9.6 During the month of July, an auto manufacturer gives its production employees a
vacation period so it can tool up for the new model run. In surveying a simple random
sample of 200 production workers, the personnel director finds that 38% of them plan to
vacation out of state for at least one week during this period. Is this a point estimate or an
interval estimate? Explain
This is a point estimate because the estimate is given as a single exact value of 38%.
There is no inference about the population proportion as an interval.
9.14 The following data values are a simple random sample from a population that is
normally distributed, with σ2 = 25.0; 47, 43, 33, 42, 34, and 41. Construct and interpret
the 95% and 99% confidence intervals for the population mean.
x
47  43  33  42  34  41
 40
6
95% confidence interval:
x  1.96

n
   x  1.96
99% confidence interval:
x  2.33

n
   x  2.33

n

n
 40  1.96
5
 40  2.33
5
6
6
   40  1.96
5
   40  2.33
5
6
6
 36    44
 35.24    44.76
9.28 For df =25, determine the value of A that corresponds to each of the following
probabilities:
a. P(t ≥ A) = 0.025
A = 2.06
b. P(t ≤ A) = 0.10
A = -1.316
c. P(-A ≤ t ≤ A) = 0.99
A = 2.787
9.30 Given the following observations in a simple random sample from a population that
is approximately normally distributed, construct and interpret the 90% and 95%
confidence intervals for the mean:
67 79 71 98 74 70 59 102 92 96
x  80.8
s  15.02
df = n - 1 = 10 - 1 = 9
90% confidence interval:
x  1.8331
s
n
   x  1.8331
s
n
 80.8  1.8331
15.02
10
   80.8  1.8331
15.02
10
 72.1    89.5
95% confidence interval:
x  2.2622
s
n
   x  2.2622
s
 80.8  2.2622
n
15.02
10
   80.8  2.2622
15.02
10
 70.1    91.5
9.32 A consumer magazine has contacted a simple random sample of 33 owners of a
certain model of automobile and asked each owner how many defects had to be corrected
within the first 2 months of ownership. The average number of defects was x¯ = 3.7, with
a standard deviation of 1.8 defects.
a. Use the t distribution to construct a 95% confidence interval for µ = the average
number of defects for this model.
x  2.0369
s
n
   x  2.0369
s
 3.7  2.0369
n
1.8
33
   3.7  2.0369
1.8
33
 3.06    4.34
b. Use the z distribution to construct a 95% confidence interval for µ = the average
number of defects for this model.
x  1.96
s
n
   x  1.96
s
 3.7  1.96
n
1.8
33
   3.7  1.96
1.8
33
 3.09    4.31
c. Given that the population standard deviation is not known, which of these two
confidence intervals should be used as the interval estimate for µ?
t-distribution should be used.
9.44 An airline has surveyed a simple random sample of air travelers to find out whether
they would be interested in paying a higher fare in order to have access to e-mail during
their flight. Of the 400 travelers surveyed, 80 said e-mail access would be worth a slight
extra cost. Construct a 95% confidence interval for the population proportion of air
travelers who are in favor of the airline's e-mail idea.
pˆ 
x 80

