Inferential Statistics

INFERENTIAL STATISTICS Hypothesis Testing Testing Meaning of Hypothesis

 A hypothesis hypothesis is a tentative tentative explanation explanation for certain events, phenomena phenomena or  behav behavior iors. s. In statis statistic tical al langu languag age, e, a hypot hypothe hesi sis s is a state statemen mentt of pred predict iction ion or  relation relationshi ship p between between or among variables. variables. Plainly Plainly stated, stated, a hypothe hypothesis sis is the most specific statement statement of a problem. It is a requirement that that these variables variables are related. Furthermore, the hypothesis is testable which means that the relationship between the variables can be put into test on the data gathered about the variables. Null and Alternative Hypotheses

There are two ways of stating stating a hypothesis. A hypothesis hypothesis that is intended for  statistical statistical test is generally stated in the null form. eing the starting starting point of the testing process, it serves as our wor!ing hypothesis. A null hypothesis (H o )  ) expresses the idea of non" non"sig signi nific ficanc ance e of diffe differen rence ce or non" non" signi signific ficanc ance e of relat relatio ionsh nship ip betw between een the variables under study. It is so stated for the purpose of being accepted or re#ected. If the null hypothesis is re#ected, the alternative hypothesis (H a )  ) is accepted. This is the researche researcher$s r$s way of stating stating his research hypothesis hypothesis  in an operational manner. The research hypothesis is a statement of the expectation derived from the theory under study. study. If the related literature literature points to the findings that that a certain technique technique of  teaching for example, example, is effective, effective, we have to assume the same prediction. prediction. This is our  alternative alternative hypothesis. hypothesis. %e cannot cannot do otherwise since since there is no scientific basis basis for  such prediction. &hain of reasoning for inferential statistics '. (ample)s (ample)s** must must be randomly randomly selected selected +. (ample estimate estimate is is compared compared to underlying underlying distribution distribution of the same sie sie sampling distribution -. etermine the probability probability that a sample estimate reflects the population population parameter The four possible outcomes in hypothesis testing Actual Population &omparison /ull 0y 0yp. True /ull 0y 0yp. Fa False 1&I(I2/ )there is no )there is a difference* difference* 3e#ected /ull &orrect ecision Type I error  0yp (alpha alpha id not 3e#ect &orrect ecision Type Ty pe II Error  /ull )Alpha 4 probability of ma!ing a Type I error* 3egardless of whether statistical tests are conducted by hand or through statistical software, there is an implicit understanding that systematic steps are being followed to determine statistical significance. These general steps are described on the following page page and and includ include e '* assump assumptio tions ns,, +* state stated d hypo hypothe thesis sis,, -* re#ect re#ection ion criter criteria ia,, 5*

computation of statistics, and 6* decision regarding the null hypothesis. The underlying logic is based on re#ecting a statement of no difference or no association, called the null hypothesis. The null hypothesis is only re#ected when we have evidence beyond a reasonable doubt that a true difference or association exists in the population)s* from which we drew our random sample)s*. 3easonable doubt is based on probability sampling distributions and can vary at the researcher7s discretion. Alpha .86 is a common benchmar! for reasonable doubt. At alpha .86 we !now from the sampling distribution that a test statistic will only occur by random chance five times out of '88 )69 probability*. (ince a test statistic that results in an alpha of .86 could only occur by random chance 69 of the time, we assume that the test test stati statist stic ic resul resulte ted d beca because use there there are true true diffe differe renc nces es betw between een the the popul populat ation ion parameters, not because we drew an extremely biased random sample. %hen %hen learn learning ing statis statistic tics s we gener generall ally y conduc conductt stati statist stica icall tests tests by hand. hand. In these these situations, we establish before the test is conducted what test statistic is needed )called the !riti!al value * to claim statistical significance. (o, if we !now for a given sampling distribution that a test statistic of plus or minus '.:; would only occur 69 of the time randomly, any test statistic that is '.:; or greater in absolute value would be statistically significant. In an analysis where a test statistic was exactly '.:;, you would have a 69 chance of being wrong if you claimed statistical significance. If the test statistic was -.88, statistical significance could also be claimed but the probability of being wrong would be much less )about .88+ if using a +"tailed test or two"tenths of one percent< 8.+9*. oth .86 and .88+ are !nown as alpha< the probability of a Type I error. %hen conducting statistical tests with computer software, the exact probability of a Type I error is calculated. It is presented in several formats but is most commonly reported as "p #" or "Sig$" or "Signif$" or "Signifi!an!e$" =sing >p ?> as an example, if a priori you established a threshold for statistical significance at alpha .86, any test statistic with significance at or less than .86 would be considered statistically significant and you would be required to re#ect the null hypothesis of no difference. The following table lin!s p values with a benchmar! alpha of .86@

