The following calendar-year information is taken from the December.docx

The following calendar-year information is taken from the
December 31, 2011, adjusted trial balance and other records of
Azalea Company.
1. Each team member is to be responsible for computing one of
the following amounts. You are not to duplicate your
teammates' work. Get any necessary amounts from teammates.
Each member is to explain the computation to the team in
preparation for reporting to class.
a. Materials used.
b. Factory overhead.
c. Total manufacturing costs.
d. Total cost of goods in process.
e. Cost of goods manufactured.
2. Check your cost of goods manufactured with the instructor. If
it is correct, proceed to part (3).
3. Each team member is to be responsible for computing one of
the following amounts. You are not to duplicate your
teammates' work. Get any necessary amounts from teammates.
Each member is to explain the computation to the team in
preparation for reporting to class.
a. Net sales.
b. Cost of goods sold.
c. Gross profit.
d. Total operating expenses.
e. Net income or loss before taxes.
CALCULATE T TEST
Calculate the “t” value for independent groups for the following
data using the formula provided in the attached word
document. Using the raw measurement data presented,
determine whether or not there exists a statistically significant
difference between the salaries of female and male human

resource managers using the appropriate t-test. Develop
a testable hypothesis, confidence level, and degrees of
freedom. Report the required “t” critical values based on the
degrees of freedom. Show calculations.
Answer
The null hypothesis tested is
H0: There is no significant difference between the average
salaries of female and male human resource managers. (µ1= µ2)
The alternative hypothesis is
H1: There is significant difference between the average salaries
of female and male human resource managers. (µ1≠ µ2)
The test statistic used is
12
12
2
~
NN
DM
MM
tt
S
+-
-
=
Where
22

1122
1212
(1)(1)
11
2
DM
NsNs
S
NNNN
éùéù
-+-
=+
êúêú
+-
ëûëû
Here M1 = 62,200, M2 = 63,700
s1 = 9330.95, s2 = 6912.95
N1 = 10, N2 = 10 (See the excel sheet)
Then,
(
)
(
)
22
(101)9330.95(101)6912.95
11
101021010
DM
S
éù
-+-
éù

=+
êú
êú
+-
ëû
êú
ëû
= 3672.267768
Therefore test statistic,
62,20063,700
3672.267768
t
-
=
= -0.408466946
Degrees of freedom = N1 + N2 – 2 = 10 + 10 – 2 = 18
Let the significance level be 0.05.
Rejection criteria: Reject the null hypothesis, if the calculated
value of t is greater than the critical value of t at 0.05
significance level.
The critical values can be obtained from the student’s t tables
with 18 d.f. at 0.05 significance level.
Upper critical value = 2.1
Lower critical value = -2.1
0
.
4

0
.
3
0
.
2
0
.
1
0
.
0
X
D
e
n
s
i
t
y
-
2
.
1
0
0
.
0
2
5
2
.
1
0
0
.

0
2
5
0
D
i
s
t
r
i
b
u
t
i
o
n
P
l
o
t
T
,
d
f
=
1
8
Conclusion: Fails to reject the null hypothesis. The sample does
not provide enough evidence to support the claim that there is
significant difference between the salaries of female and male
human resource managers.
_1320131899.unknown
_1320132612.unknown

_1320133261
_1320132305.unknown
_1320131785.unknown
Hypothesist Test for Differences in Two
MeansDataHypothesized Difference0Level of
Significance0.05Population 1 SampleSample Size10Sample
Mean62200Sample Standard Deviation9330.95Population 2
SampleSample Size10Sample Mean63700Sample Standard
Deviation6912.95Intermediate CalculationsPopulation 1 Sample
Degrees of Freedom9Population 2 Sample Degrees of
Freedom9Total Degrees of Freedom18Pooled
Variance67427752.8025Difference in Sample Means-1500t Test
Statistic-0.408466946Two-Tail TestLower Critical Value-
2.1009220369Upper Critical Value2.1009220369p-
Value0.687749195Do not reject the null hypothesis
OutputDescriptive statisticsFEMALE MALE count10 10
mean62,200.00 63,700.00 sample variance87,066,666.67
47,788,888.89 sample standard deviation9,330.95 6,912.95
Sheet1FEMALEMALE5000058000750006900072000730006700
06700054000550005800063000520005300068000700007100069
00055000600006220063700Mean9330.9520771826912.9508090
893S.D.
Sheet2
Sheet3
T-test (1 of 2)
Difference testing is used primarily to identify if there is a
detectable difference between products, services, people, or
situations. These tests are often conducted in business situations
to:
· Ensure a change in formulation or production introduces no

