CHAPTER 9 RESERVE PROBLEMS
The following problems have been reserved for your use in assignments and testing and do not
appear in student versions of the text.
Reserve Problems Chapter 9 Section 1 Problem 1
A frequenter of a pub had observed that the new barman poured in average 0.47 liters of beer
into the glass with a standard deviation equal to 0.19 liters instead of a half a liter with the same
standard deviation. The frequenter had used a random sample of 45 glasses of beer in his
experiment.
Consider the two-sided hypothesis test for volume of beer in a glass:
0: 0.5H
=
against
1: 0.5H
.
Determine the P-value of this test.
SOLUTION
Reserve Problems Chapter 9 Section 1 Problem 2
At the shooting competitions in case of a miss, a shooter gets zero points. A good sniper always
hits the bull’s-eye, but random circumstances cause a bullet to deviate from the target. Assume
that the deviations introduced by random circumstances in horizontal and vertical directions are
independent and normally distributed with the zero mean and standard deviation
0.4 m
=
. The
diameter of the target is
1 mL=
.
What is the probability that a good sniper will score zero points for three shots and will be
considered a poor shooter?
SOLUTION
Let
denote the deviation from the center of the target.
Reserve Problems Chapter 9 Section 1 Problem 3
The heat evolved in calories per gram of a cement mixture is approximately normally distributed.
The mean is thought to be 70, and the standard deviation is 2. You wish to test
0: 70H
=
versus
1: 70H
with a sample n = 5 specimens.
(a) If the acceptance region is defined as
68.5 71.5x
, find the type I error probability
.
(b) Find
for the case in which the true mean heat evolved is 73.
(c) Find
for the case where the true mean heat evolved is 75.
Compound this value of
with the one found in part (b). Explain the difference if it exists.
SOLUTION
(c)
Reserve Problems Chapter 9 Section 1 Problem 4
The heat evolved in calories per gram of a cement mixture is approximately normally distributed.
The mean is thought to be 110, and the standard deviation is 5.
Find the boundary of the critical region
if the type I error probability is
.
(a)
0.01
=
and
3n=
______
X
_______.
(b)
0.05
=
and
3n=
_______
X
_______.
(c)
0.01
=
and
5n=
_______
X
_______.
(d)
0.05
=
and
5n=
_______
X
_______.
SOLUTION
Problems Chapter 9 Section 1 Problem 5
The heat evolved in calories per gram of a cement mixture is approximately normally distributed.
The mean is thought to be 110, and the standard deviation is 5.
Calculate the probability of a type II error if the true mean heat evolved is 113:
(a) α = 0.01 and n = 3 β =
(b) α = 0.01 and n = 5 β =
(c) Compare the values of
calculated in previous parts. What conclusions can you draw?
SOLUTION
(a)
Reserve Problems Chapter 9 Section 1 Problem 6
A consumer products company is formulating a new shampoo and is interested in foam height
(in millimeters). Foam height is approximately normally distributed and has a standard deviation
of 20 millimeters. The company wishes to test
0: 175H
=
millimeters versus
1: 175H
millimeters, using the results of
10n=
samples.
(a) Find the type I error probability
if the critical region is
185x
.
(b) What is the probability of type II error if the true foam height is 185 millimeters?
(c) Find
for the true mean of 195 millimeters.
SOLUTION
(a)
(b)
(c)
Reserve Problems Chapter 9 Section 1 Problem 7
A manufacturer is interested in the output voltage of a power supply used in a PC. Output
voltage is assumed to be normally distributed with standard deviation 0.3 volt, and the
manufacturer wishes to test
0:5H
=
volts against
1:5H
volts, using n=10 units.
(a) The acceptance region is
4.85 5.15X
. Find the value of
.
(b) Find the power of the test for detecting a true mean output voltage of 4.8 voltage.
SOLUTION
(a)
( )
4.85 | 5 ( 5.15 | 5)
P X P X
= = + = =
Reserve Problems Chapter 9 Section 1 Problem 8
A manufacturer is interested in the output voltage of a power supply used in a PC. Output
voltage is assumed to be normally distributed with standard deviation 0.15 volt, and the
manufacturer wishes to test
0:5H
=
volts against
1:5H
volts.
