Solution manual for statistics for business economics 14th by anderson by marilyn.roundtree366

Solution Manual for Statistics for Business & Economics 14th by Anderson

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Chapter 7 Sampling and Sampling Distributions

Learning Objectives

1. Understand the importance of sampling and how results from samples can be used to provide estimates of population characteristics such as the population mean, the population standard deviation, or the population proportion.

2. Know what simple random sampling is and how simple random samples are selected.

3. Understand the concept of a sampling distribution.

4. Understand the central limit theorem and the important role it plays in sampling.

5. Specifically know the characteristics of the sampling distribution of the sample mean ( x ) and the sampling distribution of the sample proportion ( p ).

6. Learn about a variety of sampling methods, including stratified random sampling, cluster sampling, systematic sampling, convenience sampling, and judgment sampling.

7. Know the definition of the following terms:

parameter

target population sampled population

sample statistic

simple random sampling

sampling without replacement

sampling with replacement

point estimator

point estimate

Solutions

sampling distribution

finite population correction factor

standard error

central limit theorem

unbiased

relative efficiency

consistency

1. a. AB, AC, AD, AE, BC, BD, BE, CD, CE, DE

b. With 10 samples, each has a 1/10 probability.

c. E and C because 8 and 0 do not apply; 5 identifies E; 7 does not apply; 5 is skipped because E is already in the sample; 3 identifies C; 2 is not needed because the sample of size 2 is complete.

2. Using the last three digits of each five-digit grouping provides the random numbers: 601, 022, 448, 147, 229, 553, 147, 289, 209

Numbers greater than 350 do not apply, and the 147 can only be used once. Thus, the simple random sample of four includes 22, 147, 229, and 289.

3. 459, 147, 385, 113, 340, 401, 215, 2, 33, 348

4. a. 5, 0, 5, 8 Vale SA, Ford, General Electric

b. !10!3,628,800 120 !()!3!(103)!(6)(5040)

nNn

5. 283, 610, 39, 254, 568, 353, 602, 421, 638, 164

6. 2782, 493, 825, 1807, 289

7. 108, 290, 201, 292, 322, 9, 244, 249, 226, 125, (continuing at the top of column 9) 147, and 113.

8. Random numbers used: 13, 8, 27, 23, 25, 18

The second occurrence of the random number 13 is ignored.

Companies selected: Goldman Sachs, Coca Cola, UnitedHealth, Pfizer, Travelers Companies Inc, and JP Morgan Chase

9. 102, 115, 122, 290, 447, 588, 351, 157, 835, 498, 55, 801

===

10. a. Finite population. A frame could be constructed obtaining a list of licensed drivers from the New York State driver’s license bureau.

b. Infinite population. Sampling from a process. The process is the production line producing boxes of cereal.

c. Infinite population. Sampling from a process. The process is one of generating arrivals to the Golden Gate Bridge.

d. Finite population. A frame could be constructed by obtaining a listing of students enrolled in the course from the professor.

e. Infinite population. Sampling from a process. The process is one of generating orders for the mail-order firm.

11. a. x x n i = = =  / 54 6 9 b. s x x n i = ( )2 1 ( ) x x i 2 = (–4)2 + (–1)2 + 12 (–2)2 + 12 + 52 = 48 s = 48 6 1 31 = 12. a. p = 75/150 = .50 b. p = 55/150 = .3667 13. a. x x n i = = =  / 465 5 93

Two of the 40 stocks in the sample received a five-star rating.

b. Seventeen of the 40 stocks in the sample are rated Above Average with respect to risk.

c. There are eight stocks in the sample that are rated one star or two stars.

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. b. xi ( ) x x i ( ) x x i 2 94 +1 1 100 +7 49 85 –8 64 94 +1 1 92 –1 1 Totals 465 0 116 s x x n i = = = ( ) . 2 1 116 4 539 14. a.

2 .05 40 p ==

17 .425 40 p ==

8 .20 40 p == 15. a. 984 82 12 ix x n  === b. 2() 1022 9.6389 1121 i xx s n − ===

16. a. The sampled population is U. S. adults that are 50 years of age or older.

b. We would use the sample proportion for the estimate of the population proportion. 350

c. The sample proportion for this issue is .74, and the sample size is 426. The number of respondents citing education as “very important” is (.74) 426 = 315.

d. We would use the sample proportion for the estimate of the population proportion.

e. The inferences in parts b and d are being made about the population of U.S. adults who are age 50 or older. So the population of U.S. adults who are age 50 or older is the target population. The target population is the same as the sampled population. If the sampled population was restricted to members of AARP who were 50 years of age or older, the sampled population would not be the same as the target population. The inferences made in parts b and d would only be valid if the population of AARP members age 50 or older was representative of the U.S. population of adults age 50 and older

17. a. p = 2,977/4,135 = .7200

b. p = 2,770/4,135 = .6699

c. p = 2,233/4,135 = .5400

18. a. E x( ) = =  200

b.   x n = = = / 50/ 100 5

c. Normal with E ( x ) = 200 and  x = 5

d. It shows the probability distribution of all possible sample means that can be observed

.8216 426

426 p ==

354 .8310

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. with random samples of size 100. This distribution can be used to compute the probability that x is within a specified + from μ. 19. a. The sampling distribution is normal with E () x = μ = 200 /50/1005 x n === For +5, 195205 x  Using standard normal probability table: At x = 205, 5 1 5 x x z   === (1)Pz  = .8413 At x = 195, 5 1 5 x x z   ===− (1)Pz − = .1587 (195205)Px = .8413 – .1587 = .6826 b. For + 10, 190210 x  Using standard normal probability table: At x = 210, z x x = = =   10 5 2 (2)Pz  = .9772 At x = 190, 10 2 5 x x z   ===− (2)Pz − = .0228 (190210)Px = .9772 – .0228 = .9544 20.   x n = / 25/503.54 x  == 25/1002.50 x  ==

The standard error of the mean decreases as the sample size increases.

