Chapter 9_Tutorial_W24_Solution
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Course
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Subject
Mathematics
Date
Apr 3, 2024
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Department of Mathematics and Statistics
STAT2910-01: Statistics for Sciences
Faculty of Science
University of Windsor
Tutorial for Chapter 9 with solution
Exercise 1.
Identify the choice that best completes the statement or answers the question.
1.
Suppose that we reject the null hypothesis
H
0
:
µ
1
−
µ
2
= 0 at the 0
.
05 level of significance. For which
of the following
α
values do we also reject the null hypothesis?
(a)
0.02
(b)
0.04
(c)
0.06
2.
If you wish to estimate the difference between two population means using two independent large
samples, the 90%confidence interval estimate can be constructed using which of the following critical
values?
(a)
2.33
(b)
1.96
(c)
1.645
(d)
1.28
3.
In testing
H
0
:
µ
1
−
µ
2
= 5 vs.
H
a
:
µ
1
−
µ
2
>
5 the test statistic value
z
is found to be 1
.
69. What is
the p-value of the test?
(a)
0.0455
(b)
0.0910
(c)
0.1977
(d)
0.3023
4.
When testing
H
0
:
µ
1
−
µ
2
= 0 vs.
H
a
:
µ
1
−
µ
2
<
0, the observed value of the
z
−
score was found to
be
−
2
.
15. What would the p-value for this test be?
(a)
0.0158
(b)
0.0316
(c)
0.9684
(d)
0.9842
5.
When testing
H
0
:
µ
1
−
µ
2
= 0 vs.
H
a
:
µ
1
−
µ
2
̸
= 0, the observed value of the
z
−
score was found to
be
−
2
.
15. What would the p-value for this test be?
(a)
0.0158
(b)
0.0316
(c)
0.9684
(d)
0.9842
1
Solution 1.
(1)
(c)
(2)
(c)
(3)
(a)
(4)
(a)
(5)
(b)
Exercise 2.
Develop the null and appropriate hypotheses that are most appropriate for each of the following situations:
(a)
A meteorologist claims that the average high temperature for the month of August in Montreal is
27
◦
C. If the residents of Montreal do not believe this to be true, what hypotheses should they test?
(b)
A car manufacturing plant is acting in accordance with the public’s interest in making cars that
have fuel consumption of at most 8.2 liters per 100 km.
The supervisor will let the cars off the
manufacturing floor only if the fuel consumption is less than 8.8 L/100 km. What hypotheses should
the plant test?
(c)
A spokesperson for the Health Department reports that a fish is unsafe for human consumption if
the polychlorinated biphenyl (PCB) concentration exceeds 5 ppb. The Carlson family is interested
in the mean PCB concentration in a fish from the lake on which they live. What hypotheses should
they test?
(d)
An Internet survey revealed that 50% of Internet users received more than 10 email messages per
day. A similar study on the use of email was repeated. The purpose of the study was to see whether
use of email has increased.
Solution 2.
(a)
The hypotheses are:
H
0
:
µ
= 27
vs
H
a
:
µ
̸
= 27
(b)
The hypotheses are:
H
0
:
µ
= 8
.
2
vs
H
a
:
µ >
8
.
2
(c)
The hypotheses are:
H
0
:
µ
= 5
vs
H
a
:
µ >
5
(d)
The hypotheses are:
H
0
:
p
= 0
.
5
vs
H
a
:
p >
0
.
5
Exercise 3.
A sample of size 150 is to be used to test the hypotheses
H
0
:
µ
= 3
.
75 kg vs.
H
a
:
µ
̸
= 3
.
75 kg, where
µ
is the average weight of a newborn Canadian baby. Give the appropriate rejection region associated with
each of the following significance levels
(a)
α
= 0
.
01
(b)
α
= 0
.
05
2
(c)
α
= 0
.
1
Solution 3.
(a)
Reject
H
0
if
z > z
α
2
or
z <
−
z
α
2
, which means if
z >
2
.