 0.2
n 400
pˆ (1  pˆ )
pˆ (1  pˆ )
0.2(1  0.2)
0.2(1  0.2)
   pˆ  1.96
 0.2  1.96
   0.2  1.96
n
n
400
400
 0.1608    0.2392
pˆ  1.96
9.52 Based on its 1999 survey, Student Monitor reports that 20% of U.S. college students
used the Internet for job hunting during the month preceding the survey. Assuming this
finding to be based on a simple random sample of 1600 college students, construct and
interpret the 90% confidence interval for the population proportion of college students
who used the Internet for job hunting during this period.
pˆ (1  pˆ )
pˆ (1  pˆ )
0.2(1  0.2)
0.2(1  0.2)
   pˆ  1.645
 0.2  1.645
   0.2  1.645
n
n
1600
1600
 0.1836    0.2165
pˆ  1.645
10.2 For each of the following pairs of null and alternative hypotheses, determine
whether the pair would be appropriate for a hypothesis test. If a pair is deemed
inappropriate, explain why.
a. H0: µ ≥ 10, H 1: µ < 10
ok.
b. H0: µ = 30, H1: µ ≠ 30
ok.
c. H0: µ > 90, H1: µ ≤ 90
Not appropriate, alternative can not have equality.
d. H0: µ ≤ 75, H1: µ ≤ 85
Not appropriate, alternative can not have equality and the same direction with the
null.
e. H0: ¯x ≥ 15, H1: ¯x < 15
Not appropriate, it must be a population parameter.
f. H0: ¯x = 58, H1: ¯x ≠ 58
Not appropriate, it must be a population parameter.
10.10 For each of the following statements, formulate appropriate null and alternative
hypotheses. Indicate whether the appropriate test will be one-tail or two-tail, then sketch
a diagram that shows the approximate location of the "rejection" region(s) for the test.
a. "The average college student spends no more than $300 per semester at the university's
bookstore."
H0: µ ≤ 300, H1: µ > 300............................one tailed test
b. "The average adult drink's 1.5 cups of coffee per day."
H0: µ = 1.5, H1: µ ≠ 1.5............................two tailed test
c. "The average SAT score for entering freshmen is at least 1200."
H0: µ ≥ 1200, H 1: µ < 1200............................one tailed test
d. The average employee put in 3.5 hours of overtime last week."
H0: µ = 3.5, H1: µ ≠ 3.5............................two tailed test
10.26 For each of the following tests and z values, determine the p-value for the test:
a. Left-tail test and z = -1.62
p-value = 0.0526
b. Right-tail test and z = 1.43
p-value = 0.0764
c. Two-tail test and z = 1.27
p-value = 0.2041
10.28. Fro a sample of 12 items from a normally distributed population for which the
standard deviation is σ = 17.0, the sample mean is 230.8. At the 0.05 level of
significance, test H0: µ ≤ 220 versus H1: µ > 220. Determine and interpret the p-value for
the test.
z
x
/ n

230.8  220
17.0 / 12
 2.20
p-value = 0.0139
We reject the null hypothesis since the p-value is less than the significance level (0.05).
10.44 The International Coffee Association has reported the mean daily coffee
consumption for U.S. residents as 1.65 cups. Assume that a sample of 38 people from a
North Carolina city consumed a mean of 1.84 cups of coffee per day, with a standard
deviation of 0.85 cups. In a two-tail test at the 0.05 level, could the residents of this city
be said to b significantly different form their counterparts across the nation?
H0: µ = 1.65, H1: µ ≠ 1.65
t
x
s/ n

1.84  1.65
0.85 / 38
 1.38
p-value = 0.1676
We can not reject the null hypothesis since the p-value is greater than the significance
level (0.05). We don't have sufficient evidence to conclude that the residents of this city
are significantly different than their counterparts across the nation.
10.48 During 2002, college work-study students earned a mean of $1252. Assume that a
sample consisting of 45 of the work-study students at a large university was found to
have earned a mean of $1277 during the year, with a standard deviation of $210. Would a
one-tail test at the 0.05 level suggest the average earnings of this university's work-study
students were significantly higher than the national mean?
H0: µ ≤ 1.65, H1: µ > 1.65
t
x
s/ n

1277  1252
210 / 45
 0.80
p-value = 0.2119
We can not reject the null hypothesis since the p-value is greater than the significance
level (0.05). We don't have sufficient evidence to conclude that the average earnings of
this university's work-study students were significantly higher than the national mean.
10.62 For a simple random sample, n = 700 and p = 0.63. At the 0.025 level, test H0: π ≤
0.60 versus H1: π > 0.60.
z
pˆ  p
p(1  p)
n

