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 and / 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 sample statistic simple random sampling sampling without replacement sampling with replacement point estimator point estimate

sampling distribution finite population correction factor standard error central limit theorem unbiased relative efficiency consistency

7-1

Chapter 7

Solutions: 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 since E is already in the sample; 3 identifies C; 2 is not needed since the sample of size 2 is complete.

2.

Using the last 3-digits of each 5-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. 4.

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

6, 8, 5, 4, 1 Nasdaq 100, Oracle, Microsoft, Lucent, Applied Materials

b.

N! 10! 3, 628,500 = = = 252 n !( N − n)! 5!(10 − 5)! (120)(120)

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.

13, 8, 23, 25, 18, 5 The second occurrences of random numbers 13 and 25 are ignored. Maryland, Iowa, Florida State, Virginia, Pittsburgh, Oklahoma

9.

102, 115, 122, 290, 447, 351, 157, 498, 55, 165, 528, 25

10.

finite, infinite, infinite, infinite, finite

11. a.

x = Σxi / n =

b.

s=

54 =9 6

Σ( xi − x ) 2 n −1

Σ( xi − x ) 2 = (-4)2 + (-1)2 + 12 (-2)2 + 12 + 52 = 48

s=

48 = 31 . 6 −1

7-2

Sampling and Sampling Distributions 12. a. b. 13. a.

p = 75/150 = .50 p = 55/150 = .3667 x = Σxi / n =

465 = 93 5

b.

Totals

s=

xi

( xi − x )

94 100 85 94 92 465

+1 +7 -8 +1 -1 0

( xi − x ) 2 1 49 64 1 1 116

Σ( xi − x ) 2 116 = = 5.39 n −1 4

14. a.

149/784 = .19

b.

251/784 = .32

c.

Total receiving cash = 149 + 219 + 251 = 619 619/784 = .79

15. a.

b.

x = Σxi / n =

s=

70 = 7 years 10

Σ( xi − x ) 2 20.2 = = 1.5 years n −1 10 − 1

16.

p = 1117/1400 = .80

17. a.

595/1008 = .59

b.

332/1008 = .33

c.

81/1008 = .08

18. a.

E ( x ) = µ = 200

b.

σ x = σ / n = 50 / 100 = 5

c.

Normal with E ( x ) = 200 and σ x = 5

d.

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

7-3

Chapter 7 19. a.

The sampling distribution is normal with E ( x ) = µ = 200

σ x = σ / n = 50 / 100 = 5 For ±5, ( x - µ ) = 5 z=

x−µ

σx

=

5 = 1 Area = .3413 5

2(.3413) = .6826 b.

For ± 10, ( x - µ ) = 10 z=

x−µ

σx

=

10 =2 5

Area = .4772

2(.4772) = .9544

σx =σ / n

20.

σ x = 25 / 50 = 3.54 σ x = 25 / 100 = 2.50 σ x = 25 / 150 = 2.04 σ x = 25 / 200 = 1.77 The standard error of the mean decreases as the sample size increases. 21. a. b.

σ x = σ / n = 10 / 50 = 141 . n / N = 50 / 50,000 = .001 Use σ x = σ / n = 10 / 50 = 141 .

c.

n / N = 50 / 5000 = .01 Use σ x = σ / n = 10 / 50 = 141 .

d.

n / N = 50 / 500 = .10 Use σ x =

N −n σ = N −1 n

500 − 50 10 = 134 . 500 − 1 50

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

7-4

Sampling and Sampling Distributions 22. a.

σ x = σ / n = 4000 / 60 = 516.40

x

51,800

E(x ) The normal distribution is based on the Central Limit Theorem. b.

For n = 120, E ( x ) remains $51,800 and the sampling distribution of x can still be approximated by a normal distribution. However, σ x is reduced to 4000 / 120 = 365.15.

c.

As the sample size is increased, the standard error of the mean, σ 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 σ x , should decrease as the sample size is increased.

23. a.

σ x = σ / n = 4000 / 60 = 516.40

x 51,300 51,800 52,300

z=

52,300 − 51,800 = +.97 Area = .3340 516.40

2(.3340) = .6680 b.

σ x = σ / n = 4000 / 120 = 36515 . z=

52,300 − 51,800 = +1.37 Area = .4147 365.15

2(.4147) = .8294

7-5

Chapter 7

24. a.

Normal distribution, E ( x ) = 4260

σ x = σ / n = 900 / 50 = 127.28 b.

Within $250 P(4010 ≤ x ≤ 4510) z=

4510 − 4260 = 1.96 127.28

Area = .4750

2(.4750) = .95 c.

