Sampling and Sampling Distributions

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 sampling distribution sample statistic finite population correction factor simple random sampling standard error sampling without replacement central limit theorem sampling with replacement unbiased point estimator relative efficiency point estimate 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. 459, 147, 385, 113, 340, 401, 215, 2, 33, 348

4. a. 6, 8, 5, 4, 1

Nasdaq 100, Oracle, Microsoft, Lucent, Applied Materials

N ! 10! 3,628,500 b.    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

54 11. a. x  x / n   9 i 6

(x  x) 2 b. s  i n 1

2 2 2 2 2 2 2 (xi  x) = (-4) + (-1) + 1 (-2) + 1 + 5 = 48

48 s =  3.1 6 1

12. a. p = 75/150 = .50

7 - 2 Sampling and Sampling Distributions

b. p = 55/150 = .3667

465 13. a. x  x / n   93 i 5 b. 2 xi (xi  x) (xi  x) 94 +1 1 100 +7 49 85 -8 64 94 +1 1 92 -1 1 Totals 465 0 116

(x  x) 2 116 s  i   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

70 15. a. x  x/ n   7 years i 10

(x  x )2 20.2 b. s i   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)    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 with random samples of size 100. This distribution can be used to compute the probability that x is within a specified  from 

19. a. The sampling distribution is normal with

7 - 3 Chapter 7

E ( x ) =  = 200

 x   / n  50 / 100  5

For 5, ( x -  ) = 5

x   5 z    1 Area = .3413  x 5

2(.3413) = .6826

b. For  10, ( x -  ) = 10

x   10 z    2 Area = .4772  x 5

2(.4772) = .9544

20.  x   / n

 x 25/ 50  3.54

 x 25/ 100  2.50

 x 25/ 150  2.04

 x 25/ 200  1.77

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

21. a.  x   / n  10 / 50  1.41

b. n / N = 50 / 50,000 = .001

Use  x   / n  10 / 50  1.41

c. n / N = 50 / 5000 = .01

Use  x   / n  10 / 50  1.41

d. n / N = 50 / 500 = .10

N  n  500  50 10 Use  x    1.34 N 1 n 500 1 50

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

22. a.

 x   / n  4000 / 60  516.40

x 51,800

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,  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.

 x   / n  4000 / 60  516.40

x 51,300 51,800 52,300 52,300 51,800 z   .97 Area = .3340 516.40

2(.3340) = .6680

b.  x   / n  4000 / 120  365.15

52,300 51,800 z   1.37 Area = .4147 365.15

2(.4147) = .8294

24. a. Normal distribution, E( x ) 4260

x   /n  900 / 50  127.28

7 - 5 Chapter 7

b. Within $250

P(4010  x  4510)

4510 4260 z   1.96 Area = .4750 127.28

2(.4750) = .95

c. Within $100

P(4160  x  4360)

4360 4260 z   .79 Area = .2852 127.28

2(.2852) = .5704

25. a. E( x ) = 1020

 x   / n  100 / 75  11.55

1030 1020 b. z   .87 Area =.3078 11.55

1010 1020 z   .87 Area =.3078 11.55

probability = .3078 + .3078 =.6156

1040 1020 c. z   1.73 Area = .4582 11.55

1000 1020 z   1.73 Area = .4582 11.55

probability = .4582 + .4582 = .9164

x  34,000 26. a. z   / n

Error = x - 34,000 = 250

250 n = 30 z   .68 2(.2517) = .5034 2000 / 30

250 n = 50 z   .88 2(.3106) = .6212 2000 / 50

250 n = 100 z   1.25 2(.3944) = .7888 2000 / 100

250 n = 200 z   1.77 2(.4616) = .9232 2000 / 200

250 n = 400 z   2.50 2(.4938) = .9876 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  ranges from .5036 for a sample of size 30 to .9876 for a sample of size 400.

27. a. E( x ) = 982

x   /n  210 / 40  33.2

x   100 z    3.01 Area = .4987  /n 210 / 40

2(.4987) = .9974

x   25 b. z    .75 Area = .2734  /n 210 / 40

2(.2734) = .5468

c. 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.