% # Alpha %ro&a&ility of Ty Type I Error .86 .86 69 ch chance difference is is no not significant .'8 .86 '89 ch chance di difference is is no not significant .8' .86 '9 ch chance difference is is no not significant .:; .86 :;9 ch chance di difference is is no not significant

Final 'e!ision (tatistically significant /ot statistically significant (tatistically significant /ot statistically significant

(teps to 0ypothesis Testing Testing 0ypothesis testing is used to establish whether the differences exhibited by random samples can be inferred to the populations from which the samples originated. eneral Assu)ptions •

Population is normally distributed

•

3andom sampling

•

utually exclusive comparison samples

•

ata characteristics match statistical technique

For interval  / ratio data use t"tests, Pearson correlation, A/2BA, regression 

For nominal / ordinal  data use ifference of proportions, chi square and related measures of association State the Hypothesis 

/ull 0ypothesis )0o*@ There is no difference between CCC and CCC.  Alternative  Alternative 0ypothesis 0ypothesis )0a*@ )0a*@ There There is a differen difference ce between between CC and and CC. /ote@ The alternative hypothesis will indicate whether a '"tailed or a +"tailed test is utilied to re#ect the null hypothesis. 0a for '"tail tested@ The CC of CC is greater )or less* than the CC of CC. Set the Re*e!tion Criteria This determines how different the parameters andDor statistics must be before the null hypothesis hypothesis can be re#ected. This >region >region of re#ection> re#ection> is is based based on alpha alpha ) * "" the error associated with the confidence level. The point of re#ection is !nown as the critical value. Co)pute the Test Statisti! The collected data are converted into standardied scores for comparison with the critical value. 'e!ide Results of Null Hypothesis If the test statistic equals or exceeds the region of re#ection brac!eted by the critical value)s*, the null hypothesis is re#ected. In other words, the chance that the difference exhibited between the sample statistics is due to sampling error is remote""there is an actual difference in the population.

Eet us consider an experiment involving two groups, an experimental group and a control group. The experimenter li!es to test whether whether the treatment )values clarification lessons* will improve the self"concept of the experimental group. The same treatment is not given given to the control control group. group. It is presume presumed d that any differenc difference e betwee between n the two groups after the treatment can be attributed to the experimental treatment with a certain degree of confidence. The hypothesis for this experiment can be stated in various ways@ a* /o existence existence or existence existence of a differen difference ce between between groups groups 0o@

There There is no significa significant nt differenc difference e in self"con self"concept cept between between the group exposed to values clarification lessons and the group not exposed to the same. 0a@ The self"con self"concep ceptt of the group exposed exposed to values values clarifica clarification tion lessons lessons differ significantly that of the other group. b* /o existence existence or existence existence of of an effect effect of the treatment 0o@ There is no significant effect of the values values clarification lessons on the self" concept of the students. 0a@ Balues clarification lessons have no significant effect on the self"concept of students. c* /o existen existence ce of relation relationship ship betwee between n the variabl variables es

0o@ The self"concept self"concept of the students students is not significantly significantly related to the values clarification lessons conducted on them. 0a@ The self"concept self"concept of the students students is not significantly significantly related to the values values clarification lessons they were exposed to. %ara)etri! Test Test What are the parametric tests? 

The parametric parametric tests are tests that that require require normal distributi distribution, on, the levels levels of  measur measureme ement nt of which which are are expre expresse ssed d in an inter interva vall or ratio ratio data data.. The The follow following ing parametric tests are@ t"test for one population mean compared to a sample mean t" test for Independent (amples t" test for &orrelated (amples " test, for Two (ample eans " test for 2ne (ample roup F" test )A/2BA* )A/2BA* 3 )Pearson Product oment &oefficient of &orrelation* G4aHbx )(imple Einear 3egression Analysis* G4

b0

H

b1 x1

H

b2 x 2

H... H

bn x n

)ultiple 3egression Analysis* Co)paring a %opulation Mean to a Sa)ple Mean ( T+test

1xample '@ &ompare the mean age of incoming students to the !nown mean age for all previ previous ous incomi incoming ng stude students nts.. A rand random om sampl sample e of -8 incomi incoming ng colleg college e fresh freshmen men revealed the following statistics@ mean age ':.6 years, standard deviation ' year. The college database shows the mean age for previous incoming students was '.

