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Chapter 7
Chapter 7
Sampling and Sampling Distributions
Chapter 7
Sampling and Sampling Distributions
Learning Objectives1.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 () and the sampling distribution of the sample proportion ().
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
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
b.
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.
b.
= (-4)2 + (-1)2 + 12 (-2)2 + 12 + 52 = 48
s =
12.a.
= 75/150 = .50
b.
= 55/150 = .3667
13.a.
b.
94+11
100+749
85-864
94+11
92-1 1
Totals4650116
14.a.149/784 = .19
b.251/784 = .32
c.Total receiving cash = 149 + 219 + 251 = 619
619/784 = .79
15.a.
years
b.
years
16.
= 1117/1400 = .80
17.a.595/1008 = .59
b.332/1008 = .33
c.81/1008 = .08
18.a.
b.
c.Normal with E () = 200 and = 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 is within a specified from 19.a.The sampling distribution is normal with
E () = = 200
For 5, (- ) = 5
Area = .3413
2(.3413) = .6826
b.For 10, (- ) = 10
Area = .4772
2(.4772) = .9544
20.
The standard error of the mean decreases as the sample size increases.
21.a.
b.n / N = 50 / 50,000 = .001
Use
c.n / N = 50 / 5000 = .01
Use
d.n / N = 50 / 500 = .10
Note: Only case (d) where n /N = .10 requires the use of the finite population correction factor.
22.a.
The normal distribution is based on the Central Limit Theorem.
b.For n = 120, E () remains $51,800 and the sampling distribution of can still be approximated by a normal distribution. However, is reduced to = 365.15.
c.As the sample size is increased, the standard error of the mean, , 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, should decrease as the sample size is increased.
23.a.
Area = .3340
2(.3340) = .6680
b.
Area = .4147
2(.4147) = .8294
24.a.Normal distribution,
b.Within $250
P(4010 (( 4510)
Area = .4750
2(.4750) = .95
c.Within $100
P(4160 ( ( 4360)
Area = .2852
2(.2852) = .5704
25.a.E() = 1020
b.
Area =.3078
Area =.3078
probability = .3078 + .3078 =.6156
c.
Area = .4582
Area = .4582
probability = .4582 + .4582 = .9164
26.a.
Error = - 34,000 = 250
n = 30
2(.2517) = .5034
n = 50
2(.3106) = .6212
n = 100
2(.3944) = .7888
n = 200
2(.4616) = .9232
n = 400
2(.4938) = .9876
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.
= 10994
Area = .4987
2(.4987) = .9974
b.
Area = .2734
2(.2734) = .5468
c. The sample with n = 40 has a very high probability (.9974) of providing a sample mean within ( NT$1,000. However, the sample with n = 40 only has a .5468 probability of providing a sample mean within ( NT$250. A larger sample size is desirable if the ( N$250 is needed.
28.a.
= 3909
b.Within NT$570
P(3339 ( ( 4476)
Area = .4982
2(.4982) = .9964
c.Within NT$142
P(3980 ( ( 3838)
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.
29.
( = 1.46 ( = .15
a.n = 30
P(1.43 ( ( 1.49) = P(-1.10 ( z ( 1.10) = 2(.3643) = .7286
b.n = 50
P(1.43 ( ( 1.49) = P(-1.41 ( z ( 1.41) = 2(.4207) = .8414
c.n = 100
P(1.43 ( ( 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
Without the finite population correction factor
Including the finite population correction factor provides only a slightly different value for than when the correction factor is not used.
c.
Area = .4382
2(.4382) = .8764
31.a.E () = p = .40
b.
c.Normal distribution with E () = .40 and = .0490
d.It shows the probability distribution for the sample proportion .
32.a.E () = .40
Area = .3078
2(.3078) = .6156
b.
Area = .4251
2(.4251) = .8502
33.
decreases as n increases
34.a.
Area = 2(.3078) = .6156
b.
Area = 2(.3907) = .7814
c.
