2= 3

*,=4

n4 = 3

12(13)

1.5

-2:0

T h e te st statistic, b ased o n the ran k sum s, is H =

12 n (n + \ 12

'( 4 8 ) : _ ( 2 2 ) : 1 (2 3 ): , (12 )2

- 3 ( 1 5 ) = 7 .4 3 3 3

14(15) T h e reje c tio n reg ió n w ith a = .01 and k - 1 = 3 d f is b ased on the ch i-sq u are d istrib u tio n , o r H > %20l = 11.34 . T he nuil h y p o th e sis is not re je cte d and we co n c lu d e th at th ere is in su ffic ien t e v id en c e to ind ícate a d iffe re n c e in th e d istrib u tio n o f tim es fo r the fo u r groups.

161

9780534463250