Solving &y the Step,ise Method

significant nt differen difference ce between between the mean age of past college college I$ %ro&le)- Is there a significa students and the mean age of current incoming college studentsJ II$ Hypothesis

0o@ There is no significant significant difference difference between the mean age of past past college students and the mean age of current incoming college students. 0o@

 x´ 1= x´ 2

0a@ There is a significant difference between the mean age of past college students and the mean age of current incoming college students.

III$ Level of Signifi!an!e Set the Re*e!tion Criteria (ignificance level .86 alpha, +"tailed test egrees of Freedom 4 n"' or +: &ritical value from t"distribution 4 +.856

I.$ Statisti!s Co)pute the Test Statisti!

¿

 x´ −

s √ n −1 t =¿

t =

19.5

−18

1

√ 30 −1 t =

1.5 1

√ 29 t =

1.5 .186

t 

4 .8;6

.$ 'e!ision Rule- If the t" computed value is greater than or beyond the t"tabular D  H 0 critical value, re#ect . .I Con!lusion-

iven that the test statistic ).8;6* exceeds the critical value )+.856*, the null hypothesis is re#ected in favor of the alternative. There is a statistically significant difference between the mean age of the current class of incoming students and the mean age of freshman students from past years. In other words, this year7s freshman class is on average older than freshmen from prior  years. If the results had not been significant , the null hypothesis would not have been re#ected. This would be interpreted as the following@ There is insufficient evidence to conclude there is a statistically significant difference in the ages of current and past freshman students.

What is the t-test for independent samples? 

The t"test is a test of difference between two independent groups. The means are  x´ 1  x´ being compared  against 2 .

When do we use the t-test for independent samples? 

The t"test for independent samples is used when we compare means of two independent groups. %hen the distribution is normally distributed, (! 4 8 and Ku 4 .+;6. we use interval or ratio data. the sample is less than -8.

Why do we use the t-test for independent sample? 

The t"test is used for independent sample because it is more powerful test compared with other tests of difference of two independent groups. How do we use the t-test for independent samples?  n1 +¿ n −2 2

SS1 + SS2

[

t4

¿ ¿ 1

+

1

n1 n2

¿ √ ¿  x´ − x´ ¿ 1

]

2

%here@ t 4 the t test  x´ 1 4 the mean of group '  x´ 2

 4 the mean of group +

SS 1

 4 the sum of squares of group '

SS 2

4 the sum of squares of group +

n1 n2

 4 the number of observations in group '

 4 the number of observations in group +

or 

n1 +¿ n −2 2

( n −1 ) ( s 1 ) 2 + ( n − 1 ) ( s 2 ) 2 1

t4

1

[

¿ ¿ 1

+

 1

n1 n2

¿

√ ¿

 x´ 1− x´ 2

¿

]

%here@   Typeequationhere . t 4 the t test  x´ 4 the mean of group ' or sample '  x´ 2  4 the mean of group + or sample + 1

S1

 4 the standard deviation of group ' or sample '

S2 n1 n2

4 the standard deviation of group + or sample +  4 the number of observations in group '

 4 the number of observations in group +

students in E/a)ple 0$ The following are the scores of '8 male and '8 female A spelling. Test the null hypothesis that there is no significant difference between the performance of male and female A students in the rest. =se the t"test .86 level of significance. ale )

x1

*

Female )

'5 ' 'L '; 5 '5 '+ '8 : 'L

x2

*

'+ : '' 6 '8 L + ; '-

Solution-

ale  x 1

Female  x

 2

 x 2

1

x

 2 2

'5 ':; '+ '55 ' -+5 : ' 'L +: '' '+' '; +6; 6 +6 5 '; '8 '88 '5 ':; : '+ '55 L 5: '8 '88 + 5 : ' ; -; 'L +: '';: CCCCCC CCCCCCC CCCCCCC CCCCCCC 2 2  x 1  x 2 Σ X 2  4 L-  4'-' Σ X 1 4 ':' 4L

∑ n1

∑

4 '8

n2

4 '8

 x´ 1

 x´ 2

 4 '-.'

