Download 1 Chapter 7 Problems and Solutions

1
Chapter 7 Problems and Solutions - Sampling and Sampling Distributions
13. The American Association of Individual Investors (AAII) polls its subscribers on a weekly
basis to determine the number who are bullish, bearish, or neutral on the short-term prospects
for the stock market. Its findings for the week ending March 2, 2006, are consistent with the
following sample results (AAII website, March 7, 2006).
Bullish 409
Neutral 299
Bearish 291
Develop a point estimate of the following population parameters.
a.
b.
c.
The proportion of all AAII subscribers who are bullish on the stock market.
The proportion of all AAII subscribers who are neutral on the stock market.
The proportion of all AAII subscribers who are bearish on the stock market.
16. Assume the population standard deviation is σ = 25. Compute the standard error of the mean, σx¯ ,
for sample sizes of 50, 100, 150, and 200. What can you say about the size of the standard error of the
mean as the sample size is increased?
19. In the EAI sampling problem (see Figure 7.8), we showed that for n = 30, there was .5064
probability of obtaining a sample mean within $500 of the population mean.
a. What is the probability that is within $500 of the population mean if a sample of size 60 is used?
b. Answer part (a) for a sample of size 120.
22. The mean annual cost of automobile insurance is $939 (CNBC, February 23, 2006).
Assume that the standard deviation is σ = $245.
a. What is the probability that a simple random sample of automobile insurance policies will have a
sample mean within $25 of the population mean for each of the following sample sizes: 30, 50, 100, and
400?
b. What is the advantage of a larger sample size when attempting to estimate the population mean?
24. The average score for male golfers is 95 and the average score for female golfers is 106
(Golf Digest, April 2006). Use these values as the population means for men and women and assume
that the population standard deviation is σ = 14 strokes for both. A simple random sample of 30 male
golfers and another simple random sample of 45 female golfers will be taken.
a. Show the sampling distribution of x̄ for male golfers.
b. What is the probability that the sample mean is within three strokes of the population mean for the
sample of male golfers?
c. What is the probability that the sample mean is within three strokes of the population mean for the
sample of female golfers?
d. In which case, part (b) or part (c), is the probability of obtaining a sample mean within three strokes
of the population mean higher? Why?
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Chapter 7 Solutions
13. a. 409/999 = .41
b. 299/999 = .30
c. 291/999 = .29
x  / n
16. 16.
 x  25 / 50  3.54
 x  25/ 100  2.50
 x  25/ 150  2.04
 x  25/ 200  1.77
19. a. With a sample of size 60
At x = 52,300,
z
x 
4000
60
 516.40
52,300  51,800
 .97
516.40
P( x ≤ 52,300) = P(z ≤ .97) = .8340 At x = 51,300,
z
51,300  51,800
 .97
516.40
( x < 51,300) = P(z < -.97) = .1660 P(51,300 ≤ x ≤ 52,300) = .8340 - .1660 = .6680
b.
x 
At x = 52,300,
At x = 51,300,
4000
120
 365.15
z
52,300  51,800
 1.37
365.15
z
51,300  51,800
 1.37
365.15
P( x ≤ 52,300) = P(z ≤ 1.37) = .9147
P( x < 51,300) = P(z < -1.37) = .0853
P(51,300 ≤ x ≤ 52,300) = .9147 - .0853 = .8294
P(190  x  210)
= .9772 - .0228 = .9544
22. a.
z
x  939
/ n
Within  25 means x - 939 must be between -25 and +25.
The z value for x - 939 = -25 is just the negative of the z value for x - 939 = 25. So
we just show the computation of z for x - 939 = 25.
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n = 30
n = 50
n = 100
n = 400
z
z
z
z
25
245 / 30
25
245 / 50
 .56
P(-.56 ≤ z ≤ .56) = .7123 - .2877 = .4246
 .72
25
245 / 100
25
245 / 400
 1.02
 2.04
P(-.72 ≤ z ≤ .72) = .7642 - .2358 = .5284
P(-1.02 ≤ z ≤ 1.02) = .8461 - .1539 = .6922
P(-2.04 ≤ z ≤ 2.04) = .9793 - .0207 = .9586
b. A larger sample increases the probability that the sample mean will be
within a specified distance of the population mean. In the automobile
insurance example, the probability of being within 25 of  ranges from
.4246 for a sample of size 30 to .9586 for a sample of size 400.
24. a. This is a graph of a normal distribution with E ( x ) = 95 and
 x   / n  14 / 30  2.56
b. Within 3 strokes means 92  x  98
z
98  95
 1.17
2.56
z
92  95
 1.17
2.56
P(92  x  98) = P(-1.17 ≤ z ≤ 1.17) = .8790 - .1210 = .7580
The probability the sample means will be within 3 strokes of the population mean of
95 is .7580.
c.  x   / n  14 / 45  2.09
Within 3 strokes means 103  x  109
z
109  106
 1.44
2.09
z
103  106
 1.44
2.09
P(103  x  109) = P(-1.44 ≤ z ≤ 1.44) = .9251 - .0749 = .8502
The probability the sample means will be within 3 strokes of the population mean of
106 is .8502.
d. The probability of being within 3 strokes for female golfers is higher because the
sample size is larger.