575 or
z <
−
2
.
575. As shown in the figure
below
Normal Curve, mean = 0 , SD = 1 Shaded Area = 0.01
x
density
0.0
0.1
0.2
0.3
0.4
-2.575
2.575
(b)
Reject
H
0
if
z > z
α
2
or
z <
−
z
α
2
, which means if
z >
1
.
96 or
z <
−
1
.
96.
Normal Curve, mean = 0 , SD = 1 Shaded Area = 0.05
x
density
0.0
0.1
0.2
0.3
0.4
-1.96
1.96
(c)
Reject
H
0
if
z > z
α
2
or
z <
−
z
α
2
, which means if
z >
1
.
645 or
z <
−
1
.
645.
3
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Normal Curve, mean = 0 , SD = 1 Shaded Area = 0.1
x
density
0.0
0.1
0.2
0.3
0.4
-1.645
1.645
Exercise 4.
Transport Quebec repaired hundreds of bridges in 1993.
To check the average cost to repair a bridge,
a random sample of
n
= 55 bridges was chosen. The mean and standard deviation for the sample are
$25
,
788 and $1540, respectively. Records from previous years indicate an average bridge repair cost was
$25
,
003. Use the sample data to test that the 1993 mean
µ
is greater than $25
,
003. Use
µ
= 0
.
05.
Solution 4.
This question can be answered following the four steps shown in class room
1. Hypotheses
H
0
:
µ
= 25003
vs
H
a
:
µ >
25003
2. Test statistic
z
0
=
¯
x
−
µ
0
s/
√
n
=
25788
−
25003
1540
/
√
55
= 3
.
78
3. Decision rule
i.
Rejection region approach : Reject
H
0
if
z
0
> z
α
, which implies, if
z
0
>
1
.
645.
4
Normal Curve, mean = 0 , SD = 1 Shaded Area = 0.05
x
density
0.0
0.1
0.2
0.3
0.4
0
1.645
ii.
P-value approach:
pv
=
P
(
z > z
0
) =
P
(
z >
3
.
78) = 1
−
P
(
Z
≤
3
.
78)
≈
1
−
1 = 0
Since
pv < alpha
, we reject the null hypothesis.
4. Conclusion: We conclude that the average bridge repair cost in 1993 is greater than $25
,
003.
Exercise 5.
The manufacturer of a particular battery pack for laptop computers claims its battery pack can function
for eight hours, on average, before having to be recharged. A random sample of 36 battery packs was
selected and placed on test. The mean functioning time before having to be recharged was 7.2 hours with
a standard deviation of 1.9 hours.
(a)
A competitor claims that the manufacturer’s claim is too high.
Perform the appropriate test of
hypothesis to determine whether the competitor is correct. Test using
µ
= 0
.
05.
(b)
Find the p-value for this test.
Solution 5.
(a)
This question can be answered following the four steps shown in class room
(a) Hypotheses
H
0
:
µ
= 8
vs
H
a
:
µ <
8
(b) Test statistic
z
0
=
¯
x
−
µ
0
s/
√
n
=
7
.
2
−
8
1
.
9
/
√
36
=
−
2
.
53
(c) Decision rule: Reject
H
0
if
z
0
<
−
z
α
, which implies, if
z
0
<
−
1
.
645.
5
Normal Curve, mean = 0 , SD = 1 Shaded Area = 0.05
x
density
0.0
0.1
0.2
0.3
0.4
-1.645
0
(d) Conclusion:
We conclude that the true average functioning time of the battery pack before
having to be recharged is less than eight hours (i.e., can reject the manufacturer’s claim).
(b)
The p-value is given by
pv
=
P
(
z <
−
z
0
) =
P
(
z <
−
2
.
53) = 0
.
0057
Exercise 6.