4 L. n1 +¿ n −2 2

( n −1 ) ( s 1 ) 2 + ( n − 1 ) ( s 2 ) 2 1

1

¿ ¿

[

t4

1

+

 1

n1 n2

¿

]

√ ¿

 x´ 1− x´ 2

¿ 13.1 −7.8

√[

 t 4

174.9 + 129.6 10 + 10 −2

][

1 10

+

1 10

]

5.3

t

4

√[

304.5 18

](

.1 + .1 )

5.3

t

4 √ ( 16.92 ) ( .2 ) 5.3

t

4

√ 3.384 5.3

t 4

1.8395

t 1 2$33 Solving &y the Step,ise Method I$

%ro&le)Is there a significant difference between the performance of the male and female students in spellingJ

II$

Hypothesis H 0 :  H 0 :  H a :  H a :

There is no significant difference between the performance of the male and the female A students in spelling.  x´ 1=´  x 2 There is a significant difference between the performance of  the male and the female A students in spelling.  x´ 1=´  x 2

III$ Level of Signifi!an!eα =.05 df 4

n1 + n2

"+

= '8 H '8 M + = ' t..86 = +.'8' t"tabular value at .86

t"test for two independent (amples

I.$

Statisti!s-

.$

'e!ision Rule-

.I$

Con!lusion(ince the t"computed value of +. is greater than t"tabular value of +.'8' at .86 level of significance with ' degrees of freedom, the null hypothesis is re#ected in favor of the research hypothesis. This means that there is a significant difference between the performance of the male and female A students in spelling, implying that the male students performed better than the female students considering that the meanDaverage score of the male students of '-.' is greater compared to the average score of female students of only L..

 If the t" computed value is greater than or beyond the t"  H 0 tabular D critical value, re#ect .

1xample +. Two groups of experimental rats were in#ected with tranquilier at '.8 mg. And '.6 mg dose respectively. The time given in seconds that too! them to fall asleep in hereby given. =se the t"test for independent samples at .8' to test null hypothesis that the difference in dosage has no effect on the length of time it too! them to fall asleep.

'.8 mg dose@ :., '-.+, ''.+, :.6, '-.8, '+.', :., '+.-, L.:, '8.+, :.L. '.6 mg dose@ '+.8, L.5, :., ''.6, '-.8, '+.6, :., '8.6, '-.6.

(olution@ )'.8 mg dose*

)'.6 mg dose*

N'

N'+

N+

:. '-.+ ''.+ :.6 '-.8 '+.' :. '+.L.: '8.+ :.L

:;.85 'L5.+5 '+6.55 :8.+6 ';:.88 '5;.5' :;.85 '6'.+: ;+.5' '85.85 :5.8:

'+.8 L.5 :. ''.6 '-.8 '+.6 :. '8.6 '-.6 ΣN+ 4'88 n +4 :

N++ '55.88 65.L; :;.85 '-+.+6 ';:.88 '6;.+6 :;.85 ''8.+6 '+.+6 ΣN++ 4 ''58.5

ΣN' 4 ''.L  x´ 1=11.11

ΣN'+ 4'-8:.+6

/' 4 '' N' 4'8.L: SS 1

1−¿

 4 " Σ

( ∑  x ¿

2

2

1

 X ¿

SS 2

n1

 4 Σ X   "

2

2

118.7 ¿

SS 1

100 ¿

¿ ¿ ¿

4 '-8:.+6"

4 ''58.5 M

28.37

29.73 SS

 4

2

n1 +¿ n −2 2

SS1 + SS2

t4

[

¿ ¿ 1

+

1

n1 n2

¿

]

√ ¿

 x´ 1− x´ 2

¿

10.79 −11.11

√[

4

28.37 + 29.73 11+ 9 −2

][

1

+

4

√[

4

−0 . 32 √ ( 3 . 23 ) (. 2)

58 . 10 18

](

. 09 + . 11)

−0 . 32 4 √ . 646

−0 . 32 . 8037

t 4 "8.58 Solving &y the Step,ise Method

1

11 9

−0 . 32

4

2 2

]

4

¿ ¿ ¿

(∑  x n2

2

2

¿

- Is there a significant difference brought about by the dosages on the length of time it too! for the rats to fall asleep

I$

%ro&le)

II$

Hypothesis-

 H 0 :

  There is no significant differences brought about by the

dosages on the length of time it too! the rats to fall asleep.  H a : There is a significant difference brought

about by the

dosages on the length of time it too! for the rats to fall asleep. III$

Level of Signifi!an!e-

α =.01

n1 + n2

df 4

"+

= '' H : M + = ' t.05 = "+. t"test for two independent samples

I.$

Statisti!s-

.$

'e!ision Rule- If the t" computed value is greater than or  H 0 tabular D critical value, re#ect .