Area = 2(.4744) = 0.9488
d.
Area = 2(.4971) = .9942
e.With a larger sample, there is a higher probability will be within .04 of the population proportion p.
35.a. 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 .40) = ?
Area = .4854
2(.4854) = .9708
c.P (.25 .35) = ?
Area = .3621
2(.3621) = .7242
36.a.Normal distribution,= .49
b.Within .03
Area = .3485
2(.3485) = .6970
c.For n = 600,
Area = .4292
2(.4292) = .8584
For n = 1000,
Area = .4713
2(.4713) = .942637.a.Normal distribution
E () = .50
b.
2(.4738) = .9476
c.
2(.4279) = .8558
d.
2(.3340) = .6680
38.a.Normal distribution
E() = .25
b.
Area = .3365
2(.3365) = .6730
c.
Area = .4484
2(.4484) = .8968
39.a.Normal distribution with E() = p = .25 and
b.
P(.22 ( ( .28) = P(-2.19 ( z ( 2.19) = 2(.4857) = .9714
c.
P(.22 ( ( .28) = P(-1.55 ( z ( 1.55) = 2(.4394) = .8788
40.a.E () = .76
Normal distribution because np = 400(.76) = 304 and n(1 - p) = 400(.24) = 96
b.
Area = .4192
Area = .4192
probability = .4192 + .4192 = .8384
c.
Area = .4726
Area = .4726
probability = .4726 + .4726 = .9452
41.a.E() = .25
b.
Area = .4049
Area = .4049
probability = .4049 + .4049 = .8098
c.
Area = .4678
Area = .4678
probability = .4678 + .4678 = .935642.
112, 145, 73, 324, 293, 875, 318, 618
43.a.Normal distribution
E() = 3
b.
Area = .4292
2(.4292) = .8584
44.a.Normal distribution
E() = 31.5
b.
Area = .2224
2(.2224) = .4448
c.
Area = .4616
2(.4616) = .9232
45.a.E() = 34944
Area = .3729
2(.3729) = .7458
b.
Area = .4887
2(.4887) = .9774
46.
( = 56953 ( = 16104
a.
b.
P(>56953) = P(z > 0) = .50
c.
P(53753 (( 60153) = P(-1.41 ( z ( 1.41) = 2(.4207) = .8414
d.
P(53,753((60,153) = P(-1.99 ( z ( 1.99) = 2(.4767) = .9534
47.a.
N = 2000
N = 5000
N = 10,000
Note: With n / N .05 for all three cases, common statistical practice would be to ignore the finite population correction factor and use for each case.
b.N = 2000
Area = .3925
2(.3925) = .7850
N = 5000
Area = .3907
2(.3907) = .7814
N = 10,000
Area = .3907
2(.3907) = .7814
All probabilities are approximately .78
48.a.
= 500/20 = 25 and n = (25)2 = 625
b.For 25,
Area = .3944
2(.3944) = .7888
49.
Sampling distribution of
The area between = 2 and 2.1 must be .45. An area of .45 in the standard normal table shows
z = 1.645.
Thus,
Solve for
50.
p = .305
a.Normal distribution with E() = p = .305 and
b.
P(.265 ( ( .345) = P(-1.23 ( z ( 1.23) = 2(.3907) = .7814
c.
P(.285 ( ( .325) = P(-.61 ( z ( .61) = 2(.2291) = .4582
51.
P ( .375) = ?
Area = .3461
P ( .375) = .3461 + .5000 = .8461
52.a.
Area = .4726
2(.4726) = .9452
b.
Area = .4370
P ( .65) = .5000 - .4370 = .0063
53.a.Normal distribution with E () = .15 and
b.P (.12 .18) = ?
Area = .3485
2(.3485) = .6970
54.a.
Solve for n
b.Normal distribution with E() = .25 and = .0625
c.P ( .30) = ?
Area = .2881
P ( .30) = .5000 - .2881 = .2119
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