A large university claims that the average cost of housing within 3 kilometres of the campus is $6900 per
school year. A high school student is preparing her budget for her freshman year at the university. She
is concerned that the university’s estimate is too low. Having taken statistics, she decides to perform the
following test of hypothesis:
H
0
:
µ
= 6900 vs.
H
a
:
µ >
6900, where
µ
represents the average cost of
housing per year within 3 kilometres of the university.
(a)
Describe the Type I error for this problem.
(b)
Describe the Type II error for this problem.
(c)
Which error has more serious consequences for the student? Explain.
Solution 6.
(a)
A Type I error occurs if the student rejects the null hypothesis when, in fact, it is true; that is, if the
student concludes the average cost of housing is more than $6900 when, in fact, it is not.
(b)
A Type II error occurs if the student does not reject the null hypothesis when, in fact, it is false;
that is, if the student does not conclude that the average cost of housing is more than $6900 when,
in fact, it is.
(c)
The Type II error is more serious since the student would not budget a sufficient amount for the cost
of housing.
6
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Exercise 7.
A manufacturer of copper pipes must produce pipes with a diameter of precisely 5 cm. The firm’s quality
inspector wants to test the hypothesis that pipes of the proper size are being produced. Accordingly, a
simple random sample of 100 pipes is taken from the production process.
The sample mean diameter
turns out to be 4.98 cm and the sample standard deviation 0.2 cm. Using a significance level of
α
= 0
.
05,
test the appropriate hypotheses.
(a)
State the appropriate hypotheses.
(b)
Calculate the value of the test statistic.
Solution 7.
(a)
The appropriate hypotheses are:
H
0
:
µ
= 5
vs
H
a
:
µ
̸
= 5
(b)
The test statistic is
z
0
=
¯
x
−
µ
0
s/
√
n
=
4
.
98
−
5
0
.
2
/
√
100
=
−
1
Exercise 8.
An airline company would like to know if the average number of passengers on a flight in November is
less than the average number of passengers on a flight in December. The results of random sampling are
printed below. Test the appropriate hypotheses using
α
= 0
.
01.
November
n
1
= 75
¯
x
1
= 476
s
1
= 9
December
n
2
= 80
¯
x
2
= 482
s
1
= 7
Solution 8.
This question can be answered following the four steps shown
1. Hypotheses
H
0
:
µ
1
−
µ
2
= 0
vs
H
a
:
µ
1
−
µ
2
<
0
2. Test statistic
z
0
=
(¯
x
1
−
¯
x
2
)
q
s
2
1
n
1
+
s
2
2
n
2
=
476
−
482
1
.
30
=
−
4
.
62
3. Decision rule Rejection region approach : Reject
H
0
if
z
0
< z
α
, which implies, if
z
0
<
−
2
.
33.
Since
z
0
<
−
2
.
33, we reject the null hypothesis
4. Conclusion: We conclude that the average number of passengers on a November flight is significantly
less than that on a December flight.
7
Exercise 9.
Independent random samples of 35 and 50 observations are drawn from two quantitative populations, 1
and 2, respectively. The sample data summary is shown here:
Sample 1
Sample 2
Sample size
36
50
Sample mean
1.28
1.35
Sample variance
0.058
0.056
Do the data present sufficient evidence to indicate that the mean for population 1 is smaller than
the mean for population 2? Use the p-value approach and the critical value approach and explain your
conclusion.
Solution 9.
This question can be answered following the four steps shown
1. Hypotheses
H
0
:
µ
1
−
µ
2
= 0
vs
H
a
:
µ
1
−
µ
2
<
0
2. Test statistic
z
0
=
(¯
x
1
−
¯
x
2
)
q
s
2
1
n
1
+
s
2
2
n
2
=
1
.
28
−
1
.
35
q
0
.
058
36
+
0
.
056
50
=
−
1
.
34
3. Decision rule:
p-value approach: Calculate p-value =
P
(
z <
−
1
.
34) = 0
.
0901. Since this p-value is greater
than 0.05, the null hypothesis is not rejected.