.I$

critical Con!lusion- (ince the t"computed value of D".58D is within the value of D"+.D at .8' level of significance with ' degrees of freedom, the null hypothesis is not re#ected. This means that no significant difference was brought about by the dosages on the length of time it too! for the rats to fall asleep.

beyond the t"

E/a)ple 4. To find out whether a new serum would arrest leu!emia, '; patients on the advanced stage of the disease were selected. 1ight patients received the treatment and eight did not. The survival was ta!en from the time the experiment was conducted.

/o treatment N'+

N'

∑ x

%ith Treatment N++

+

N

2$0

5$50

5$2

06$75

4$2

08$25

9$0

27$80

4$8

:$88

9$8

29$88

2$3

6$35

5$7

20$07

2$0

5$50

4$:

09$20

0$2

0$55

5$4

03$5:

0$3

4$25

9$2

26$85

0$:

4$70

4$:

09$20

 4 '.'

1

n1

 x´ 1

2 1

Σ X 

4 55.':

∑ x

4  4 +.+

Solution-

(olve for

∑ x

1

,

2

2

2

 4 -;+ n2

4

 x´ 2

4 5.6+

Σ X 1 and

 x´ 1

Σ X 2  4 ';6.L;

li!ewise for

∑ x

2

2

, Σ X 2

and

 x´ 2 .

 Then solve for the sum of the squares of group ', and sum of squares of group +

SS 1

 and

SS 2 .

SS 1

(∑  x

2 1

 4 " Σ X  "

¿ ¿ ¿

¿2

(∑  x ¿

2

SS 2

n1

18.7

4 55.': "

1

2

¿

2 2

 4 Σ X   "

36.2 ¿

4 ';6.L; M

4 55.': M 58.:6

2

n2

2

¿ ¿ ¿

4';6.L; M ';-.8

SS 1=¿

SS 2

 44 3.24

4 0$:7

n1 +¿ n −2 2

SS1 + SS2

t4

[

¿ ¿ 1

+

]

1

n1 n2

¿

√ ¿

 x´ 1− x´ 2

¿ − 4.52

2.26

4

√[

+ 8 + 8 −2

3 . 24 1 . 96

][ + ] 1

1

8

8

−2.26

4

√[ ][ ]

4

−2.26 √ ( .37 ) ( .25 )

5.2

2

14

8

−2.26

4 4

√ .0925

−2.26 .3041

t 4 -7.43 Solving &y the Step,ise Method I$

%ro&le)- %ill the new serum arrest leu!emia for the  reached the advanced stageJ

II$

Hypothesis-

patients who had

 H 0 :

The new serum will not arrest the leu!emia of the  patients who

had reached the advanced stage of the disease.  H 1 : The new serum will arrest the leu!emia of the  patients who had reached an advanced stage of the disease. III$

Level of Signifi!an!eα =.05 df 4

t .05  I.$

Statisti!s-

n1 + n2

"+

=  H  M + = '5 = "+.'56

t"test for independent samples

.$ 'e!ision Rule-  If the t"computed value is greater than or beyond the critical or tabular   H 0. value, re#ect .I$

The t"computed value is D"L.5-D which is beyond the critical value of  Con!lusionsD"+.'56D at .86 level of significance with '5 degrees of freedom. The null hypothesis therefore is disconfirmed in favor of the research hypothesis. This means that the new serum will arrest the advanced stage of leu!emia. What is the t-test correlated samples? 

The t"test for correlated samples is another parametric test applied to one group of samples. It can be used in the evaluation of a certain program or treatment. (ince this is another parametric test, conditions must be met li!e the normal distribution and the use of interval or ratio data. When do we use the t-test for correlated samples? 

The test for correlated samples is applied when the mean before and the mean after are being compared. The pretest )mean before* is measured, the treatment of the intervention is applied and then the posttest )mean after* is li!ewise measured. Then the two means )pretest vs. the posttest* are compared. Why do we use the t-test for correlated samples? 

The t"test for correlated samples is used to find out if a difference exists between the before and after means. If there is a difference in favor of the posttest then the treatment or intervention is effective. 0owever, if there is no significant difference then the treatment is not effective. This is the appropriate test for evaluation of government programs. This is used in an experimental design to test the effectiveness of a certain technique or  method or program that had been developed. How do we use the t-test for correlated samples? 