0.63  0.60
0.60(1  0.60)
700
 1.62
p-value = 0.0526
We can not reject the null hypothesis since the p-value is greater than the significance
level (0.025).
10.64 The director of admissions at a large university says that 15% of high school
juniors to whom she sends university literature eventually apply for admission. In a
sample of 300 persons to who materials were sent, 30 students applied for admission. In a
two-tail test at the 0.05 level of significance, should we reject the director's claim?
Ho: π = 0.15, Ha: π ≠ 0.15
x 30

 0.10
n 300
pˆ  p
0.10  0.15
z

 2.43
p(1  p)
0.15(1  0.15)
n
300
pˆ 
p-value = 0.0151
We reject the null hypothesis since the p-value is less than the significance level (0.05).
We can conclude that the director's claim is not true.
11.2 An educator is considering two different videotapes for use in a half-day session
designed to introduce students to the basics of economics. Students have been randomly
assigned to two groups, and they all take the same written examination after viewing the
videotape. The scores are summarized here. Assuming normal populations with equal
standard deviations, does it appear that the two videotapes cold be equally effective?
What is the most accurate statement that could be made about the p-value for the test?
Videotape 1: ¯x1 = 77.1 s1 = 7.8 n1 = 25
Videotape 2: x¯2 = 80.0 s2 = 8.1 n2 = 25
11.6 A sample of 40 investment customers serviced by an account manager are found to
have had an average of $23,000 in transactions during the past year, with a standard
deviation of $8500. A sample of 30 customers serviced by another account manger
averaged $28,000 in transaction, with a standard deviation of $11,000. Assuming the
population standard deviations are equal, use the 0.05 level of significance in testing
whether the population means could be equal for customers serviced by the two accounts
manages. Using the appropriate statistical table, what is the most accurate statement we
can make about the p-value for this test? Construct and interpret the 95% confidence
interval for the difference between the population means.
11.18 In two independent samples from populations that are normally distributed, x¯1
=35.0, s1 = 5.8, n1 =12 and x¯2 = 42.5, s2 = 9.3, n2 = 14. Using the 0.05 level of
significance, test H0: µ1 = µ2 versus H1: µ1 ≠ µ2.
11.32 For the following independent random samples, use the z-test and the 0.01 level of
significance in testing as an approximation to the unequal variances t-test when
comparing two sample means?
H0: µ1 = µ2 versus H1: µ1 ≠ µ2.
x¯1 = 33.5 s1 = 6.4 n1 = 31
x¯2 = 27.6 s2 = 2.7 n2 = 30
11.54 A bank manager has been presented with a new brochure that was designed to be
more effective in attracting current customers to a personal financial counseling session
that would include an analysis of additional baking services that could be advantageous
to both the bank and the customer. The manager's assistant, who created the new
brochure, randomly selects 400 current customers, then randomly chooses 200 to receive
the standard brochure that has been used in the past, with the other 200 receiving the
promising new brochure that he has developed. Of those receiving the standard brochure,
35% call for more information about the counseling session, while 42% of those
receiving the new brochure call for more information. Using the 0.10 level of
significance, is it possible that the superior performance of the new brochure was just due
to chance and that the new brochure might really be no better than the old ones?
11.66 For two samples from what are assumed to be normally distributed populations, the
sample sizes and standard deviations are n1 = 10, s1 = 23.5, n2 = 9 and s2 = 10.4. At the
0.10 level of significance, test the null hypothesis that the population variances are equal.
Would your conclusion be different if the test had been conducted at the 0.05 level? At
the 0.02 level?
12.20 In a one-way ANOVA, there are three independent samples, with n1 8, n2 = 10, and
n3 = 7. The calculated F-statistic is F = 3.95. At the 0.05 level of significance, what
conclusion would be reached? Based on the F distribution tables, what is the most
accurate statement that can be made about the p-value for this test?