Within $100 P(4160 ≤ x ≤ 4360) z=

4360 − 4260 = .79 127.28

Area = .2852

2(.2852) = .5704 25. a.

E( x ) = 1020

σ x = σ / n = 100 / 75 = 1155 . b.

z=

1030 − 1020 = .87 11.55

z=

1010 − 1020 = −.87 Area =.3078 11.55

Area =.3078

probability = .3078 + .3078 =.6156 c.

z=

1040 − 1020 = 1.73 11.55

z=

1000 − 1020 = −1.73 Area = .4582 11.55

Area = .4582

probability = .4582 + .4582 = .9164

7-6

Sampling and Sampling Distributions

26. a.

z=

x − 34, 000

σ/ n

Error = x - 34,000 = 250 n = 30

z=

n = 50

z=

n = 100

z=

n = 200

z=

n = 400

z=

250 2000 / 30 250 2000 / 50

= .68

2(.2517) = .5034

= .88

2(.3106) = .6212

250 2000 / 100

= 1.25

2(.3944) = .7888

= 1.77

2(.4616) = .9232

= 2.50

2(.4938) = .9876

250 2000 / 200 250 2000 / 400

b. A larger sample increases the probability that the sample mean will be within a specified distance from the population mean. In the salary example, the probability of being within ±250 of µ ranges from .5036 for a sample of size 30 to .9876 for a sample of size 400. 27. a.

E( x ) = 982

σ x = σ / n = 210 / 40 = 33.2 z=

x −µ

σ/ n

=

100 210 / 40

= 3.01 Area = .4987

2(.4987) = .9974 b.

z=

x −µ

σ/ n

=

25 210 / 40

= .75 Area = .2734

2(.2734) = .5468 c.

28. a.

The sample with n = 40 has a very high probability (.9974) of providing a sample mean within ± $100. However, the sample with n = 40 only has a .5468 probability of providing a sample mean within ± $25. A larger sample size is desirable if the ± $25 is needed. Normal distribution, E ( x ) = 687

σ x = σ / n = 230 / 45 = 34.29

7-7

Chapter 7 b.

Within $100 P(587 ≤ x ≤ 787) z=

787 − 687 = 2.92 34.29

Area = .4982

2(.4982) = .9964 c.

Within $25 P(662 ≤ x ≤ 712) z=

712 − 687 = .73 34.29

Area = .2673

2(.2673) = .5346 d.

Recommend taking a larger sample since there is only a .5346 probability n = 45 will provide the desired result.

µ = 1.46 σ = .15

29. a.

n = 30 z=

x −µ

σ/ n

=

.03 .15 / 30

= 1.10

P(1.43 ≤ x ≤ 1.49) = P(-1.10 ≤ z ≤ 1.10) = 2(.3643) = .7286 b.

n = 50 z=

x −µ .03 = = 1.41 σ / n .15 / 50

P(1.43 ≤ x ≤ 1.49) = P(-1.41 ≤ z ≤ 1.41) = 2(.4207) = .8414 c.

n = 100 z=

x −µ

σ/ n

=

.03 .15 / 100

= 2.00

P(1.43 ≤ x ≤ 1.49) = P(-2 ≤ z ≤ 2) = 2(.4772) = .9544 d.

A sample size of 100 is necessary.

7-8

Sampling and Sampling Distributions 30. a. b.

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

σx =

N −n σ = N −1 n

4000 − 40 8.2 = 129 . 4000 − 1 40

Without the finite population correction factor

σ x = σ / n = 130 . Including the finite population correction factor provides only a slightly different value for σ x than when the correction factor is not used. c.

z=

2 x−µ = = 154 . 130 . 130 .

Area = .4382

2(.4382) = .8764 31. a.

E ( p ) = p = .40 p (1 − p ) .40(.60) = = .0490 100 n

b.

σp =

c.

Normal distribution with E ( p ) = .40 and σ p = .0490

d.

It shows the probability distribution for the sample proportion p .

32. a.

E ( p ) = .40

σp =

z=

p (1 − p ) .40(.60) = = .0346 200 n

p− p

σp

=

.03 = .87 Area = .3078 .0346

2(.3078) = .6156 b.

z=

p− p

σp

=

.05 = 1.44 Area = .4251 .0346

2(.4251) = .8502 33.

σp =

p(1 − p) n

σp =

(.55)(.45) = .0497 100

7-9

Chapter 7

σp =

(.55)(.45) = .0352 200

σp =

(.55)(.45) = .0222 500

σp =

(.55)(.45) = .0157 1000

σ p decreases as n increases 34. a.