The formula is

2

 D ¿

¿

¿ t4

n n ( n −1)

¿  D −¿ √ ¿ ´  D ¿ 2

´  D 4 the eman difference between the pretest

%here@

and the posttest. Σ D

2

4 the sum of the squares of the difference

between the pretest and the posttest. Σ 4 the summation of the difference between

the pretest and the posttest. n 4 the sample sie E/a)ple 0$ An experiment study was conducted on the effect of programmed materials in 1nglish on the performance of +8 selected college students. efore the program was implemented the pretest was administered and after 6 months the same instrument was used to get the posttest result. The following is the result of the experiment. =se α  at .86 level.

Pretest N' +8 -8 '8 '6 +8 '8 ' '5 '6 +8 ' '6 +8 ' 58 '8 '8 '+ +8

Posttest D

+ '



+6 -6 +6 +6 +8 +8 ++ +8 +8 '6 -8 '8 +6 '8 56 '6 '8 ' +6

"6 "6 "'6 "'8 8 "'8 "5 "; "6 6 "'+ 6 "6  "6 "6 8 "; "6

N

CCCCCCCCCCCCC

Σ 4 "'

´  D 4

−81 20

2

+6 +6 ++6 '88 8 '88 '; -; +6 +6 '55 +6 ' ;5 +6 +6 8 -; +6 CCCCCCCCCCCCC C 

Σ D

2

 4 :5

´  D  4 " 5.86

What are the steps in using the t-test for correlated samples? 

(ubtract the difference between the two observations before and after the treatment. Find the summation of the difference algebraically Σ )consider the H and  M sign*

´  D  the mean difference, Σ

&ompute

n (quare the difference between the before and after observation and get 2 the summation that is Σ D etermine the number of observation that is small letter n. If the t"compound value is greater than or beyond the critical value disconfirm the null hypothesis and confirm the research or alternative hypothesis. 2

 D ¿

¿

¿

n n ( n −1 )

¿  D −¿ √ ¿ ´  D ¿

t4

2

2

−81 ¿ ¿ ¿

4

4

20

( −1 ) ¿ 947−¿ √ ¿ −4.05 ¿

20 20

√

−4.05 947 −328.05 20 ( 19)

−4.05

4

√

618.95 380

−4.05

4

√ 1.6288

−4.05

4

1.2762

t 1 +4$06 Solving &y the Step,ise Method I$

Is there a significant difference between the pretest and the %ro&le)posttest on the use of programmed materials in 1nglishJ

II$

Hypotheses H 0 :

  There is no significant difference between the pretest and

the posttest, or the use of the programmed materials did not affect the students$ performance in 1nglish.  H 1 :

III$

 The posttest result is higher than the pretest result.

Level of Signifi!an!e-

α =.05

df 4 n"' = +8"' = '5 t.05 =  "'.L+:

t"test for correlated samples

I.$

Statisti!s-

.$

'e!ision Rule-

.I$

Con!lusion @ The t"compound value of -.'L is beyond the t"critical value of D"'.L-D at .86 level of significance with ': degrees of  freedom. The null hypothesis is therefore re#ected in favor of  the research hypothesis. This means that the posttest result is higher than the pretest result. It implies that the use of the programmed materials in 1nglish is effective.

If the t"computed value is greater than or   H 0. beyond the critical value, re#ect

What is the z-test? 

The "test is another test under parametric statistics which requires normality of  distribution. It uses the two population parameters µ and σ. It is used to compare two means, the sample mean, and the perceived population mean. It is also used to compare the two sample means ta!en from the same population. It is used when the samples are equal to or greater than -8. The " test can be applied in two ways@ the 2ne"(ample ean Test and the Two"(ample ean Test.

The tabular value of the "test at .8' and .86 level of significance is shown below.

Test

Eevel of (ignificance .8'

.86

2ne"tailed

O +.--

O '.;56

Two"tailed

O +.6L6

O '.:;

What is the z-test for one sample group? 

The "test for one sample group is used to compare the perceived ´ population mean µ against the sample mean,  X  When is the z-test for a one-sample group? 

The one"sample group test is used when the sample is being compared to the perceived population mean. 0owever if the population standard deviation is not !nown the sample standard deviation can be used as a substitute.

Why is the z-test used for a one-sample group? 

The "test is used for a one"sample group because this is appropriate for  ´ comparing the perceived population mean µ against the sample mean  X  . %e are interested if significant difference exists between the population µ against the sample mean. For instance a certain tire company would claim that the life span odd its product will last +6,888 !ilometers. To chec! the claim, sample tires will ´ be tested by getting sample mean  X  . How do we use the z-test for a one-sample group? 