Note: Only case d where n /N = .10 requires the use of the finite population correction factor.

The normal distribution for x is based on the central limit theorem.

b. For n = 120, E ( x ) remains $71,800, and the sampling distribution of x can still be

x  ==

25/1502.04

25/2001.77

21. a.   x n = = = / 10/ 50 141

n / N = 50 / 50,000 = .001 Use   x n = = = / / . 10 50 141 c. n / N = 50 / 5000 = .01 Use   x n = = = / / . 10 50 141 d. n / N = 50 / 500 = .10 Use   x N n N n = = = 1 500 50 500 1 10 50 134

22. a. E( x ) = 71,800 and   x n = = = / 4000/ 60 51640

approximated by a normal distribution. However,  x is reduced to 4000 120 / = 365.15.

c. As the sample size is increased, the standard error of the mean,  x , is reduced. This appears logical from the point of view that larger samples should tend to provide sample means that are closer to the population mean. Thus, the variability in the sample mean, measured in terms of  x , should decrease as the sample size is increased.

23. a. With a sample of size 60 4000 516.40 60 x  == At x = 72,300, 72,30071,800 .97 516.40 z == P( x ≤ 72,300) = P(z ≤ .97) = .8340 At x = 71,300, 71,30071,800 .97 516.40 z ==− P( x < 71,300) = P(z < –.97) = .1660 P(71,300 ≤ x ≤ 72,300) = .8340 – .1660 = .6680 b. 4000 365.15 120 x  == At x = 72,300, 72,30071,800 1.37 365.15 z == P( x ≤ 72,300) = P(z ≤ 1.37) = .9147 At x = 71,300, 71,30071,800 1.37 365.15 z ==− P( x < 71,300) = P(z < –1.37) = .0853 P(71,300 ≤ x ≤ 72,300) = .9147 – .0853 = .8294 24

Ex =

. a. Normal distribution, ()17.5

25.

c. The probability of being within 10 of the mean on the evidence-based reading and writing portion of the test is exactly the same as the probability of being within 10 on the Mathematics portion of the SAT. This is because the standard error is the same in

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. /4/50.57 x n === b. Within one week means 16.5  x  18.5. At x = 18.5, 18.517.5 1.75 .57 z == P(z ≤ 1.75) = .9599 At x = 16.5, z = –1.75. P(z < –1.75) = .0401 So P(16.5 ≤ x ≤ 18.5) = .9599 – .0401 = .9198 c. Within 1/2 week means 17.0 ≤ x ≤ 18.0 At x = 18.0, 18.017.5 .88 .57 z == P(z ≤ .88) = .8106 At x = 17.0, z = –.88 P(z < –.88) = .1894 P(17.0 ≤ x ≤ 18.0) = .8106 – .1894 = .6212

/100/9010.54 x n ===

a and b. a. 543533 0.87 10.54 z == P(z ≤ .95) = .8289 523533 0.87 10.54 z ==− P(z < –.95) = .1711 Probability = .8289

.1711 =.6578 b. 537527 1.73 10.54 z == P(z ≤ .95) = .8289 517527 1.73 10.54 z ==− P(z < –.95) = .1711

=.6578

This value for the standard error can be used for parts

–

Probability = .8289 – .1711

both cases. The fact that the means differ does not affect the probability calculation.

Within  200 means x – 16,642 must be between –200 and +200.

value for x – 16,642 = 200. So we just show the computation of z for x – 16,642 = 200.

The z value for x – 16,642 = –200 is the negative of the

b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean. In this instance, the probability of being within +200 of μ ranges from .3544 for a sample of size 30 to .9050 for a sample of size 400.

26. a. 16642 / x z n  =

n = 30 200 .46 2400/30 z == P(–.46 ≤ z ≤ .46) = .6772 – .3228 = .3544 n = 50 200 .59 2400/50 z == P(–.59 ≤ z ≤ .59) = .7224 – .2776 = .4448 n = 100 200 .83 2400/100 z == P(–.83 ≤ z ≤ .83) = .7967 – .2033 = .5934 n = 400 200 1.67 2400/400 z == P(–.1.67 ≤ z ≤ .1.67) = .9525 – .0475 = .9050

27. a. /4.60/50.6505 x n === At x = 38.39, 38.3937.39 1.54 .6505 / x z n   === P(z ≤ 1.54) = .9382 At x = 36.39, z = –1.54 P(z < –1.54) = .0618, thus P(36.39 ≤ x ≤ 38.39) = .9382 – .0618 = .876

/4.10/50.5798

n ===

c. In part b we have a higher probability of obtaining a sample mean within $1.00 of the population mean because the standard error for female graduates (.5798) is smaller than the standard error for male graduates (.6505).

The probability the sample mean will be within 1 inch of the population mean of 22 is .8294.