σp =

z=

(.30)(.70) = .0458 100

p− p

σp

=

.04 = .87 .0458

Area = 2(.3078) = .6156 b.

σp =

z=

(.30)(.70) = .0324 200

p− p

σp

=

.04 = 1.23 .0324

Area = 2(.3907) = .7814 c.

σp =

z=

(.30)(.70) = .0205 500

p− p

σp

=

.04 = 1.95 .0205

Area = 2(.4744) = 0.9488 d.

σp =

z=

(.30)(.70) = .0145 1000

p− p

σp

=

.04 = 2.76 .0145

Area = 2(.4971) = .9942 e.

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

7 - 10

Sampling and Sampling Distributions 35. a.

σp =

p(1 − p) .30(.70) = = .0458 n 100

p .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) = ? z=

.40 − .30 = 2.18 Area = .4854 .0458

2(.4854) = .9708 c.

P (.25 ≤ p ≤ .35) = ? z=

.35 − .30 = 1.09 Area = .3621 .0458

2(.3621) = .7242 36. a.

Normal distribution, E ( p ) = .56

σp = b.

p(1 − p ) .56(1 − .56) = = .0287 300 n

Within ± .03 z=

.59 − .56 = 1.05 .0287

Area = .3531

2(.3531) = .7062 c.

For n = 600, σ p =

z=

.59 − .56 = 1.48 .0203

.56(1 − .56) = .0203 600

Area = .4306

2(.4306) = .8612

7 - 11

Chapter 7

For n = 1000, σ p =

z=

.59 − .56 = 1.91 .0157

.56(1 − .56) = .0157 1000

Area = .4719

2(.4719) = .9438 37. a.

Normal distribution E ( p ) = .50

σp =

b.

z=

p (1 − p) = n

p− p

σp

=

(.50)(1 − .50) = .0206 589

.04 = 1.94 .0206

2(.4738) = .9476 c.

z=

p− p

σp

=

.03 = 1.46 .0206

2(.4279) = .8558 d.

z=

p− p

σp

=

.02 = .97 .0206

2(.3340) = .6680 38. a.

Normal distribution E( p ) = .25

σp =

b.

z=

p (1 − p) = n

(.25)(.75) = .0306 200

.03 = .98 Area = .3365 .0306

2(.3365) = .6730 c.

z=

.05 = 1.63 Area = .4484 .0306

2(.4484) = .8968

7 - 12

Sampling and Sampling Distributions 39. a.

Normal distribution with E( p ) = p = .25 and

σp =

b.

z=

p (1 − p) .25(1 − .25) = = .0137 1000 n

p− p

σp

=

.03 = 2.19 .0137

P(.22 ≤ p ≤ .28) = P(-2.19 ≤ z ≤ 2.19) = 2(.4857) = .9714 c.

p− p

z=

.25(1 − .25) 500

=

.03 = 1.55 .0194

P(.22 ≤ p ≤ .28) = P(-1.55 ≤ z ≤ 1.55) = 2(.4394) = .8788 40. a.

E ( p ) = .76

σp =

p (1 − p) .76(1 − .76) = = .0214 n 400

Normal distribution because np = 400(.76) = 304 and n(1 - p) = 400(.24) = 96 b.

z=

.79 − .76 = 1.40 .0214

Area = .4192

z=

.73 − .76 = −1.40 .0214

Area = .4192

probability = .4192 + .4192 = .8384 c.

σp =

p (1 − p) .76(1 − .76) = = .0156 n 750

z=

.79 − .76 = 1.92 .0156

Area = .4726

z=

.73 − .76 = −1.92 .0156

Area = .4726

probability = .4726 + .4726 = .9452 41. a.

E( p ) = .17

σp =

p (1 − p) = n

(.17)(1 − .17) = .0133 800

7 - 13

Chapter 7

b.

z=

.19 − .17 = 1.51 .0133

Area = .4345

z=

.15 − .17 = −1.51 .0133

Area = .4345

probability = .4345 + .4345 = .8690 c.

p (1 − p) = n

σp =

(.17)(1 − .17) = .0094 1600

z=

.19 − .17 = 2.13 .0094

Area = .4834

z=

.15 − .17 = −2.13 .0094

Area = .4834

probability = .4834 + .4834 = .9668 42.

112, 145, 73, 324, 293, 875, 318, 618

43. a.

Normal distribution E( x ) = 3

σx =

b.

z=

σ n

1.2

=

x −µ

σ/ n

= .17

50 =

.25 1.2 / 50

= 1.47 Area = .4292

2(.4292) = .8584 44. a.

Normal distribution E( x ) = 31.5

σx =

b.

z=

σ n

=

12 50

1 = .59 1.70

= 170 .