significant change in the end product or service.
· Substantiate a claim of a new or improved product or service
· Confirm that a new ingredient/supplier does not affect the
perceived attributes of the product or service.
· Track changes during shelf life of a product or the length of
time of a service.
Differences Between Two Independent Sample Means:
Coke vs. Pepsi. Independent sample t-tests are used to compare
the means of two independently sampled groups (ex., do those
drinking Coke differ on a performance variable, or the numbers
of cans consumed in one week) compared to those drinking
Pepsi. The individuals are randomly assigned to the Coke and
Pepsi groups. With a confidence interval of ≤.05 (corresponding
probability level of 95%) the researcher concludes the two
groups are significantly different in their means (average
consumption rate of Coke and Pepsi over a one week period of
time) if the t-test value meets or exceeds the required critical
value. If the tvalue does not meet the critical t value required
then the research investigator simply concludes that no
differences exist. Further explanation is not required. Presented
below is a more useable situation.
Using the raw data and formula above to calculate the actual t-
test value, when calculated properly, is 2.43. Always remember
that S = Standard deviation and that the mean is oftentimes
shown by the capital letter M rather than a bar mark over a
capital X. By going to the appropriate t-tables in your textbook
find the critical value for t at the .05 confidence interval. The
value you should find is 1.761
T-test (2 of 2)
Differences Between Two Means of Correlated Samples

Correlated t-test statistical processes are used to determine
whether or not there is a relationship of a particular
measurement variable on a pre- and post-test basis. Oftentimes
when there exists a statistically significant relationship on a
pre- and post-test basis the business manager can use the first
measurement values to predict the second in future situations
without having to present a post-test situation.
Example: Using the same data presented above let us assume
that there are not two independent groups but the same group
under two different conditions—noise production environment
and non-noise production environment.
Using the raw data and formula above to calculate the t-test
value the actual t-test, when calculated properly, is 3.087. By
going to the appropriate t tables in your text book find the
critical value for t at the .05 confidence interval. The value you
should find is 1.895.
Using the navigation on the left, please proceed to the next
page.
CHAPTER 9: Hypothesis Testing
Chapter Outline
9.1The Null and Alternative Hypotheses and Errors in
Hypothesis Testing

9.2z Tests about a Population Mean: σ Known
9.3t Tests about a Population Mean: σ Unknown
9.4z Tests about a Population Proportion
9.5Type II Error Probabilities and Sample Size Determination
(Optional)
9.6The Chi-Square Distribution (Optional)
9.7Statistical Inference for a Population Variance (Optional)
Hypothesis testing is a statistical procedure used to provide
evidence in favor of some statement (called a hypothesis). For
instance, hypothesis testing might be used to assess whether a
population parameter, such as a population mean, differs from a
specified standard or previous value. In this chapter we discuss
testing hypotheses about population means, proportions, and
variances.
In order to illustrate how hypothesis testing works, we revisit
several cases introduced in previous chapters and also introduce
some new cases:
The Payment Time Case: The consulting firm uses hypothesis
testing to provide strong evidence that the new electronic
billing system has reduced the mean payment time by more than
50 percent.
The Cheese Spread Case: The cheese spread producer uses
hypothesis testing to supply extremely strong evidence that
fewer than 10 percent of all current purchasers would stop
buying the cheese spread if the new spout were used.
The Electronic Article Surveillance Case: A company that sells
and installs EAS systems claims that at most 5 percent of all
consumers would never shop in a store again if the store
subjected them to a false EAS alarm. A store considering the
purchase of such a system uses hypothesis testing to provide
extremely strong evidence that this claim is not true.
The Trash Bag Case: A marketer of trash bags uses hypothesis
testing to support its claim that the mean breaking strength of
its new trash bag is greater than 50 pounds. As a result, a
television network approves use of this claim in a commercial.