Critical value approach:
The rejection region, with
α
= 0
.
05, is
z
0
>
−
1
.
645.
Since the
observed value of
z
0
does not fall in the rejection region,
H
0
is not rejected.
4. Conclusion: There is insufficient evidence to indicate that the mean for population 1 is smaller than
the mean for population 2.
Exercise 10.
To test the theory that the consumption of red meat in Canada has decreased over the past 10 years,
a researcher decides to select hospital nutrition records for 400 subjects surveyed 10 years ago and to
compare their average amount of beef consumed per year to amounts consumed by an equal number of
subjects interviewed this year. The data are given in the table.
Ten years ago
This year
Sample mean
34.1 kg
30.9 kg
Sample standard deviation
12.73 kg
14.09 kg
(a)
Do the data present sufficient evidence to indicate that per capita beef consumption has decreased
in the past 10 years? Test at the 1% level of significance.
(b)
Perform the appropriate test of hypothesis to determine whether the proportion of homes heated by
gas has changed. Test using
α
= 0
.
05.
(c)
Find the p-value for this test.
8
Solution 10.
(a)
This question can be answered following the four steps shown
1. Hypotheses
H
0
:
µ
1
−
µ
2
= 0
vs
H
a
:
µ
1
−
µ
2
>
0
2. Test statistic
z
0
=
(¯
x
1
−
¯
x
2
)
q
s
2
1
n
1
+
s
2
2
n
2
=
34
.
1
−
30
.
9
q
(12
.
73)
2
400
+
(14
.
09)
2
400
= 3
.
37
3. Decision rule:
The rejection region, with
α
= 0
.
05, is
z
0
>
2
.
33.
Since the observed value of
z
0
>
2
.
33 ,
H
0
is rejected.
4. Conclusion: There is sufficient evidence to indicate that
µ
1
−
µ
2
>
0 or
µ
1
> µ
2
, that is, the
average per capita beef consumption has decreased in the past 10 years
(b)
The p-value is given by
pv
=
P
(
Z > z
0
) =
P
(
Z >
3
.
37) = 1
−
P
(
Z
≤
3
.
37)
≈
0
.
00037
.
Exercise 11.
A gas company president for a particular city is interested in the proportion of homes heated by gas.
Historically, the proportion of homes heated by gas has been 0.65. A sample of 75 homes was selected
and it was found that 44 of them heat with gas.
(a)
Perform the appropriate test of hypothesis to determine whether the proportion of homes heated by
gas has changed. Test using
α
= 0
.
05.
(b)
Find the p-value for this test.
Solution 11.
(a)
The sample proportion of homes heated by gas is ˆ
p
=
x/n
= 44
/
75 = 0
.
587.
Let p = the true
proportion of homes heated by gas.
1. The hypotheses to be tested are
H
0
:
p
= 0
.
65
vs
H
1
:
p
̸
= 0
.
65
.
2. The value of test statistic is
z
0
=
ˆ
p
−
p
0
q
p
0
q
0
n
=
(0
.
587–0
.
65)
/
0
.
0551
= –1
.
14
.
3. Decision rule: Since
−
1
.
96
< z <
1
.
96, we fail to reject the null hypothesis.
4. Conclusion:
Therefore, we cannot conclude that the proportion of homes heated by gas has
changed.
(b)
The p-value is given by
pv
=
P
(
z <
–1
.
14) +
P
(
z >
1
.
14) = 2
P
(
z <
−
1
.
14) = 0
.
2542
9
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Exercise 12.
A random sample of 150 observations was selected from a binomial population, and 87 successes were
observed. Do the data provide sufficient evidence to indicate that the population proportion p is greater
than 0.5? Use the critical value approach and the p-value approach.
Solution 12.
The sample proportion of homes heated by gas is ˆ
p
=
x/n
= 90
/
150 = 0
.
58. Let p = the true population
proportion.
