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