sampling and sampling distributions chap 7

8/13/2019 sampling and sampling distributions chap 7

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7 - 1

Chapter 7

Sampling and Sampling Distributions

Learning Objectives

1. Understand the importance of sampling and how results from samples can be used to provideestimates of population characteristics such as the population mean, the population standarddeviation 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 thesampling 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 distributionsample statistic finite population correction factorsimple random sampling standard error

sampling without replacement central limit theoremsampling with replacement unbiasedpoint estimator relative efficiencypoint estimate consistency


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Chapter 7

7 - 2

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 isalready 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, thesimple 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

b.! 10! 3,628,500

252!( )! 5!(10 5)! (120)(120)

N

n N n= = =

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 x ni

= = = /54

69

b. s x xn

i=

( )

2

1

( )x xi

2 = (-4)2+ (-1)2+ 12(-2)2+ 12+ 52 = 48

s=48

6 131

= .


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7 - 3

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

b. p = 55/150 = .3667

13. a. x x ni

= = = /465

593

b.

xi

( )x xi ( )x xi

2

94 +1 1

100 +7 4985 -8 6494 +1 192 -1 1

Totals 465 0 116

s x x

n

i=

= =

( ).

2

1

116

4539

14. a. 149/784 = .19

b. 251/784 = .32

c. Total receiving cash = 149 + 219 + 251 = 619

619/784 = .79

15. a.70

/ 710

ix x n= = = years

b.2

( ) 20.2 1.51 10 1

ix xsn

= = =

years

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 randomsamples of size 100. This distribution can be used to compute the probability that x is within a

specified from .


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Chapter 7

7 - 4

19. a. The sampling distribution is normal with

E(x ) = = 200

x n= = =/ /50 100 5

For 5, (x - ) = 5

z x

x

=

= =

5

51 Area = .3413

2(.3413) = .6826

b. For 10, (x - ) = 10

z x

x

=

= =

10

52 Area = .4772

2(.4772) = .9544

20. x

n= /

25 / 50 3.54x

= =

25 / 100 2.50x = =

25 / 150 2.04x

= =

25 / 200 1.77x

= =

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

21. a. x

n= = =/ / .10 50 141

b. 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

50134.

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


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7 - 5

x n= = =/ / .4000 60 516 40

x

x n= = =/ / .4000 60 516 40

22. a.

( )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, 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 ofx , should

decrease as the sample size is increased.

23. a.

52,300 51,800.97

516.40z

= = + Area = .3340

2(.3340) = .6680

b. x

n= = =/ / .4000 120 36515

52,300 51,8001.37

365.15z

= = + Area = .4147

2(.4147) = .8294

51,800

51,80051,300 52,300

x


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Chapter 7

7 - 6

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

/ 900 / 50 127.28x n = = =

b. Within $250

P(4010 x4510)

4510 42601.96

127.28z

= = Area = .4750

2(.4750) = .95

c. Within $100

P(4160 x4360)

4360 4260

.79127.28z

= = Area = .2852

2(.2852) = .5704

25. a. E(x ) = 1020

x n= = =/ / .100 75 1155

b.1030 1020

.8711.55

z

= = Area =.3078

1010 1020

.8711.55z

= = Area =.3078

probability = .3078 + .3078 =.6156

c.1040 1020

1.7311.55

z

= = Area = .4582

1000 10201.73

11.55z

= = Area = .4582

probability = .4582 + .4582 = .9164


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7 - 7

26. a.34,000

/

xz

n

=

Error = x - 34,000 = 250

n= 30

250

.682000 / 30z= = 2(.2517) = .5034

n= 50250

.882000 / 50

z= = 2(.3106) = .6212

n= 100250

1.252000 / 100

z= = 2(.3944) = .7888

n= 200250

1.772000 / 200

z= = 2(.4616) = .9232

n= 400 250 2.502000 / 400

z= = 2(.4938) = .9876

b. A larger sample increases the probability that the sample mean will be within a specifieddistance 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

/ 210 / 40 33.2x

n = = =

1003.01

/ 210 / 40

xz

n

= = = Area = .4987

2(.4987) = .9974

b.25

.75/ 210 / 40

xz

n

= = = Area = .2734

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

/ 230 / 45 34.29x

n = = =


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Chapter 7

7 - 8

b. Within $100

P(587 x787)

787 6872.92

34.29z

= = Area = .4982

2(.4982) = .9964

c. Within $25

P(662 x712)

712 687.73

34.29z

= = Area = .2673

2(.2673) = .5346

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

29. = 1.46 = .15

a. n= 30

.031.10

/ .15 / 30

xz

n

= = =

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

b. n= 50

.031.41

/ .15 / 50

xz

n

= = =

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

c. n= 100

.032.00

/ .15 / 100

xz

n

= = =

P(1.43 x 1.49) = P(-2 z2) = 2(.4772) = .9544

d. A sample size of 100 is necessary.