1. The hypotheses to be tested are
H
0
:
p
= 0
.
5
vs
H
1
:
p >
0
.
5
.
2. The value of test statistic is
z
0
=
ˆ
p
−
p
0
q
p
0
q
0
n
=
(0
.
58–0
.
5)
0
.
0408
= 1
.
96
.
3. Decision rule:
Critical value approach:
The rejection region is one-tailed,
z
0
>
1
.
645 with
α
= 0
.
05 or
z
0
>
2
.
33 with
α
= 0
.
01
.
Hence,
H
0
is rejected at the 5% level, but not at the 1% level.
p
−
valueapproach
: Calculate p-value =
P
(
z >
1
.
96) = 0
.
025. Since this p-value is between 0.01
and 0.05,
H
0
is rejected at the 5% level, but not at the 1% level.
4. Conclusion: At the 5% significance level, we conclude that
p >
0
.
5.
Exercise 13.
A union composed of several thousand employees is preparing to vote on a new contract.
A random
sample of 500 employees yielded 320 who planned to vote yes. It is believed that the new contract will
receive more than 60% yes votes.
(a)
State the appropriate null and alternative hypotheses.
(b)
Can we infer at the 5% significance level that the new contract will receive more than 60% yes votes?
Justify your conclusion.
(c)
Find the p-value for this test.
Solution 13.
(a)
The hypotheses are
H
0
:
p
= 0
.
60
vs
H
1
:
p >
0
.
60
.
(b)
The value of test statistic is
z
0
=
ˆ
p
−
p
0
q
p
0
q
0
n
= 1
.
83
.
Since
z
0
> z
α/
2
= 1
.
645,
H
0
is rejected. Yes, we can infer at the 5% significance level that the new
contract will receive more than 60% yes votes.
(c)
The p-value is given by
pv
=
P
(
z >
1
.
83) = 0
.
0336
.
10
Exercise 14.
A group in favour of freezing production of nuclear weapons believes that the proportion of individuals
in favour of a nuclear freeze is greater for those who have seen the movie “The Day After” (population 1)
than those who have not (population 2). In an attempt to verify this belief, random samples of size 500
are obtained from the populations of interest. Among those who had seen “The Day After,” 228 were in
favour of a freeze. For those who had not seen the movie, 196 favoured a freeze.
(a)
Set up the appropriate null and alternative hypotheses.
(b)
Set up the rejection region for this test using
α
= 0
.
05
(c)
Find the appropriate test statistic.
(d)
State and interpret your conclusion.
Solution 14.
(a)
The hypotheses are
H
0
:
p
1
−
p
2
= 0
vs
H
1
:
p
1
−
p
2
>
0
.
(b)
Reject the null hypothesis is
z >
1
.
645
(c)
Given that ˆ
p
1
= 228
/
500 = 0
.
456, ˆ
p
2
= 196
/
500 = 0
.
392 and the pooled estimate for
p
required for
the standard error is ˆ
p
=
x
1
+
x
2
n
1
+
n
2
=
228+196
1000
= 0
.
424. Then, the test statistic is
z
0
=
ˆ
p
1
−
ˆ
p
2
r
ˆ
p
ˆ
q
1
n
1
+
1
n
2
= 2
.
045
.
(d)
Since
z
0
>
1
.
645, we reject the null hypothesis. Based on this data, the proportion in favour of a
freeze who have seen the movie is greater than the proportion in favour of a freeze who have not seen
the movie.
Exercise 15.
An experiment was conducted to test the effect of a new drug on a viral infection.
The infection was
induced in 100 mice, and the mice were randomly split into 2 groups of 50. The first group, the
control
group
, received no treatment for the infection. The second group received the drug. The proportions of
survivors, ˆ
p
1
and ˆ
p
2
in the 2 groups after a 30-day period, were found to be 0.40 and 0.64, respectively.
(a)
Is there sufficient evidence to indicate that the drug is effective in treating the viral infection? Use
α
= 0
.
