Aczel Solution 005

Chapter 05 - Sampling and Sampling Distributions

CHAPTER 5

SAMPLING AND SAMPLING DISTRIBUTIONS 5.1.

Parameters are numerical measures of populations. Sample statistics are numerical measures of samples. An estimator is a sample statistic used for estimating a population parameter.

5-2.

x = 97.9225 s = 51.8303 s2 = 2,686.38

5-3.

(estimate of  ) (estimate of  ) (estimate of  2—the population variance)

ˆ = x/n = 5/12 = 0.41667 p

(5 out of 12 accounts are over $100.) 5-4.

x = 2121.667 s = 1737.714 Basic Statistics from Raw Data Measures of Central tendency Mean 2121.6667

Median

Measures of Dispersion If the data is of a Sample Population Variance 3019651.52 St. Dev. 1737.71445

5.5.

average price = 4.367

standard deviation = 0.3486

Basic Statistics from Raw Data Measures of Central tendency Mean 4.3676471

Median

Measures of Dispersion If the data is of a Sample Population Variance 0.12154412 St. Dev. 0.34863178

5-1

Chapter 05 - Sampling and Sampling Distributions

5.6.

ˆ = x/n = 11/18 = 0.6111, where x = the number of users of the product. p

5.7.

We need 25 elements from a population of 950 elements. Use the rows of Table 5-1, the rightmost 3 digits of each group starting in row 1 (left to right). So we skip any such 3-digit number that is either > 950 or that has been generated earlier in this list, giving us a list of 25 different numbers in the desired range. The chosen numbers are: 480, 11, 536, 647, 646, 179, 194, 368, 573, 595, 393, 198, 402, 130, 360, 527, 265, 809, 830, 167, 93, 243, 680, 856, 376.

5.8.

We will use again Table 5-1, using columns this time. We will use right-hand columns, first 4 digits from the right (going down the column): 4,194 3,402 4,830 3,537 1,305.

5.9

We will use Table 5-1, sets of 2 columns using all 5 digits from column 1 and the first 3 digits from column 2, continuing by reading down in these columns. Then we will continue to the set: column 3 and first 3 digits column 4. We skip any numbers that are > 40,000,000. The resulting voter numbers are: 10,480,150 22,368,465 24,130,483 37,570,399 1,536,020.

5.10.There are 7 x 24 x 60 minutes in one week: (7)(24)(60) = 10,080 minutes. We will use Table 5-1 Start in the first row and go across the row, then to the next row (left to right using all 5 digits in each set), discarding any of the resulting 5-digit numbers that are > 10,080. The resulting minute numbers are: 1,536 2,011 6,243 7,856 6,121 6,907 5-11.

A sampling distribution is the probability distribution of a sample statistic. The sampling distribution is useful in determining the accuracy of estimation results.

5.12.Only if the population is itself normal. 5-13.

E  X    = 125

SE  X    / n  20/ 5 = 8.944

5.14.The fact that, in the limit, the population distribution does not matter. Thus the theorem is very general. 5.15.When the population distribution is unknown. 5.16.The Central Limit Theorem does not apply.

5-2

Chapter 05 - Sampling and Sampling Distributions

5.17. Pˆ is binomial. Since np = 1.2, the Central Limit Theorem does not apply and we cannot use the normal distribution. 5.18.  = 1,247

 2 = 10,000

n = 100



1,230  1,247 



100 / 10

P( X < 1,230) = P  Z 

 = P(Z < –1.7) = .5 – .4554 = 0.0446 

Sampling Distribution of Sample Mean Population Distribution Mean Stdev 1247 100 Sample Size n 100

5.19.P 

Sampling Distribution of X-bar

Mean 1247

P(X 3.6) = P  Z 

3.6  3.4 

 = P(Z > 1.333) = 0.0912 1.5 / 100  Sampling Distribution of Sample Mean 

Population Distribution Mean Stdev 3.4 1.5 Sample Size n 100

x 3.6

Sampling Distribution of X-bar

Mean 3.4

Stdev 0.15

P(X>x) 0.0912

5-3

Chapter 05 - Sampling and Sampling Distributions

 12  13.1

15  13.1 

Z   5.21.P(12 < X < 15) = P  1.2 / 36   1.2 / 36

= P(–5.5 < Z < 9.5) = 2 (.5) = 1.000 (approximately) (Use template: Sampling Distribution.xls, sheet: x-bar) Sampling Distribution of Sample Mean Population Distribution Mean 13.1

Is the population normal?

Stdev 1.2

Sample Size n 36

x1 12

Sampling Distribution of X-bar Mean Stdev 13.1 0.2 P(x1x) 0.0000

x 3

5-26.

Is the population normal?

n = 16  = 1.5



=2



0  1.5 



2 / 16 

P( X > 0) = P  Z 

 = P(Z > -3) = .5 + .4987 = 0.9987

Sampling Distribution of Sample Mean Population Distribution Mean Stdev 1.5 2 Sample Size n 16

x 0

Sampling Distribution of X-bar

Mean 1.5

Stdev 0.5

P(X>x) 0.9987

5-5

Chapter 05 - Sampling and Sampling Distributions

5.27.

p = 1/7 

.10  .143

ˆ < .10) = P  Z  P( P 

 



(1 / 7)(6 / 7) / 180 

= P(Z < 1.648) = 0.5  0.4503 =

0.0497, a low probability. The sample size, along with np and n(1 – p), are large enough here that the sample distribution (over all the different samples of 180 people in the population) of the proportion of people who get hospitalized during the year is going to be pretty close to normal. Therefore, any one such sample proportion will be close to the predicted mean 1/7 with reasonable probability, and 1/10 is far enough away from that mean given our estimated sample standard deviation that the probability of falling even farther away than that from the mean is small. 5-28.