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7 - 9

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

b. With the finite population correction factor

x

N n

N n=

=

=

1

4000 40

4000 1

8 2

40129

..

Without the finite population correction factor

x

n= =/ .130

Including the finite population correction factor provides only a slightly different value for x than

when the correction factor is not used.

c. z x

=

= =

130

2

130154

. .. Area = .4382

2(.4382) = .8764

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

b.(1 ) .40(.60)

.0490100

p

p p

n

= = =

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

(1 ) .40(.60).0346

200p

p p

n

= = =

.03.87

.0346p

p pz

= = = Area = .3078

2(.3078) = .6156

b..05

1.44.0346

p

p pz

= = = Area = .4251

2(.4251) = .8502

33. pp p

n=

( )1

(.55)(.45).0497

100p = =


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Chapter 7

7 - 10

(.55)(.45).0352

200p

= =

(.55)(.45).0222

500p = =

(.55)(.45).0157

1000p = =

p

decreases as nincreases

34. a.(.30)(.70)

.0458100

p = =

.04.87

.0458p

p pz

= = =

Area = 2(.3078) = .6156

b.(.30)(.70)

.0324200

p = =

.041.23

.0324p

p pz

= = =

Area = 2(.3907) = .7814

c. (.30)(.70) .0205500

p = =

.041.95

.0205p

p pz

= = =

Area = 2(.4744) = 0.9488

d.(.30)(.70)

.01451000

p = =

.042.76.0145

p

p pz

= = =

Area = 2(.4971) = .9942

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

proportion p.


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7 - 11

35. a.

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

b. P(.20 p .40) = ?

.40 .30 2.18.0458

z = = Area = .4854

2(.4854) = .9708

c. P(.25 p .35) = ?

.35 .301.09

.0458z

= = Area = .3621

2(.3621) = .7242

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

(1 ) .56(1 .56).0287

300p

p p

n

= = =

b. Within .03

.59 .561.05

.0287z

= = Area = .3531

2(.3531) = .7062

c. For n= 600, .56(1 .56) .0203600

p= =

.59 .561.48

.0203z

= = Area = .4306

2(.4306) = .8612

.30

(1 ) .30(.70).0458

100p

p p

n

= = =

p


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Chapter 7

7 - 12

For n= 1000,.56(1 .56)

.01571000

p

= =

.59 .561.91

.0157z

= = Area = .4719

2(.4719) = .9438

37. a. Normal distribution

E( p ) = .50

(1 ) (.50)(1 .50).0206

589p

p p

n

= = =

b..04

1.94.0206

p

p pz

= = =

2(.4738) = .9476

c..03

1.46.0206

p

p pz

= = =

2(.4279) = .8558

d..02

.97.0206

p

p pz

= = =

2(.3340) = .6680


E( p ) = .25

(1 ) (.25)(.75).0306

200p

p p

n

= = =

b..03

.98.0306

z= = Area = .3365

2(.3365) = .6730

c..05

1.63.0306

z= = Area = .4484

2(.4484) = .8968


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7 - 13

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

(1 ) .25(1 .25).0137

1000p

p p

n

= = =

b. .03 2.19.0137

p

p pz= = =

P(.22 p .28) = P(-2.19 z2.19) = 2(.4857) = .9714

c..03

1.55.0194.25(1 .25)

500

p pz

= = =

P(.22 p .28) = P(-1.55 z1.55) = 2(.4394) = .8788

40. a. E( p ) = .76

(1 ) .76(1 .76).0214

400p

p p

n

= = =

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

b..79 .76

1.40.0214

z

= = Area = .4192

.73 .761.40

.0214z

= = Area = .4192

probability = .4192 + .4192 = .8384

c.(1 ) .76(1 .76)