Find the boundary of the critical region for the Type I error probability and sample size provided.
(a)
0.01
=
and
7n=
______
X
_____.
(b)
0.05
=
and
7n=
______
X
_____.
(c)
0.01
=
and
10n=
_____
X
______.
(d)
0.05
=
and
10n=
______
X
_____.
SOLUTION
Reserve Problems Chapter 9 Section 1 Problem 9
A manufacturer is interested in the output voltage of a power supply used in a PC. Output
voltage is assumed to be normally distributed with standard deviation 0.25 volt, and the
manufacturer wishes to test
0:4H
=
volts against
1:4H
volts, using n=8 units. Calculate
the P-value if the observed statistic is
(a)
4.1x=
(b)
4.2x=
(c)
3.8x=
SOLUTION
(a)
Reserve Problems Chapter 9 Section 1 Problem 10
A manufacturer is interested in the output voltage of a power supply used in a PC. Output
voltage is assumed to be normally distributed with standard deviation 0.20 volt, and the
manufacturer wishes to test
0:4H
=
volts against
1:4H
volts.
Calculate the probability of a type II error if the true mean output is 4.05 volts and round your
answers to three decimal places (e.g. 98.765).
(a)
0.01 and 10n
==
(b)
0.01 and 14n
==
(c) Compare the values of
calculated in the previous parts. What conclusion can you draw?
SOLUTION
(a)
(b)
/2 /2 /2 /2
( ) ( ) ( )
/ / / / /
N
X
P z z P Z z P Z z
n n n n n
−
= − + + = + − − + =
Reserve Problems Chapter 9 Section 1 Problem 11
A quality-control inspector is testing a batch of printed circuit boards to see wheather they are
capable of performing in a high temperature environment. He knows that the boards that will
survive will pass all five of the tests with probability 95%. They will pass at least four tests with
probability 99%, and they always pass at least three. On the other hand, the boards that will not
survive sometimes pass the tests as well. In fact, 5% pass all five tests, and another 20% pass
exactly four. The rest pass at most three tests. The inspector decides that if a board passes all five
tests, he will classify it as “good.” Otherwise, he’ll classify it as “bad.”
(a) What does a type I error mean in this context?
(b) What is the probability of a type I error?
(c) What does a type II error mean here?
SOLUTION
(a)
Reserve Problems Chapter 9 Section 1 Problem 12
A quality-control inspector is testing a batch of printed circuit boards to see wheater they are
capable of performing in a high temperature environment. He knows that the boards that will
survive will pass all five of the tests with probability 98%. They will pass at least four tests with
probability 99%, and they always pass at least three. On the other hand, the boards that will not
survive sometimes pass the tests as well. In fact, 3% pass all five tests, and another 20% pass
exactly four. The rest pass at most three tests. The inspector decides that if a board passes all five
tests, he will classify it as “good.” Otherwise, he’ll classify it as “bad.” The manager says that
the probability of a type I error must be no larger than 0.01.
(a) How does this change the rule deciding whether a board is “good”?
(b) How does this affect the type II error?
(c) Do you think this reduction in type I error is justified?
SOLUTION
(a)
Type I error can be improved if a board is only required to pass at least 4 of the 5 tests. With this
(b)
Reserve Problems Chapter 9 Section 2 Problem 1
A children’s goods company is formulating a new diapers filler and is interested in the volume of
absorbed liquid (in milliliters). The volume of absorbed liquid is approximately normally
distributed and has a standard deviation of 80 milliliters. The company wishes to test
0: 500 mlH
=
versus
1: 500 mlH
, using the results of
20n=
samples.
(a) Find type I error probability if the critical region is
450x
.
(b) What is the probability of type II error if the true mean volume of absorbed liquid is 460
milliliters?
(c) Find
for the true mean of 480 milliliters.
SOLUTION
(a)
( )
450 500
450, when 500 / 90 / 20
X
P X P
n
−−
= = =
Reserve Problems Chapter 9 Section 2 Problem 2
A new technology for the production of liquid-crystal displays (LCD) is being mastered. The
technology will be rejected if the percentage of defective LCD will exceed 1%. 33.2 Megapixels
are located on the LCD. The LCD is defective if there are more than 300 malfunctioning pixels.