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. At x = 28.83, 28.8327.83 1.72 .5798 / x z n   === P(z ≤ 1.72) = .9573 At x = 26.83, z = –1.72, P(z < –1.72) = .0427, thus P(26.83 ≤ x ≤ 28.83) = .9573 – .0427 = .915

n === At x = 27.23, 27.2327.83 1.60 .3743 z ==− P( x < 27.23) = P(z < –1.60) = .055 28. a. This is a

of a normal distribution

()Ex = 22 and /4/30.7303 x n === .

With n = 120, /4.10/120.3743 x

graph

with

21  x  23 2322 1.37 .7303 z == 2122 1.37 .7303 z ==− P(21  x  23) = P(–1.37 ≤ z ≤ 1.37) = .9147 – .0853 = .8294

b. Within

inch means

=== Within 1 inch means 41  x  43 4342 1.68 .5963 z == 4142 1.68 .5963 z ==−

c. /4/45.5963 x n

The probability the sample mean will be within 1 inch of the population mean of 42 is .9070.

The probability of being within 1 inch is greater for New York in part c because the sample size is larger.

100

d. None of the sample sizes in parts a, b, and c is sufficiently large. The sample size will need exceed n = 100, which was used in part c.

30. a. n / N = 40 / 4000 = .01 < .05; therefore, the finite population correction factor is not necessary.

b. With the finite population correction factor

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. P(41  x  43) = P(–1.68 ≤ z ≤ 1.68) = .9535 – .0465 = .9070

=

=

a. n = 30 Within 16 means 257289 x  16 .88 /100/30 x z n   === P(257  x  289) = P(–.88  z  .88) = .8106 – .1894 = .6212

n = 50 Within 16 means 257289 x  16 1.13 /100/50 x z n   === P(257  x  289) = P(–1.13  z  1.13) = .8708 – .1292 = .7416

n = 100 Within 16 means 257289 x  16 1.60 /100/100 x z n   === P(257  x  289) = P(–1.60  z  1.60) = .9452 – .0548 = .8904

29.

273

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.   x N n N n = = = 1 4000 40 4000 1 82 40 129 .

correction factor   x n = = / 130

population correction factor

different value for  x than when the correction factor is not used.

z x = = =  130 2 130 154 . P(z ≤ 1.54) = .9382

(z < –

Probability

–

31. a. E( p ) = p = .40

(1).40(.60) .0490 100 p pp n  ===

Without the finite population

Including the finite

provides only

slightly

1.54)

.0618

= .9382

.0618 = .8764

and 

.0490.

32. a. E( p ) = .40 (1).40(.60) .0346 200 p pp n  === Within ± .03 means .37 ≤ p ≤ .43 .03 .87 .0346 p pp z  === P(z ≤ .87) = .8078 P(z < –.87) = .1922 P(.37 ≤ p ≤ .43) = .8078 – .1922 = .6156

c. Normal distribution with

(

)

.40

p =

shows the probability distribution for the sample proportion p .

The standard error of the proportion, ,

decreases as

increases.

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. b. .05 1.44 .0346 p pp z  === P(z ≤ 1.44) = .9251 P(z < –1.44) = .0749 P(.35 ≤ p ≤ .45) = .9251 – .0749 = .8502 33.  p p p n = ( ) 1 (.55)(.45) .0497 100 p  == (.55)(.45) .0352 200 p  == (.55)(.45) .0222 500 p  == (.55)(.45) .0157 1000 p  ==

p 

34. a. (.30)(.70) .0458 100 p  == Within ± .04 means .26 ≤ p ≤ .34. .04 .87 .0458 p pp z  === P(z ≤ .87) = .8078 P(z < –.87) = .1922 P(.26 ≤ p ≤ .34) = .8078 – .1922 = .6156 b. (.30)(.70) .0324 200 p  ==

e. With a larger sample, there is a higher probability p will be within + .04 of the population proportion p.

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. .04 1.23 .0324 p pp z  === P(z ≤ 1.23) = .8907 P(z < –1.23) = .1093 P(.26 ≤ p ≤ .34) = .8907 – .1093 = .7814 c. (.30)(.70) .0205 500 p  == .04 1.95 .0205 p pp z  === P(z ≤ 1.95) = .9744 P(z < –1.95) = .0256 P(.26 ≤ p ≤ .34) = .9744 – .0256 = .9488 d. (.30)(.70) .0145 1000 p  == .04 2.76 .0145 p pp z  === P(z ≤ 2.76) = .9971 P(z < –2.76) =

P(.26 ≤ p ≤ .34) = .9971 – .0029 = .9942

.0029

35. a. .30

The normal distribution is appropriate because np = 100(.30) = 30 and n(1 – p) = 100(.70) = 70 are both greater than 5.

b. P (.20 ≤ p ≤ .40) = ?

36.

a. This is a graph of a normal distribution with a mean of ()Ep = .55 and

b. Within ± .05 means .50

c. This is a graph of a normal distribution with a mean of ()Ep = .45 and

be scanned, copied or duplicated, or

publicly

website,

part.

posted

accessible

in whole or in

.40.30 2.18 .0458 z == P(z ≤ 2.18) = .9854 P(z < –2.18) = .0146 P(.20 ≤ p ≤ .40) = .9854 – .0146 = .9708

P (.25 ≤ p ≤ .35) = ? .35.30 1.09 .0458 z == P(z ≤ 1.09) = .8621 P(z < –1.09) = .1379 P(.25 ≤ p ≤

= .8621 – .1379 = .7242

.35)

(1).55(1.55) .0352 200 p pp n  ===

≤ p ≤ .60 .60.55 1.42 .0352 p pp z  === .50.55 1.42 .0352 p pp z  ===− P(.50  p  .60) = P(–1.42 ≤ z ≤ 1.42) = .9222

.0778 = .8444

–

(1).45(1.45) .0352 200 p pp n  ===

e. No, the probabilities are exactly the same. This is because p  , the standard error, and the width of the interval are the same in both cases. Notice the formula for computing the standard error. It involves (1)pp . So whenever p = 1 – p, the standard error will be the same. In part b, p = .45 and 1 – p = .55. In part d, p = .55 and 1

The probability is larger than in part b This is because the larger sample size has reduced the standard error from .0352 to .0249.