.0156750

p

p p

n

= = =

.79 .761.92

.0156z

= = Area = .4726

.73 .761.92

.0156z

= = Area = .4726

probability = .4726 + .4726 = .9452

41. a. E( p ) = .17

(1 ) (.17)(1 .17).0133

800p

p p

n

= = =


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Chapter 7

7 - 14

b..19 .17

1.51.0133

z

= = Area = .4345

.15 .171.51

.0133z

= = Area = .4345

probability = .4345 + .4345 = .8690

c.(1 ) (.17)(1 .17)

.00941600

p

p p

n

= = =

.19 .172.13

.0094z

= = Area = .4834

.15 .172.13

.0094z

= = Area = .4834

probability = .4834 + .4834 = .9668

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


E(x ) = 3

1.2.17

50x

n

= = =

b..25

1.47

/ 1.2 / 50

xz

n

= = = Area = .4292

2(.4292) = .8584


E(x ) = 31.5

xn

= = =12

50170.

b.1

.591.70

z= = Area = .2224

2(.2224) = .4448

c. z= =3

170177

.. Area = .4616

2(.4616) = .9232


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7 - 15

45. a. E(x ) = $24.07

4.80.44

120x

n

= = =

.501.14

.44z= = Area = .3729

2(.3729) = .7458

b.1.00

2.28.44

z= = Area = .4887

2(.4887) = .9774

46. = 41,979 = 5000

a. 5000 / 50 707x = =

b.0

0707

x

xz

= = =

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

c.1000

1.41707

x

xz

= = =

P(40,979 x42,979) = P(-1.41 z1.41) = 2(.4207) = .8414

d. 5000 / 100 500x

= =

10002.00

500x

xz

= = =

P(40,979 x42,979) = P(-2 z2) = 2(.4772) = .9544

47. a.

x

N n

N n=

1

N = 2000

x=

=

2000 50

2000 1

144

502011.

N = 5000

x =

=

5000 50

5000 1

144

502026.


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Chapter 7

7 - 16

N = 10,000

x=

=

10 000 50

10 000 1

144

502031

,

,.

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

the finite population correction factor and use x = =144

502036. for each case.

b. N= 2000

251.24

20.11z= = Area = .3925

2(.3925) = .7850

N= 5000

251.23

20.26z= = Area = .3907

2(.3907) = .7814

N= 10,000

251.23

20.31z= = Area = .3907

2(.3907) = .7814

All probabilities are approximately .78

48. a.

xn n

= = =500

20

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

b. For 25,

251.25

20z= = Area = .3944

2(.3944) = .7888


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7 - 17

49. Sampling distribution of x

=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 showsz = 1.645.

Thus,

2.1 2.01.645

/ 30

= =

Solve for .

(.1) 30.33

1.645= =

50. p= .305

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

(1 ) .305(1 .305).0326

200p

p p

n

= = =

b..04

1.23.0326

p

p pz

= = =

P(.265 p .345) = P(-1.23 z1.23) = 2(.3907) = .7814

c.

.02

.61.0326p

p p

z

= = =

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

1.9 2.1

0.05 0.05

x

xn

= =30


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Chapter 7

7 - 18

51.(1 ) (.40)(.60)

.0245400

p

p p

n

= = =

P( p .375) = ?

.375 .40 1.02.0245

z = = Area = .3461

P( p .375) = .3461 + .5000 = .8461

52. a.(1 ) (.71)(1 .71)

.0243350

p

p p

n

= = =

.052.06

.0243p

p pz

= = = Area = .4803

2(.4803) = .9606

b..75 .71

1.65.0243

p

p pz

= = = Area = .4505

P( p .75) = .5000 - .4505 = .0495

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

(1 ) (.15)(.85).0292

150p

p p

n

= = =

b. P(.12 p .18) = ?

.18 .151.03

.0292z

= = Area = .3485

2(.3485) = .6970

54. a. pp p

n n=

= =

( ) . (. ).

1 25 750625

Solve for n

n= =. (. )

(. )

25 75

062548

2

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


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7 19

c. P( p .30) = ?

.30 .25.80

.0625z

= = Area = .2881

P( p .30) = .5000 - .2881 = .2119

sampling and sampling distributions chap 7

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