Assume that the probability of appearance of malfunctioning pixels is subject to the Poisson
distribution.
(a) Determine the probability of Type I Error with the critical region
1.03
, considering that the
company uses the results of testing 33 monitors.
(b) Determine the probability of Type II Error if the true mean
267
=
.
(c) Determine the probability of Type II Error if the true mean
273
=
.
SOLUTION
Because the probability of appearance of malfunctioning pixels is subject to the Poisson
distribution, the variance of this process is equal to the mean.
Reserve Problems Chapter 9 Section 2 Problem 3
A hypothesis will be used to test that a population mean equals 10 against the alternative that the
population mean is more than 10 with known variance
. What is the critical value for the
statistic
0
Z
for the following significance levels?
(a)
0.01
=
(b)
0.0005
=
(c)
0.0025
=
SOLUTION
(a)
0.01
=
, then the critical value equals
2.33z
.
0.0005
=
0.0025
=
Reserve Problems Chapter 9 Section 2 Problem 4
For the hypothesis test
0:4H
=
against
1:4H
and variance known, calculate the P-value
for each of the following test statistics.
(a)
01.91z=
(b)
01.79z=−
(c)
00.52z=
SOLUTION
01.91z=
01.79z=−
00.52z=
00.52 0.70P value Z− = =
Reserve Problems Chapter 9 Section 2 Problem 5
Output from a software package follows:
One-Sample z:
Test of μ = 35 vs ≠ 35
The assumed standard deviation = 1.8
Variable
N
Mean
StDev
SE Mean
Z
P
x
10
35.710
1.325
?
?
?
(a) Fill the missing items.
Variable
N
Mean
StDev
SE Mean
Z
P
x
10
35.710
1.325
(b) Use the normal table and the preceding data to construct a 95% two-sided CI on the mean.
(c) What would the P-value be if the alternative hypothesis is
1: 35H
?
SOLUTION
(a) The missing items are
Reserve Problems Chapter 9 Section 2 Problem 6
Output from a software package follows:
One-Sample z:
Test of
0
H
:
15.5
=
versus
1
H
:
15.5
.
The assumed standard deviation = 1.4
Variable
N
Mean
StDev
SE Mean
Z
P
x
25
16.108
1.025
?
?
?
(a) Fill the missing items.
Variable
N
Mean
StDev
SE Mean
Z
P
x
25
16.108
1.025
(b) Use the normal table and the preceding data to construct a 95% two-sided CI on the mean.
(c) What would the P-value be if the alternative hypothesis is
1: 14.5H
?
SOLUTION
(a) The missing items are
(b)
(c)
Reserve Problems Chapter 9 Section 2 Problem 7
An engineer who is studying the tensile strength of a steel alloy intended for use in golf club
shafts knows that tensile strength is approximately normally distributed with
60 psi
=
. A
random sample of 12 specimens has a mean tensile strength of
3450 psix=
.
(a) If the mean strength is 3500 psi, what is the smallest level of significance at which you would
be willing to reject the null hypothesis?
(b) What is the β-error if the true mean is 3470 and
0.01
=
?
(c) Suppose that you wanted to reject the null hypothesis with probability at least 0.8 if mean
strength
3470 psi
=
. What sample size should be used?
SOLUTION
(a)
( )
Smallest level of significance P-value 2 1 2.89
= = − =
So the power is
1 0.1982
−=
.
(c)
Reserve Problems Chapter 9 Section 2 Problem 8
Humans are known to have a mean gestation period of 280 days (from last menstruation) with a
standard deviation of about 9 days. A hospital wondered whether there was any evidence that
their patients were at risk for giving birth prematurely. In a random sample of 70 women, the
average gestation time was 274.7 days.
(a) Is the alternative hypothesis one- or two-sided?
(b) What is the P-value of the test statistic?
SOLUTION
(a)
The alternative hypothesis is one-sided since we are testing whether the mean gestation period is
Reserve Problems Chapter 9 Section 3 Problem 1
We have 5-year statistics of the average amount of wheat crop (tons) harvested from
2
1 km
per
year, the results are as follows:
560, 525, 496, 543, 499.