525(.10) = 52.5 and n(1 – p) = 525(1 – .10) = 472.5, so p is normal distributed with

(1).45(1.45) .0352 200 p pp n  === Within ± .05 means .40 ≤ p ≤ .50 .50.45 1.42 .0352 p pp z  === .40.45 1.42 .0352 p pp z  ===− P(.40  p  .50) = P(–1.42 ≤ z ≤ 1.42) = .9222 – .0778 = .8444

–

f. For n = 400, (1).55(1.55) .0249 400 p pp n  === Within ± .05 means .50 ≤ p ≤ .60. .60.55 2.01 .0249 p pp z  === .50.55 2.01 .0249 p pp z  ===− P(.50  p  .60) = P(–2.01 ≤ z ≤ 2.01) = .9778 – .0222 = .9556

p = .45.

37. a. np

().10Ep

(1)(0.10)(0.90) 0.0131 525 p pp n  === b. 0.03 2.29 0.0131 z == P(z ≤ 2.29) = .9890

= and

It is a normal distribution with E(

d. The probabilities would increase. This is because the increase in the sample size makes the standard error, p  , smaller. 39. a. np = 200(.80) = 160 and n(1 – p) = 200(1 – .80) = 40, so p is normal distributed normal distribution with ().80Epp== and

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. P(z < –2.29) = .0110 P(.07 ≤ p ≤ .13) = .9890 – .0110 = .9780 c. 0.015 1.15 0.0131 z == P(z ≤ 1.15) = .8749 P(z < –1.15) = .1251 P(.085 ≤ p ≤ .115) = .8749 – .1251 = .7498 38. a.

(1)(.42)(.58) .0285 300 p pp n  === b. .03 1.05 .0285 p pp z  === P(z ≤ 1.05) = .8531 P(z < –1.05) = .1469 P(.39 ≤ p ≤ .44) = .8531 – .1469 = .7062 c. .05 1.75 .0285 p pp z  === P(z ≤ 1.75) = .9599 P(z < –1.75) = .0401 P(.39 ≤ p ≤ .44) = .9599 – .0401 = .9198

) = .42.

c. np = 450(.80) = 360 and n(1 – p) = 450(1 – .80) = 90, so p is normal distributed normal distribution with ().80Epp== and

e. The probability of the sample proportion being within .04 of the population mean is increased .8410 to .9660. The likelihood of the sample proportion being within 0.04 of the population proportion increases by .125 by increasing the sample size from 200 to 450. If the extra cost of using the larger sample size is not too great, we should probably do so.

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (1)(0.80)(10.80) 0.0283 200 p pp n  === b. ( ) ( ) .84.800.04 1.41 0.0283 1.801.80 200 pp z pp n ==== P(z ≤ 1.41) = .9207 P(z < –1.41) = .0797 P(.76  p  .84) = P(–1.41  z  1.41) = .9207 – .0797 = .8410

(1)(0.80)(10.80)

450 p pp n  === d. ( ) ( ) .84.800.04 2.12 0.0189 1.801.80 450 pp z pp n ==== P(z ≤ 2.12) = .9830 P(z < –2.12) = .0170 P(.76  p  .84) = P(–2.12  z  2.12) = .9830 – .0170 = .9660

0.0189

40. a. E ( p ) = .76

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (1).76(1.76) .0214 400 p pp n  === Normal distribution because np = 400(.76) = 304 and n(1 – p) = 400(.24) = 96 b. .79.76 1.40 .0214 z == P(z ≤1.40) = .9192 .73.76 1.40 .0214 z ==− P(z < –1.40) = .0808 P(.73  p  .79) = P(–1.40  z  1.40) = .9192 – .0808 = .8384 c. (1).76(1.76) .0156 750 p pp n  === .79.76 1.92 .0156 z == P(z ≤ 1.92) = .9726 .73.76 1.92 .0156 z ==− P(z < –1.92) = .0274 P(.73  p  .79) = P(–1.92  z  1.92) = .9726 – .0274 = .9452 41. a. E( p ) = .17 (1)(.17)(1.17) .0133 800 p pp n  === Distribution is approximately normal because np = 800(.17) = 136 > 5 and n(1 – p) = 800(.83) = 664 > 5 b. .19.17 1.51 .0133 z == P(z ≤ 1.51) = .9345 .15.17 1.51 .0133 z ==− P(z < –1.51) = .0655 P(.15  p  .19) = P(–1.51  z  1.51) = .9345 – .0655 = .8690

42. a. ��(��)=�� =400

b. �� =��⁄√�� =100⁄√100,000=.3162

c. Normal with E( x ) = 400 and  x = .3162.

d. It shows the probability distribution of all possible sample means that can be observed with random samples of size 100,000. This distribution can be used to compute the probability that x is within a specified + from μ.