 = 700



= 100

n = 60  680  700

P(680  X  720) = P 

 100 / 60

Z

720  700  

100 / 60 

= 2TA(1.549) = 0.8786 5-29.

p =  = 0.35 P



ˆ  p  0.05 P



=

(0.35)(0.65) / 500

= 0.0213

ˆ < 0.30) + P( P ˆ > 0.40) = P( P

5-6

Chapter 05 - Sampling and Sampling Distributions

0.30  0.35  0.40  0.35    = P Z   + P Z   0.0213  0.0213    = 1 – 2TA(2.344) = 0.0190

5.30.

Estimator B is better. It has a small bias, but its variance is small. This estimator is more likely to produce an estimate that is close to the parameter of interest.

5.31.

I would use this estimator because consistency means as n   the probability of getting close to the parameter increases. With a generous budget I can get a large sample size, which will make this probability high.

5.32.

sˆ 2 = 1,287

5.33.

Advantage: uses all information in the data. Disadvantage: may be too sensitive to the influence of outliers.

5.34.

Depends also on efficiency and other factors. With respect to the bias: A has bias = 1/n B has bias = 0.01 A is better than B when 1/n < 0.01, that is, when n > 1/0.01 = 100

5.35.

Consistency is important because it means that as you get more data, your probability of getting closer to your “target” increases.

5.36.

n1 = 30, n 2 = 48, n3 = 32. The three sample means are known. The df for deviations from the three sample means are: df = n1 + n 2 + n3 – 3 = 30 + 48 + 32 – 3 = 107



n    n 1 

s2 = 

 100   1,287 = 1,300  99 

sˆ 2 = 

5-7

Chapter 05 - Sampling and Sampling Distributions

5.37.

a) the mean is the best number to use. mean = Sample 34 51 40 38 47 50 52 44 37

43.667 Deviation Deviation from meansquared -9.667 93.45089 7.333 53.77289 -3.667 13.44689 -5.667 32.11489 3.333 11.10889 6.333 40.10689 8.333 69.43889 0.333 0.110889 -6.667 44.44889 SSD =

358

degrees of freedom = 8 MSD = SSD / df = 358 / 8 = 44.75 b) choose the means of the respective block of numbers: 40.75, 49.667, 40.5 minimized SSD = 195.917, df = 6, MSD = 32.65283 mean = Sample 34 51 40 38 47 50 52 44 37

40.75 49.667 Deviation Deviation from meansquared -6.75 45.5625 10.25 105.0625 -0.75 0.5625 -2.75 7.5625 -2.667 7.112889 0.333 0.110889 2.333 5.442889 3.5 12.25 -3.5 12.25 SSD =

40.5

195.9167

c) Each of the numbers themselves. SSD = 0. MSD indicates that the variance is zero, which is true since we are using each of the individual numbers to reduce SSD to zero.

5-8

Chapter 05 - Sampling and Sampling Distributions

d) SSD = 719, df = 9, MSD = 79.889 mean = Sample 34 51 40 38 47 50 52 44 37

50 Deviation Deviation from meansquared -16 256 1 1 -10 100 -12 144 -3 9 0 0 2 4 -6 36 -13 169 SSD =

719

5.38.

No, because there are n – 1 = 19 – 1 = 18 degrees of freedom for these checks once you know their mean. Since 17 is on less, there is a remaining degree of freedom and you cannot solve for the missing checks.

5.39.

Yes. ( x1 +  + x18 + x19 )/19 = x . Since 18 of the x i are known and so is x , we can solve the equation for the unknown x19 .

5.40.

df = n-k as k increases, df decreases, SSD decreases, MSD decreases

5.41. 5.42.

E( X ) =  = 1,065

V( X ) =  2 /n = 5002/100 = 2,500

 2 = 1,000,000 Want SD( X )  25

SD( X ) =  / n = 1,000 / n 1,000 / n  25 n

n 5.43.

 1,000/25 = 40

 1,600. The sample size must be at least 1,600.

 = 53



E( X ) =  = 53

= 10 SE( X ) =

n = 400

/

n = 10 /

5-9

400 = 0.5

Chapter 05 - Sampling and Sampling Distributions

Sampling Distribution of Sample Mean Population Distribution Mean Stdev 53 10

Sample Size n

5-44.

Sampling Distribution of X-bar

400

p = 0.5

Mean 53

Stdev 0.5

n = 120

ˆ )= SE( P

p (1  p ) = n

(.5)(.5) = 0.0456 120

p (1  p ) = n

(.2)(.8) = 0.04216 90

ˆ ) = p = 0.2 5.45.E( P ˆ )= SE( P

ˆ. 5.46.P = 0.5 maximizes the variance of P p (1  p ) ˆ )= V( P n

Proof:

ˆ) dV ( P 1 d 1 = (pp 2) = (1 – 2p) dp dp n n

Set the derivative to zero: 1 (1 – 2p) = 0 1 = 2p p = 1/2 n The assertion may also be demonstrated by trying different values of p. 5.47.P(0.72 < X < 0.82) = P(–10.95 < Z