28. a. Normal distribution, E( x ) = 687

x   /n  230 / 45  34.29

b. Within $100

P(587  x  787)

7 - 7 Chapter 7

787 687 z   2.92 Area = .4982 34.29

2(.4982) = .9964

c. Within $25

P(662  x  712)

712 687 z   .73 Area = .2673 34.29

2(.2673) = .5346

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

29.  = 1.46  = .15

a. n = 30

x   .03 z    1.10  /n .15/ 30

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

b. n = 50

x   .03 z    1.41  /n .15/ 50

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

c. n = 100

x   .03 z    2.00  /n .15/ 100

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

d. A sample size of 100 is necessary.

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

b. With the finite population correction factor

N  n  4000  40 8.2  x    1.29 N 1 n 4000 1 40

Without the finite population correction factor

 x   / n  1.30

Including the finite population correction factor provides only a slightly different value for  x than when the correction factor is not used.

x   2 c. z    1.54 Area = .4382 1.30 1.30

2(.4382) = .8764

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

p(1 p ) .40(.60) b.     .0490 p n 100

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

d. It shows the probability distribution for the sample proportion p .

32. a. E ( p ) = .40

p(1 p ) .40(.60)     .0346 p n 200

p p .03 z    .87 Area = .3078  p .0346

2(.3078) = .6156

p p .05 b. z    1.44 Area = .4251  p .0346

2(.4251) = .8502

p(1 p) 33.   p n

(.55)(.45)    .0497 p 100 (.55)(.45)    .0352 p 200

7 - 9 Chapter 7

(.55)(.45)    .0222 p 500

(.55)(.45)    .0157 p 1000

 p decreases as n increases

(.30)(.70) 34. a.    .0458 p 100

p p .04 z    .87  p .0458

Area = 2(.3078) = .6156

(.30)(.70) b.    .0324 p 200

p p .04 z    1.23  p .0324

Area = 2(.3907) = .7814

(.30)(.70) c.    .0205 p 500

p p .04 z    1.95  p .0205

Area = 2(.4744) = 0.9488

(.30)(.70) d.    .0145 p 1000

p p .04 z    2.76  p .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.

35. a.

p(1 p ) .30(.70)     .0458 p n 100

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) = ?

.40 .30 z   2.18 Area = .4854 .0458

2(.4854) = .9708

c. P (.25  p  .35) = ?