The standard error of the mean , x  decreases and becomes extremely small as the sample size becomes huge

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. c. (1)(.17)(1.17) .0094 1600 p pp n  === .19.17 2.13 .0094 z == P(z ≤ 2.13) = .9834 .15.17 2.13 .0094 z ==− P(z < –2.13) = .0166 P(.15  p  .19) = P(–2.13  z  2.13) = .9834 – .0166 = .9668

43.   x n = / 25/500,000.0354 x  == 25/1,000,000.0250 x

25/5,000,000.0111 x

25/10,000,000.0079 x

25/100,000,000.0025



44. a. E(��) = p = .75

(1).75(.25) .0014 100,000

c. Normal distribution with E(��) = .75 and  p = .0014

d. It shows the probability distribution of all possible sample proportions that can be observed with random samples of size 100,000. This distribution can be used to compute the probability that �� is within a specified + from p. 45.

(.44)(.56) .00070 500,000

(.44)(.56) .00050 1,000,000

(.44)(.56) .00022 5,000,000

(.44)(.56) .00016 10,000,000

(.44)(.56) .00005 10,000,000

The standard error of the sample proportion, , p  decreases and becomes extremely small as the sample size becomes huge.

46. a. Normal distribution, ()100Ex = /48/15,000.3919

n 

===

 p

p p n = ( ) 1

p

x n ===

 x  101 At x = 101, 101100 2.55 .3919 z == P(z ≤ 2.55) = .9946 At x = 99, 99100 2.55 .3919 z ==− P(z < –2.55) = .0054

Within ±1 hour means 99

So P(99 ≤ �� ≤ 101) = .9946 – .0054 = .9892.

c. In part a we found that P(99 ≤ �� ≤ 101) = .9892, so the probability the mean annual number of hours of vacation time earned for a sample of 15,000 blue collar and service employees who work for small private establishments and have at least 10 years of service differs from the population mean μ by more than 1 hour is 1 – . 9892 – .0108, or approximately 1%.

Because these sample results are unlikely if the population mean is 100, the sample results suggest either (a) the mean annual number of hours of vacation time earned by blue-collar and service employees who work for small private establishments and have at least 10 years of service is not 100 or (b) the sample is not representative of the population of blue-collar and service employees who work for small private establishments and have at least 10 years of service. The sampling procedure should be carefully reviewed.

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not be scanned,

n ===

a and b. a. P (25.15 ≤ �� ≤ .25.25) = ? 25.2525.2 2.20 .0227 z == P(z ≤ 2.20) = .9863 25.1525.2 2.20 .0227 z ==− P(z < –2.20) = .0137 P (25.15 ≤ �� ≤ .25.25) = .9863 – .0137 = .973 b. P (25.19 ≤ �� ≤ .25.21) = ? 25.2125.2 0.44 .0227 z == P(z ≤ .44) = .6700

47. /6/70,000.0227 x

This value for the standard error can be used for parts

For samples of the same size, the probability of being within 05 of the mean miles per gallon is much higher than the probability of being within .01 miles per gallon.

The standard error is now smaller because the sample size is larger.

The probability is larger here than it is in part b because the larger sample size has made the standard error smaller.

d. P (24.2 ≤ �� ≤ .26.2) = ?

Because the probability the mean miles per gallon for a sample of 70,000 new cars and light trucks sold in the United States in 2017 differs from the population mean μ by less than 1 gallon is approximately 1.0000, the probability the mean miles per gallon for a

(25.19 ≤ �� ≤ .25.21) = .6700 – .3300 =.3400

/6/90,0000.0200 x n ===

P (25.19 ≤ �� ≤ .25.21) =

25.2125.2 0.50 0.0200 z == P(z ≤ 0.50) = .6915 25.1925.2 0.50 0.0200 z ==− P(z < –0.50) = .3085

(25.19 ≤ �� ≤ .25.21) = .6915 – .3085 =.3830

26.225.2 24.20 .0227 z == P(z ≤ 24.20) ≈ 1.0000 22.225.2 24.20 .0227 z ==− P(z < –24.20) ≈ .0000 P (24.2 ≤ �� ≤ .26.2) ≈ 1.0000 – .0000 ≈ 1.000

48. a.

sample of 70,000 new cars and light trucks sold in the United States in 2017 differs from the population mean μ by more than 1 gallon is approximately 1 – 1.0000 = .0000. These sample results are unlikely if the population mean is 25.2, so these sample results would suggest either (1) the population mean miles per gallon for new cars and light trucks sold in the United States in 2017 is not 25.2 or (2) the sample is not representative of the population of new cars and light trucks sold in the United States in 2017. The sampling procedure should be carefully reviewed.

The normal distribution is appropriate because np = 108,700(.42) = 45,654 and n(1 – p) =108,700(.58) = 63,046 are both greater than 5. b. P (.419 ≤ �� ≤ .421) = ?

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.421.42 0.67 .0015 z == P(z ≤ 0.67) = .7486 .419.42 0.67 .0015 z ==− P(z < –0.67) = .2514 P(.419 ≤ ��≤ .421) = .7486 – .2514 = .4972 c. P (.4175 ≥ �� ≥ .4225) = ? .4225.42 1.67 .0015 z == P(z ≤ 1.67) = .9525 .30 .42

For samples of the same size, the probability of being within 1% of the population proportion of repeat purchasers is much smaller than the probability of being within .25% of the population proportion of repeat purchasers

Because the probability the proportion of orders placed by repeat customers for a sample of 108,700 orders from the past six months differs from the population proportion p by less than 1% is approximately 1.0000, the probability the proportion of orders placed by repeat customers for a sample of 108,700 orders from the past six months differs from the population proportion p by more than 1% is approximately 1 –1.0000 = .0000.