12.68 In a two-way ANOVA experiment, factor A is operating on 3 levels, factor B is
operating on 2 levels, and there are 2 replications per cell. If MSA/MSE = 5.35, and
MSB/MSE = 5.72, and MSAB/MSE = 6.75 and using the 0.05 level of significance, what
conclusions would be reached regarding the respective null hypotheses for this
experiment?
13.6 For df = 5 and the constant A, identify the value of A such that
a. P(x2 > A) = 0.90
b. P(x2 > A) = 0.10
c. P(x2 > A) = 0.95
d. P(x2 > A) = 0.05
e. P(x2 < A) = 0.975
f. P(x2 < A) = 0.025
13.18 It has been reported that 10.3% of U.S. households do not own a vehicle, with
34.2% owning 1 vehicle, 38.4% owning 2 vehicles, and 17.1% owning 3 or more
vehicles. The data for a random sample of 100 households in a resort community are
summarized in the frequency distribution below. At the 0.05 level of significance, can we
reject the possibility that the vehicle-ownership distribution in this community differs
from that of the nation as a whole?
# of Vehicles Owned # of Households
0 20
1 35
2 23
3 or more 22
100
13.32 In a test of the independence of two variables, one of the variables has two possible
categories and the other has three possible categories. What will be the critical value of
chi-square if the test is to be carried out at the 0.025 level? At the 0.05 level?
13.34 A research organization has collected the following data on household size and
telephone ownership for 200 U.S. households. At the 0.05 level, are the two variables
independent? Based on the chi-square table, what is the most accurate statement that can
be made about the p-value for the test?
Telephones Owned
≤2≥3
Person ≤ 2 49 18 13 80
In the 3-4 40 27 21 88
Household ≥5 11 13 8 32
100 58 42 200
13.44 For three independent samples, each with n = 100, the respective sample
proportions are 0.30, 0.35, and 0.25. Use the 0.05 level in testing whether the three
population proportions could be the same.
13.54 A random sample of 10 observations has been drawn from a normal population,
and the sample variance is found to be s2 = 19.5. Determine the 98% confidence interval
for σ2.
13.60 A random sample of n = 12 is drawn from a population that is normally distributed,
and the sample variance is s2 = 19.3. Use ? = 0.025 in testing H0: σ2 ≤ 9.4 versus H1: σ2 >
9.4.
15.10 The following data represent x = boat sales and y = boat trailer sales from 1995
through 2000.
Year Boat Sales Boat Trailer Sales
Year (Thousands) (Thousands)
1995 649 207
1996 619 194
1997 596 181
1998 576 174
1999 585 168
2000 574 159
a. Determine the least-squares regression line and interpret its slope.
b. Estimate, for a year during which 500,000 boats are sold, the number of boat trailers
that would be sold.
c. What reasons might explain why the number of boat trailers sold per year is less than
the number of boats sold per year?
15.36 For a set of data, the total variation or sum of squares for y is SST = 143.0, and the
error sum of squares is SSE = 24.0. What proportion of the variation in y is explained by
the regression equation?
15.38 The following data show U.S. production of motor vehicles versus tons of domestic
steel shipped for motor vehicle manufacture.
X = U. S. Production y = Tons of
Year of Motor Vehicles Domestic Steel
2000 12.83 million 16.06 million
2001 11.52 14.06
2002 12.33 14.00
2003 12.15 15.88
2004 12.02 13.86
a. Determine the lest-squares regression line and calculate r.
b. What proportion of the variability in steel shipments for motor vehicles is explained by
the regression equation?
c. During a year in which U.S. production of motor vehicles is 12.0 million, what would
be the prediction for the number of tons of domestic steel used for vehicle production?
15.50 For the regression line developed in the above problem
a. Use the 0.05 level in testing whether the population coefficient of correlation could be
zero.
b. Use the 0.05 level in testing whether the population regression equation could have a
slope of zero.
c. Construct the 95% confidence interval for the slope of the population regression
equation.
15.54 In a regression analysis, the sum of the squared deviations between y and y is SST
= 200.0. If the sum of the squared deviation about the regression line is SSE = 40.0, what
is the coefficient of determination?
R2  1
SSE
40.0
 1
 0.8
SST
200