Test the hypothesis that the mean wheat crop is 550 tons per
2
1 km
per year
( )
0.05
=
and
choose the correct answer.
Determine a 95% confidence interval on the mean wheat crop.
Determine whether the hypothesis that the mean wheat crop is 550 tons per
2
1 km
per year
( )
0.05
=
is true based on the 95% confidence interval?
SOLUTION
Reserve Problems Chapter 9 Section 3 Problem 2
A group of 15 students has performed an experiment, they measured the coefficient of thermal
expansion for aluminum. The results are as follows:
22.0, 25.4, 25.6, 22.7, 22.7, 25.2, 24.9, 21.2, 26.1, 24.0, 23.5, 20.4, 21.4, 23.1, 20.4.
At what level of significance do the students fail to reject the hypothesis that the mean
coefficient value is the same as the table value of
61
22.2 10 K
−−
?
SOLUTION
The parameter of interest is the true mean of the coefficient of thermal expansion for aluminum
.
Reserve Problems Chapter 9 Section 3 Problem 3
A hypothesis will be used to test that a population mean equals 5 against the alternative that the
population mean is less than 5 with unknown variance. What is the critical value for the test
statistic
0
T
for the following significance levels?
(a)
0.01
=
and
23n=
.
(b)
0.05
=
and
12n=
.
(c)
0.10
=
and
14n=
.
SOLUTION
Reserve Problems Chapter 9 Section 3 Problem 4
Consider the following computer output.
One-Sample T:
Test of mu = 12 vs not =12
Variable
N
Mean
StDev
SE Mean
T
P
x
12
12.564
?
0.296
?
?
(a) How many degrees of freedom are there on the t-test statistic?
(b) Find the missing values. You may calculate bounds on the P-value. Is it possible to reject the
null hypothesis at the 0.05 level of significance?
(c) Is this a one-sided or a two-sided test?
(d) Construct a 95% two-sided CI on the mean.
(e) If the hypothesis had been
: 12
o
H
=
versus
1: 12H
.
(f) If the hypothesis had been
: 11.5
o
H
=
versus
1: 11.5H
.
SOLUTION
(a)
(c)
Reserve Problems Chapter 9 Section 3 Problem 5
During the 1999 and 2000 baseball seasons, there was much speculation that the unusually large
number of home runs hit was due at least in part to a livelier ball. One way to test the “liveliness”
of a baseball is to launch the ball at a vertical surface with a known velocity
L
V
and measure the
ratio of the outgoing velocity
O
V
of the ball to
L
V
. The ratio
/
OL
R V V=
is called the coefficient
of restitution. Following are measurements of the coefficient of restitution for 40 randomly
selected baseballs. The balls were thrown from a pitching machine at an oak surface.
0.6248 0.6237 0.6118 0.6159 0.6298 0.6192
0.6520 0.6368 0.6220 0.6151 0.6121 0.6548
0.6226 0.6280 0.6096 0.6300 0.6107 0.6392
0.6230 0.6131 0.6223 0.6297 0.6435 0.5978
0.6351 0.6275 0.6261 0.6262 0.6262 0.6314
0.6128 0.6403 0.6521 0.6049 0.6170
0.6134 0.6310 0.6065 0.6214 0.6141
(a) Test the hypothesis
0: 0.635H
=
versus
1: 0.635H
. Use
0.05
=
.
Is it possible to reject
0
H
hypothesis at the 0.05 level of significance? Find the P-value.
(b) Check the normality assumption, using the normal probability plot.
(c) Compute the power of the test if the true mean coefficient of restitution is as high as 0.64.
(d) What sample size would be required to detect a true mean coefficient of restitution as high as
0.638 if you wanted the power of the test to be at least 0.75?
(e) Which confidence interval should be used to consider
0
H
hypothesis? Find lower confidence
bound for this interval.
SOLUTION
(a)
In order to use t statistics in hypothesis testing, we need to assume that the underlying
distribution is normal.
1) The parameter of interest is the true mean coefficient of restitution,
.