These sample results are unlikely if the population proportion is .42, so these sample results would suggest either (1) the population proportion of orders placed by repeat customers from the past six months is not 42% or (2) the sample is not representative of the population of orders placed by repeat customers from the past six months. The sampling procedure should be carefully reviewed.

49. a. For a sample of 207,000 U.S. homes, the normal distribution is appropriate because np

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. .4175.42 1.67 .0015 z ==− P(z < –1.67) = .0475 P(.25 ≤ �� ≤ .35) = .9525 – .0475 = .9050

d. P (.41 ≤ �� ≤ .43) = ? .43.42 6.68 .0015 z == P(z ≤ 6.68) ≈ 1.0000 .41.42 6.68 .0015 z ==− P(z < –6.68) ≈ .0000 P(.41 ≤ �� ≤ .43) ≈ 1.0000 – .0000 = .0000

= 207,000(.492) = 101,844 and n(1

) =207,000(.508) = 105156 are both greater than

5. The mean is ()Ep = .492 and the standard error is

c. For a sample of 86,800 U.S. homes, the normal distribution is appropriate because np = 86,800(.492) = 42,705.6 and n

are both greater than 5. The mean is ��(��)= .492 and the standard error is

e. Yes, the probability in d is smaller than the probability in b because the sample size in d is smaller than the sample size in b. 50. a. Sorting the list of companies and random numbers to identify the five companies associated with the five smallest random numbers provides the following sample.

– p

(1).492(1.508) .0011 207,000 p pp n  ===

± .002 means .490 ≤ �� ≤ .494. .494.492 1.82 .0011 p pp z  === .490.492 1.82 .0011 p pp z  ===− P(.490  �� .494) = P(–1.82 ≤ z ≤ 1.82) = .9656 – .0344 = .9322

b. Within

– p) =86,800(.508)

44,094.4

(1).492(1.508) .0017 86,800 p pp n  ===

.490 ≤ �� ≤ .494 .494.492 1.18 .0017 p pp z  === .490.492 1.18 .0017 p pp z  ===− P(.490  ��  .494) = P(–1.18 ≤ z ≤ 1.18) = .8810 – .1190 = .7620

d. Within ± .002 means

Company Random Number

b. Step 1: Generate a new set of random numbers in column B. Step 2: Sort the random numbers and corresponding company names into ascending order and then select the companies associated with the five smallest random numbers. It is extremely unlikely that you will get the same companies as in part a 51

With n = 100, we can approximate the sampling distribution with a normal distribution having

The probability that the sample mean will be within $200 of the population mean is .5762.

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. LMI Aerospace .008012 Alpha & Omega Semiconductor .055369 Olympic Steel .059279 Kimball International .144127 International Shipholding .227759

. a.

E( x ) = 10,348 2500 250 100 x n   === b. 200 .80 /2500/100 x z n   === P(z ≤ .80) = .7881 200 .80 /2500/100 x z n   ===− P(z < –.80) = .2119 P(10,148  x  10,548) = P(–.80  z  .80) = .7881 – .2119 = .5762

c. At 12,000, 12,00010,348 6.61 2500/100 z ==

Yes, the research firm should be questioned. A sample mean this large is extremely unlikely (almost 0 probability) if a simple random sample is taken from a population with a mean of $10,348.

Yes, this is an unusually low performing group of 64 stores. The probability of a sample mean annual sales per square foot of $380 or less is only .0047.

53. With n = 60, the central limit theorem allows us to conclude the sampling distribution for part a and part b is approximately normal with

b. Within ± 20 of the population mean μ means 300.51

340.51.

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. P( x ≥ 11,000) = 1 – P( x ≤ 11,000) = 1 – P(z ≤ 6.61)  0

52. a. Normal distribution with E() x = 406 80 10 64 x n   === b. 15 1.50 /80/64 x z n   === P(z ≤ 1.50) = .9332 15 1.50 /80/64 x z n   ===− P(z < –1.50) = .0668 P(391  x  421) = P(–1.50  z  1.50) = .9332 – .0668 = .8664 c. At x = 380, 380406 2.60 /80/64 x z n   ===− P( x ≤ 380) = P(z ≤ –2.60) = .0047

E() x = 320.51 80 10.3280 60 x n   ===



 x

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. At x = 340.51, 340.51320.51 1.94 10.3280 z == P(z ≤ 1.94) = .9738 At x = 300.51, 300.51320.51 1.94 10.3280 z ==− P(z < –1.94) = .0262 P(300.51  x  340.51) = P(–1.94  z  1.94) = .9738 – .0262 = .9476 c. Within ± 10 of the population mean μ means 310.51  x  330.51 At x = 330.51, 330.51320.51 .97 10.3280 z == P(z ≤ .97) = .8340 At x = 310.51, 310.51320.51 .97 10.3280 z ==− P(z < –.97) = .1660 P(310.51  x  330.51) = P(–.97  z  .97) = .8340 – .1660 = .6680 54. = 27,175= 7400 a. 7400/60955 x  == b. 0 0 955 x x z   === P( x > 27,175) = P(z > 0) = .50 Note:

of the sampling distribution of x c. 1000 1.05 955 x x z   === P(z ≤ 1.05) = .8531 P(z < –1.05) = .1469 P(26,175  x  28,175) = P(–1.05  z  1.05) = .8531 – .1469 = .7062 d. 7400/100740 x  == 1000 1.35 740 x x z   === P(z ≤ 1.35) = .9115

This could have been answered easily without any calculations; 27,175 is the expected value

Note: With n / N ≤ .05 for all three cases, common statistical practice would be to ignore the

correction factor and use

for each case.