.35 .30 z   1.09 Area = .3621 .0458

2(.3621) = .7242

36. a. Normal distribution, E( p ) = .56

p(1 p ) .56(1  .56)     .0287 p n 300

b. Within ± .03

.59 .56 z   1.05 Area = .3531 .0287

2(.3531) = .7062

.56(1 .56) c. For n = 600,    .0203 p 600

.59 .56 z   1.48 Area = .4306 .0203

2(.4306) = .8612

.56(1 .56) For n = 1000,    .0157 p 1000

7 - 11 Chapter 7

.59 .56 z   1.91 Area = .4719 .0157

2(.4719) = .9438

37. a. Normal distribution

E ( p ) = .50

p(1 p ) (.50)(1  .50)     .0206 p n 589

p p .04 b. z    1.94  p .0206

2(.4738) = .9476

p p .03 c. z    1.46  p .0206

2(.4279) = .8558

p p .02 d. z    .97  p .0206

2(.3340) = .6680

E( p ) = .25

p(1 p ) (.25)(.75)     .0306 p n 200

.03 b. z   .98 Area = .3365 .0306

2(.3365) = .6730

.05 c. z   1.63 Area = .4484 .0306

2(.4484) = .8968

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

p(1 p ) .25(1  .25)     .0137 p n 1000

p p .03 b. z    2.19  p .0137

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

p p .03 z    1.55 c. .25(1 .25) .0194 500

P(.22  p  .28) = P(-1.55  z  1.55) = 2(.4394) = .8788

40. a. E ( p ) = .76

p(1 p ) .76(1  .76)     .0214 p n 400

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

.79 .76 b. z   1.40 Area = .4192 .0214

.73 .76 z   1.40 Area = .4192 .0214

probability = .4192 + .4192 = .8384

p(1 p ) .76(1  .76) c.     .0156 p n 750

.79 .76 z   1.92 Area = .4726 .0156

.73 .76 z   1.92 Area = .4726 .0156

probability = .4726 + .4726 = .9452

41. a. E( p ) = .17

p(1 p ) (.17)(1  .17)     .0133 p n 800

.19 .17 b. z   1.51 Area = .4345 .0133

7 - 13 Chapter 7

.15 .17 z   1.51 Area = .4345 .0133

probability = .4345 + .4345 = .8690

p(1 p ) (.17)(1  .17) c.     .0094 p n 1600

.19 .17 z   2.13 Area = .4834 .0094

.15 .17 z   2.13 Area = .4834 .0094

probability = .4834 + .4834 = .9668

42. 112, 145, 73, 324, 293, 875, 318, 618

E( x ) = 3

 1.2  x    .17 n 50

x   .25 b. z    1.47 Area = .4292  /n 1.2 / 50

2(.4292) = .8584

E( x ) = 31.5

 12  x    1.70 n 50

1 b. z   .59 Area = .2224 1.70

2(.2224) = .4448

3 c. z   1.77 Area = .4616 1.70

2(.4616) = .9232

45. a. E( x ) = $24.07

 4.80  x    .44 n 120

.50 z   1.14 Area = .3729 .44

2(.3729) = .7458

1.00 b. z   2.28 Area = .4887 .44

2(.4887) = .9774

46.  = 41,979  = 5000

a.  x 5000 / 50  707

x   0 b. z    0  x 707

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

x   1000 c. z    1.41  x 707

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

d.  x 5000 / 100  500

x   1000 z    2.00  x 500

P(40,979  x  42,979) = P(-2  z  2) = 2(.4772) = .9544

N  n  47. a.  x  N 1 n

N = 2000

2000  50 144  x   20.11 2000 1 50

N = 5000

5000  50 144  x   20.26 5000 1 50 N = 10,000

7 - 15 Chapter 7

10,000  50 144  x   20.31 10,000 1 50

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

b. N = 2000

25 z   1.24 Area = .3925 20.11

2(.3925) = .7850

N = 5000

25 z   1.23 Area = .3907 20.26

2(.3907) = .7814

N = 10,000

25 z   1.23 Area = .3907 20.31

2(.3907) = .7814

All probabilities are approximately .78

 500 48. a.  x    20 n n

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

b. For  25,

25 z   1.25 Area = .3944 20

2(.3944) = .7888

49. Sampling distribution of x

   x   n 30

0.05 0.05

x  1.9 2.1

1.9 + 2.1  = = 2 2

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

2.1 2.0    1.645  / 30

Solve for 

(.1) 30    .33 1.645

50. p = .305

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

p(1 p ) .305(1  .305)     .0326 p n 200

p p .04 b. z    1.23  p .0326

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

p p .02 c. z    .61  p .0326

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

7 - 17 Chapter 7

p(1 p ) (.40)(.60) 51.     .0245 p n 400

P ( p  .375) = ?

.375 .40 z   1.02 Area = .3461 .0245

P ( p  .375) = .3461 + .5000 = .8461

p(1 p ) (.71)(1  .71) 52. a.     .0243 p n 350

p p .05 z    2.06 Area = .4803  p .0243

2(.4803) = .9606

p p .75  .71 b. z    1.65 Area = .4505  p .0243

P ( p  .75) = .5000 - .4505 = .0495

53. a. Normal distribution with E ( p ) = .15 and

p(1 p ) (.15)(.85)     .0292 p n 150

b. P (.12  p  .18) = ?

.18 .15 z   1.03 Area = .3485 .0292

2(.3485) = .6970

p(1 p) .25(.75) 54. a.    .0625 p n n

Solve for n

.25(.75) n   48 (.0625) 2

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

c. P ( p  .30) = ?

.30 .25 z   .80 Area = .2881 .0625

P ( p  .30) = .5000 - .2881 = .2119

7 - 19