05
(b)
Use a 95% confidence interval to estimate the actual difference in the cure rates for the treated versus
the control groups.
Solution 15.
(a)
Following the four steps:
(1)
The hypotheses are
H
0
:
p
1
−
p
2
= 0
vs
H
1
:
p
1
−
p
2
<
0
.
11
(2)
Given that ˆ
p
1
= 0
.
40, ˆ
p
2
= 0
.
64, then, ˆ
p
=
x
1
+
x
2
n
1
+
n
2
=
20+32
100
= 0
.
52. Then, the test statistic is
z
0
=
ˆ
p
1
−
ˆ
p
2
r
ˆ
p
ˆ
q
1
n
1
+
1
n
2
=
0
.
40
−
0
.
64
q
(0
.
52)(0
.
48)
(
1
50
+
1
50
)
=
−
2
.
40
.
(3)
Reject the null hypothesis is
z <
−
1
.
645
(4)
Since
z
0
<
−
1
.
645, we reject the null hypothesis.
There is evidence of a difference in the
proportion of survivors for the two groups.
(b)
A approximate 95% confidence interval is given by
(ˆ
p
1
−
ˆ
p
2
)
±
z
α/
2
r
ˆ
p
1
ˆ
q
1
n
1
+
ˆ
p
2
ˆ
q
2
n
2
(0
.
40
−
0
.
64)
±
1
.
96
r
0
.
40(0
.
60)
50
+
0
.
64(0
.
36)
50
−
0
.
24
±
0
.
19
or
−
0
.
43
≤
p
1
−
p
2
≤ −
0
.
05
Exercise 16.
In testing the hypotheses
H
0
:
p
1
−
p
2
= 0 vs.
H
a
:
p
1
−
p
2
>
0, the following statistics were obtained:
n
1
= 200
, x
1
= 80
, n
2
= 400, and
x
2
= 140, where
x
1
and
x
2
represent the number of defective components
found in medical instruments in the two samples.
(a)
What conclusion can we draw at the 5% significance level?
(b)
What is the p-value of the test? Briefly explain how to use the p-value for testing the hypotheses.
(c)
Estimate with 95% confidence the difference between the two population proportions.
Solution 16.
(a)
Following the four steps:
(1)
The hypotheses are
H
0
:
p
1
−
p
2
= 0
vs
H
1
:
p
1
−
p
2
>
0
.
(2)
Given that ˆ
p
1
= 0
.
40, ˆ
p
2
= 0
.
35, then, ˆ
p
=
x
1
+
x
2
n
1
+
n
2
=
80+140
600
= 0
.
36. Then, the test statistic is
z
0
=
ˆ
p
1
−
ˆ
p
2
r
ˆ
p
ˆ
q
1
n
1
+
1
n
2
=
0
.
40
−
0
.
35
q
(0
.
36)(0
.
54)
(
1
200
+
1
400
)
= 1
.
198
.
(3)
Reject the null hypothesis is
z >
1
.
645
(4)
Since
z
0
<
1
.
645, we fail to reject the null hypothesis. There is not evidence to conclude that the
proportion of defective components found in medical instruments in population 1 greater than
the population 2.
(b)
p-value =
P
(
Z >
1
.
198) = 0
.
1151.
Since p-value = 0
.
1151
> α
= 0
.
05, we fail to reject the null
hypothesis
12
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(c)
A approximate 95% confidence interval is given by
(ˆ
p
1
−
ˆ
p
2
)
±
z
α/
2
r
ˆ
p
1
ˆ
q
1
n
1
+
ˆ
p
2
ˆ
q
2
n
2
(0
.
40
−
0
.
35)
±
1
.
96
r
0
.
40(0
.
60)
200
+
0
.
35(0
.
65)
400
0
.
05
±
0
.
0824
or
−
0
.
0324
≤
p
1
−
p
2
≤
0
.
1324
13