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. P(z < –1.35) = .0885 P(26,175  x  28,175) = P(–1.35  z  1.35) = .9115 – .0885 = .8230 55. a.   x N n N n = 1 N = 2,000  x = = 2000 50 2000 1 144 50 2011 N = 5,000  x = = 5000 50 5000 1 144 50 2026 . N = 10,000  x = = 10 000 50 10 000 1 144 50 2031 , ,

finite population

 x = = 144 50 2036

b. N = 2,000 25 1.24 20.11 z == P(z ≤ 1.24) = .8925 P(z < –1.24) = .1075 Probability = P(–1.24  z  1.24) = .8925 – .1075 = .7850 N = 5,000 25 1.23 20.26 z == P(z ≤ 1.23) = .8907 P(z < –1.23) = .1093

All

are approximately .78, which indicates that a sample of size 50 will work well for all three

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Probability = P(–1.23  z  1.23) = .8907 – .1093 = .7814 N = 10,000 25 1.23 20.31 z == P(z ≤ 1.23) = .8907 P(z < –1.23) = .1093 Probability = P(–1.23  z  1.23) = .8907 – .1093 = .7814

56. a.   x n n = = = 500 20 n = 500/20 = 25 and n = (25)2 = 625 b. For + 25, 25 1.25 20 z == P(z ≤ 1.25) = .8944 P(z < –1.25) = .1056 Probability = P(–1.25  z  1.25) = .8944 – .1056 = .7888 57. Sampling distribution of x  = 1.9 + 2.1 2 = 2 1.9 2.1 0.05 0.05  x    x n = = 30

probabilities

firms.

The area below x = 2.1 must be 1 – .05 = .95. An area of .95 in the standard normal table shows z = 1.645.

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Thus, 2.12.0 1.645 /30 z  == Solve for  (.1)30 .33 1.645  == 58. p = .15 a. This

the

normal distribution with E( p ) = p = .15 and (1).15(1.15) .0230 240 p pp n  === . b. Within ± .04 means .11 ≤ p ≤ .19 .19.15 1.74 .0230 p pp z  === .11.15 1.74 .0230 z ==− P(.11  p  .19) = P(–1.74 ≤ z ≤ 1.74) = .9591 – .0409 = .9182 c. Within ±.02 means .13 ≤ p ≤ .17 .17.15 .87 .0230 p pp z  === .13.15 .87 .0230 p pp z  ===− P(.13  p  .17) = P(–.87 ≤ z ≤ .87) = .8078 – .1922 = .6156 59. (1)(.40)(.60) .0245 400 p pp n  === P ( p ≥ .375) = ?

graph of a

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. .375.40 1.02 .0245 z ==− P(z < –1.02) = .1539 P ( p ≥ .375) = 1 – P ( p ≤ .375) = 1 – P (z ≤ –1.02) = 1 – .1539 = .8461 60. a. (1)(.40)(1.40) .0251 380 p pp n  === Within ± .04 means .36 ≤ p ≤ .44. .44.40 1.59 .0251 z == .36.40 1.59 .0251 z ==− P(.36  p  .44) = P(–1.59 ≤ z ≤ 1.59) = .9441 – .0559 = .8882 b. We want P( p .45). .45.40 1.99 .0251 p pp z  === P( p  .45) = 1 – P( p ≤ .45) = 1 – P(z ≤ 1.99) = 1 – .9767 = .0233 61. a. Normal distribution with E( p ) = .15 and (1)(.15)(.85) .0292 150 p pp n  === b. P (.12  p  .18) = ? .18.15 1.03 .0292 z == P(z ≤ 1.03) = .8485 .12.15 1.03 .0292 z ==− P(z < –1.03) = .1515 P(.12  p  .18) = P(–1.03  z  1.03) = .8485 – .1515 =.6970 62. a.  p p p n n = = = ( ) . (. ) 1 25 75 0625

© 2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Solve for n n = = (. ) (. ) 25 75 0625 48 2 b. Normal distribution with E( p ) = .25 and p  = .0625. [Note: (48)(.25) = 12 > 5, and (48)(.75) = 36 > 5.] c. P ( p ≥ .30) = ? .30.25 .80 .0625 z == P(z ≤ .80) = .7881 P ( p ≥ .30) = 1 – P ( p ≤ .30) = 1 – P(z ≤ .80) = 1 – .7881 1 – .7881 = .2119 63. a. Normal distribution, ()8,000Ex = . /480/35,0002.5657 x n === b. Within ±4 hours means 7,996  x  8,006 At x = 8,004, 8,0048,000 1.56 2.5657 z == P(z ≤ 1.56) = .9406 At x = 7,996, 87,9968,000 1.56 2.5657 z ==− P(z < –1.56) = .0594 So P(7,996 ≤ �� ≤ 8,004) = .9406 – .0594 = .8812 c. Within ±1 hour means 7,999 ≤ �� ≤ 8,001. At �� = 8,001, 8,0018,000 0.39 2.5657 z == P(z ≤ 0.39) = .6517 At �� = 7,999, 7,9998,000 0.39 2.5657 z ==− P(z < –0.39) = .3483 P(7,999 ≤ �� ≤ 8,001) = .6517 – .6483 = .3034

d. In part b we found that P(7,996 ≤ �� ≤ 8,004) = .8812, so the probability the mean life of a sample of 35,000 14-watt CFLs will differ from the population mean life μ by more than 4 hours is 1 – .8812 = .1188. If the population mean is 8,000, sample results as extreme as this would occur more than one time out of 10 samples. The results therefore do not provide strong evidence the population mean is different from 8,000.