Area = .2224

2(.2224) = .4448 c.

z=

3 = 177 Area = .4616 . 170 .

2(.4616) = .9232

7 - 14

Sampling and Sampling Distributions

45. a.

E( x ) = $24.07

σx =

z=

σ n

=

4.80 120

.50 = 1.14 .44

= .44

Area = .3729

2(.3729) = .7458 b.

z=

1.00 = 2.28 Area = .4887 .44

2(.4887) = .9774

µ = 41,979 σ = 5000

46. a.

σ x = 5000 / 50 = 707

b.

z=

x −µ

σx

=

0 =0 707

P( x > 41,979) = P(z > 0) = .50 c.

z=

x −µ

σx

=

1000 = 1.41 707

P(40,979 ≤ x ≤ 42,979) = P(-1.41 ≤ z ≤ 1.41) = 2(.4207) = .8414 d.

σ x = 5000 / 100 = 500 z=

x −µ

σx

=

1000 = 2.00 500

P(40,979 ≤ x ≤ 42,979) = P(-2 ≤ z ≤ 2) = 2(.4772) = .9544 47. a.

σx =

N −n σ N −1 n

N = 2000

σx =

2000 − 50 144 = 2011 . 2000 − 1 50

N = 5000

σx =

5000 − 50 144 = 20.26 5000 − 1 50

7 - 15

Chapter 7 N = 10,000 10,000 − 50 144 = 20.31 10,000 − 1 50

σx =

Note: With n / N ≤ .05 for all three cases, common statistical practice would be to ignore 144 the finite population correction factor and use σ x = = 20.36 for each case. 50 b.

N = 2000 z=

25 = 1.24 20.11

Area = .3925

2(.3925) = .7850 N = 5000 z=

25 = 1.23 Area = .3907 20.26

2(.3907) = .7814 N = 10,000 z=

25 = 1.23 Area = .3907 20.31

2(.3907) = .7814 All probabilities are approximately .78 48. a.

σx =

σ n

=

500 n

= 20

n = 500/20 = 25 and n = (25)2 = 625

b.

For ± 25, z=

25 = 1.25 Area = .3944 20

2(.3944) = .7888

7 - 16

Sampling and Sampling Distributions 49.

Sampling distribution of x

σx =

0.05

σ n

=

σ 30

0.05

µ

1.9

2.1

x

1.9 + 2.1 = 2 µ = 2

The area between µ = 2 and 2.1 must be .45. An area of .45 in the standard normal table shows z = 1.645. Thus,

µ=

2.1 − 2.0

σ / 30

= 1.645

Solve for σ.

σ= 50.

(.1) 30 = .33 1.645

p = .305 a.

Normal distribution with E( p ) = p = .305 and

σp =

b.

z=

p (1 − p) .305(1 − .305) = = .0326 n 200

p− p

σp

=

.04 = 1.23 .0326

P(.265 ≤ p ≤ .345) = P(-1.23 ≤ z ≤ 1.23) = 2(.3907) = .7814 c.

z=

p− p

σp

=

.02 = .61 .0326

P(.285 ≤ p ≤ .325) = P(-.61 ≤ z ≤ .61) = 2(.2291) = .4582

7 - 17

Chapter 7

σp =

51.

p (1 − p) = n

(.40)(.60) = .0245 400

P ( p ≥ .375) = ? z=

.375 − .40 = −1.02 Area = .3461 .0245

P ( p ≥ .375) = .3461 + .5000 = .8461 52. a.

σp =

z=

p (1 − p) = n

p− p

σp

=

(.71)(1 − .71) = .0243 350

.05 = 2.06 Area = .4803 .0243

2(.4803) = .9606 b.

z=

p− p

σp

=

.75 − .71 = 1.65 Area = .4505 .0243

P ( p ≥ .75) = .5000 - .4505 = .0495 53. a.

Normal distribution with E ( p ) = .15 and

σp = b.

p (1 − p) = n

(.15)(.85) = .0292 150

P (.12 ≤ p ≤ .18) = ? z=

.18 − .15 = 1.03 Area = .3485 .0292

2(.3485) = .6970 54. a.

σp =

p(1 − p) = n

.25(.75) =.0625 n

Solve for n n=

b.

.25(.75) = 48 (.0625) 2

Normal distribution with E( p ) = .25 and σ x = .0625

7 - 18

Sampling and Sampling Distributions

c.

P ( p ≥ .30) = ? z=

.30 − .25 = .80 Area = .2881 .0625

P ( p ≥ .30) = .5000 - .2881 = .2119

7 - 19