The formula is

 x´ −¿

¿

4

√ n

¿ ¿

%here@

´  X 

4 sample mean

µ 4 hypothesied value of the

population mean σ 4 population standard deviation / 4 sample sie

E/a)ple 0$ The A& company claims that the average lifetime of a certain tire as at least +,888 !m. To chec! the claim, a taxi company puts 58 of these tires on its taxis and gets a mean lifetime of +6,6;8 !m. %ith a standard deviation of ',-68 !m, is the claim trueJ =se the "test at .86. What are the steps in using the z-test for a one sample group? 

(olve for the mean of the sample and also the standard deviation if the population σ is not !nown. (ubtract the population from the sample mean and multiply it by the square root of n sample. ivide the result from step + by the population standard deviation on sample standard deviation if the population σ is not !nown.

the

&ompare the result in the table of the tabular value of the "test at .8' and .86 level of significance. Eevel of (ignificance .8' .86 O +.-O '.;56 O +.6L6 O '.:; ɑɑ

Test 2ne tailed Two tailed

If the "compound value is greater than or beyond the critical value, re#ect the null hypothesis and confirm the research hypothesis.

I$

%ro&le)- Is the claim true that the average lifetime of a certain tire is at least +,888 !mJ

II$

Hypotheses H 0 :  H 1 :

III$

The average lifetime of a certain tire is

+,888 !m.

The average lifetime of a certain tire is not +,888 !m. Level of Signifi!an!eɑ

4 .86 ± '.;56 4

I.$

Statisti!s-

"test for a one"tailed test Co)putation;   x´ −¿

¿

4

4 4 4

√ n

¿ ¿

(25,560− 28,000) √ 40 1,350

(−2,440)( 6.32) 1,350

−15,420.8 1,350

 4 "''.5+ .$

if the  computed is greater than or beyond the  'e!ision Ruletabular value, re#ect the 0 o.

.I$

(ince the  computed of "''.5+ is beyond the critical Con!lusionvalue of "'.;56 at .86 level of significance the research hypothesis is re#ected which means that the average lifetime of a certain tire is not +,888 !m.

What is the z-test for a two-sample mean test? 

The "test for a two"sample mean test is another parametric test used to compare the means of two independent groups of samples drawn from a normal population if  there are more than -8 samples for every group. When do we use the z-test for two sample mean? 

The "test for two"sample mean is used when we compare the means of samples of independent groups ta!en from a normal population.

Why do we use the z-test? 

The "test is used to find out if there is a significant difference between the two populations by only comparing the sample mean of the population. How do we use the z-test for a two-sample mean test? 

The formula is

 x´ 1−´  x 2

4

√

2

2

 s1  s 2 + n1 n2

where@  x´ 1

4 the mean of sample '

 x´ 2

4 the mean of sample +

2

s1

4 the variance of sample '

2

s2

4 the variance of sample +

n1

4 sie of sample '

n2

4 sie of sample +

1xample '.

An admission test was administered to incoming freshmen in the colleges of /ursing and Beterinary edicine with '88 students each college  x´ 1 randomly selected. The mean scores of the given samples where 4  x´ 2

:8 and

4 6 and the variances of the test scores were 58 and -6

respectively. Is there a significant difference between the two groupsJ =se .8' level of significance. What are the steps in solving for the z-test for two-sample mean? 

&ompute the sample mean of group ', +,

 x´ 1

 and also the sample mean of group

 x´ 2

&ompute the standard deviation of group ' group +

SD2

(quare the square the

SD1

  and standard deviation of 

. SD1

SD2

2

s 1   and also

of group ' to get the variance of group ' 2

 of the group + to get the variance of group +

etermine the number of observation in group ' observation in group +

n2

n1

s2 .

 and also the number of 

.

&ompare the  computed value from the " tabular value at certain level of  significance. Eevel of (ignificance Test 2ne"tailed Two"tailed

.8' O +.-O +.6L6

.86 O '.;56 O '.:;

If the computed "value is greater than or beyond the critical value re#ect the null hypothesis and do not re#ect the research hypothesis.