64. a. Normal distribution, ()17.6Ex =

/5.1/85,020.0175

b. Within ±3 minutes of the mean is the same as within 3/60 – .05 hours of the mean, i.e.,

c. In part b we found that if we assume the U.S. population mean and standard deviation are appropriate for Florida, then P(17.55 ≤ �� ≤ 17.65) = .9958, so the probability the mean for a sample of 85,020 Floridians will differ from the U.S. population mean by more than 3 minutes is 1 – .9958 = .0042. This result is would be highly unlikely and would suggest that the home Internet usage of Floridians differs from home Internet usage for the rest of the United States.

65. a.

x n ===

17.55  x  17.65 At x = 17.65, 17.6517.6 2.86 .0175 z == P(z ≤ 2.86) = .9979 At x = 17.55, z = –2.86. P(z < –2.86) = .0021 So P(17.55 ≤ �� ≤ 17.65) = .9979 – .0021 = .9958

The normal distribution is appropriate because np = 114,250(.043) = 4,912.75 and n(1

109,337.25 are both greater than 5.

For samples of the same size, the probability of being within .1% of the population proportion of undeliverable mail pieces is much larger than the probability of being within .05% of the population proportion of undeliverable mail pieces.

–p) =114,250(.957)

b. P (.042 ≤ �� ≤ .044) = ? .044.043 1.67 .0006 z == P(z ≤ 1.67) = .9525 .042.043 1.67 .0006 z ==− P(z < –1.67) = .0475 P(.042 ≤ ��≤ .044) = .9525 – .0475 = .9050 c. P (.0425 ≤ �� ≤ .0435) = ? .0435.43 0.83 .0006 z == P(z ≤ 0.83) = .7967 .0425.43 0.83 .0006 z ==− P(z < –0.83) = .2033 P(.0425 ≤ �� ≤ .0435) = .7967 – .2033 = .5934

66. a. .30 (1).043(.957) .0006 114,250 p pp n  === p .043

(1).58(.42) .0035 20,000

The normal distribution is appropriate because np = 20,000(.58) = 11,600 and n(1 – p) =20,000(.42) = 8,400 are both greater than 5. b.

c. In part b we found that P(.57 ≤ �� ≤ .59) = .9958, so the probability the proportion of a sample of 20,000 drivers that is speeding will differ from the U.S. population proportion of drivers that is speeding by more than 1% is 1 – .9958 = .0042. These sample results are unlikely if the population proportion is .58, so these sample results would suggest either (1) the population proportion of drivers who are speeding is not 58% or (2) the sample is not representative of the population of U.S. drivers. The sampling procedure should be carefully reviewed.

.59.58 2.87 .0035 z == P(z ≤ 2.87) = .9979 .57.58 2.87 .0035 z ==− P(z < –2.87) = .0021 P(.57 ≤ �� ≤ .59) = .9979 – .0021 = .9958

P (.57 ≤ �� ≤ .59) = ?

.30

n  === p 58

Case Solutions

Case Problem: Marion Dairies

1. We know that

• E(��) = μ regardless of the sample size, so the mean of all possible values of �� is equal to 80.

• The sample size is 50, so the standard error of the mean is �� = �� √�� = 25 √50 =3.5355.

• The simple random sample of 50 customers, so the central limit theorem enables us to conclude that the sampling distribution of the mean can be approximated by a normal distribution. At �� = 75, we have

The area under the curve to the left of z = –1.41 is .0793. Therefore, P(�� ≤ 75) = .0793 when n = 50. The probability the mean score of Blugurt given by a simple random sample of 50 Marion Dairies customers will be 75 or less is .0793.

2. We know that:

• E(��) = μ regardless of the sample size, so the mean of all possible values of �� is again equal to 80.

• The sample size is now 150, so the standard error of the mean is �� = �� √�� = 25 √150 = 2.0412.

• The simple random sample of 150 customers and the central limit theorem enable us to conclude that the sampling distribution of the mean can be approximated by a

�� = 75 80 35355 =

1.41

The area under the curve to the left of z = –2.45 is .0071. Therefore, P(�� ≤ 75) = .0071 when n = 150. The probability the mean score of Blugurt given by a simple random sample of 150 Marion Dairies customers will be 75 or less is .0071.

3. Whenever the sample size is increased, the standard error of the sample mean decreases. With n = 50, the standard error of the sample mean for Blugurt scores is 3.5355. With the increase in the sample size to n = 150, the standard error of the sample mean is decreased to 2.0412. Because the sampling distribution of the sample mean with n = 150 has a smaller standard error, the potential values of �� that the sample can yield have less variation and tend to be closer to the population mean than the potential values of �� when n = 50. With a smaller standard error, it is less likely that a sample will have a mean Blugurt score of 75 or lower.

2019 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. normal distribution. At �� = 75, we have �� = 75 80 2.0412 = 2.45

Solution Manual for Statistics for Business & Economics 14th by Anderson Visit TestBankBell.com to get complete for all chapters