I$

%ro&le)-  Is there a significant difference between the two groupsJ

II$

Hypotheses H 0 :

 x´ 1=´  x 2  x´ 1 ≠ ´  x2

 H 1 :

III$

Level of Signifi!an!e α   4 .8'

 4 O +.6L6

do not re#ect re#ect

 H 0

  re#ect

"+.6L6   I.$

"'

8

"'.:;

H' 8H+.6L6 H'.:;

Statisti!s- "test for two"tailed test  x´ 1−´  x 2

4

¿

√

2

2

 s1  s 2

+

n1 n2

90 −85

√

40 100

+

35 100

5

4

√

75 100 5

1

√ .75 5

1

.866

 4 6.LL5

 H 0

 H 0

.$

'e!ision Rule-   If the "computed value is greater than or beyond the  4tabular value, re#ect the null hypothesis..

.I$

Con!lusion-   (ince the "computed value of 6.LL5 is greater than the " tabular value of +.6L6 at .8' level of significance, the research hypothesis is re#ected which means that there is a significant difference between the two groups. It implies that the incoming freshmen of the &ollege of /ursing are better than the incoming freshmen of the &ollege of Beterinary edicine.

What is the F-test? 

The F"test is another parametric test used to compare the means of two or more groups of independent samples. It is also !nown as the analysis of variance, )A/2BA*. The three !inds of analysis of variance are@ one"way analysis of variance two"way analysis of variance three"way analysis of variance The F"test is the analysis of variance )A/2BA*. This is used in comparing the means of two or more independent groups. 2ne"way A/2BA is used when there is only one variable involved. The two"way A/2BA is used when two variables are involved@ the column and the row variables. The researcher is interested to !now if there are significant differences between and among columns and rows. This is also used in loo!ing at the interaction effect between the variables being analyed. Ei!e the t"test, the F"test is also a parametric test which has to meet some conditions, and the data to be analyed if they are normal are expressed in interval or ratio data. This test is more efficient than other tests of difference. Why do we use the F-test? 

The F"test is used to find out if there is a significant difference between and among the means of the two or more independent groups. When do we use F-test? 

The F"test is used when there is normal distribution and when the level of  measurement is expressed in interval or ratio data #ust li!e the t"test and the "test. How do we use the F-test? 

To get the F computed value, the following computations should be done. 2

CF =

(¿)

 N 

TSS is the total sum of squares minus the &F, the correction factor. .A TA+FACT>R AN>.A ,ith Signifi!ant Intera!tion TEACHER FACT>R (Colu)n

ethod of  Teaching FA&T23 ' )row* Total ethod of  Teaching FA&T23 + )row* Total ethod of  Teaching FA&T23 )row* Total Total

¿¿2 ¿

1882 ¿

¿ ¿ ¿ ¿

&F 1

SS T 

1

 A



&

58 5' 58 -: - ':

68 68 5 5 56 +5'

58 5' 58 - - ':L

Σ4 ;-;

58 5' -: - - ':;

56 5+ 5+ 5' 58 +'8

68 5; 555+ ++5

Σ4 ;-8

58 5-' -: - +8' 6:6

58 56 55 55 5+'; ;;L

58 5' 5' -: - ':: ;+8

2

 1 LL8:.5+

2

40

+ 412+ &+ 392 + 382 + &F

4 L:+'"LL8:.5+ 4 68.6

Σ4;';

4 '+

2

198 ¿

¿

2

196 ¿

¿

2

201 ¿

¿

SS #

2

241 ¿

 1 L:+'+

¿

2

210 ¿

¿ ¿ ¿ ¿ ¿ ¿ ¿

216

¿ 197

¿ 224

¿ 199

¿ ¿ ¿ ¿ ¿ ¿

2

¿

2

¿

2

¿

2

¿

4 L:+' M L:8. 4 '+:.+ 2 620 ¿ ¿ 2 667 ¿ +¿ SS '  1 595 ¿2 +¿  + CF ¿ ¿ 1183314

1

15

 +6368:$52

1 063$03 2 616 ¿ ¿ 2 630 ¿ +¿ 2 636 ¿ +¿  ? CF ¿ SSr =¿ 1180852

1

15

 +LL8:.5+

4 LL.L+-.5L M LL8:.5+   4'5.86 SS ' r= SSt − SS # −SS ' − SSr 4 68.6 M '+:.+ M 'L.' M '5.86 4 'L.'6 The degrees of freedom for the different parts of the problem are@

df t 

4 /"' 4 56"' 4 55

df #

  4 !)n"'* 4 :)6"'* 4 :)5* 4 -;

df '

  4 )c"'* 4 )-"'* 4 +

df r

4 )r"'* 4 )-"'* 4 +

df ' ∗r

4 )c"'*)r"'* 4

)-"'*)-"'* 4 )+*)+* 4 5 AN>.A TA