The Valentine’s Day Chocolate Case: A candy company projects
that this year’s sales of its special valentine box of assorted
chocolates will be 10 percent higher than last year. The candy
company uses hypothesis testing to assess whether it is
reasonable to plan for a 10 percent increase in sales of the
valentine box.
9.1: The Null and Alternative Hypotheses and Errors in
Hypothesis Testing
One of the authors’ former students is employed by a major
television network in the standards and practices division. One
of the division’s responsibilities is to reduce the chances that
advertisers will make false claims in commercials run on the
network. Our former student reports that the network uses a
statistical methodology called hypothesis testing to do this.
Chapter 9
To see how this might be done, suppose that a company wishes
to advertise a claim, and suppose that the network has reason to
doubt that this claim is true. The network assumes for the sake
of argument that the claim is not valid. This assumption is
called the null hypothesis. The statement that the claim is valid
is called the alternative, or research, hypothesis. The network
will run the commercial only if the company making the claim
provides sufficient sample evidence to reject the null hypothesis
that the claim is not valid in favor of the alternative hypothesis
that the claim is valid. Explaining the exact meaning of
sufficient sample evidence is quite involved and will be
discussed in the next section.
The Null Hypothesis and the Alternative Hypothesis
In hypothesis testing:
1The null hypothesis, denoted H0, is the statement being tested.
Usually this statement represents the status quo and is not
rejected unless there is convincing sample evidence that it is
false.
2The alternative, or research, hypothesis, de noted Ha, is a
statement that will be accepted only if there is convincing

sample evidence that it is true.
Setting up the null and alternative hypotheses in a practical
situation can be tricky. In some situations there is a condition
for which we need to attempt to find supportive evidence. We
then formulate (1) the alternative hypothesis to be the statement
that this condition exists and (2) the null hypothesis to be the
statement that this condition does not exist. To illustrate this,
we consider the following case studies.
EXAMPLE 9.1: The Trash Bag Case1
A leading manufacturer of trash bags produces the strongest
trash bags on the market. The company has developed a new 30-
gallon bag using a specially formulated plastic that is stronger
and more biodegradable than other plastics. This plastic’s
increased strength allows the bag’s thickness to be reduced, and
the resulting cost savings will enable the company to lower its
bag price by 25 percent. The company also believes the new bag
is stronger than its current 30-gallon bag.
The manufacturer wants to advertise the new bag on a major
television network. In addition to promoting its price reduction,
the company also wants to claim the new bag is better for the
environment and stronger than its current bag. The network is
convinced of the bag’s environmental advantages on scientific
grounds. However, the network questions the company’s claim
of increased strength and requires statistical evidence to justify
this claim. Although there are various measures of bag strength,
the manufacturer and the network agree to employ “breaking
strength.” A bag’s breaking strength is the amount of a
representative trash mix (in pounds) that, when loaded into a
bag suspended in the air, will cause the bag to rip or tear. Tests
show that the current bag has a mean breaking strength that is
very close to (but does not exceed) 50 pounds. The new bag’s
mean breaking strength μ is unknown and in question. The
alternative hypothesis Ha is the statement for which we wish to
find supportive evidence. Because we hope the new bags are
stronger than the current bags, Ha says that μ is greater than 50.

The null hypothesis states that Ha is false. Therefore, H0 says
that μ is less than or equal to 50. We summarize these
hypotheses by stating that we are testing
H0: μ ≤ 50 versus Ha: μ > 50
The network will run the manufacturer’s commercial if a
random sample of n new bags provides sufficient evidence to
reject H0: μ ≤ 50 in favor of Ha: μ > 50.
EXAMPLE 9.2: The Payment Time Case
Recall that a management consulting firm has installed a new
computer-based, electronic billing system for a Hamilton, Ohio,
trucking company. Because of the system’s advantages, and
because the trucking company’s clients are receptive to using
this system, the management consulting firm believes that the
new system will reduce the mean bill payment time by more
than 50 percent. The mean payment time using the old billing
system was approximately equal to, but no less than, 39 days.
Therefore, if μ denotes the mean payment time using the new
system, the consulting firm believes that μ will be less than 19.5
days. Because it is hoped that the new billing system reduces
mean payment time, we formulate the alternative hypothesis as
Ha: μ < 19.5 and the null hypothesis as H0: μ ≥ 19.5. The
consulting firm will randomly select a sample of n invoices and
determine if their payment times provide sufficient evidence to
reject H0: μ ≥ 19.5 in favor of Ha: μ < 19.5. If such evidence
exists, the consulting firm will conclude that the new electronic
billing system has reduced the Hamilton trucking company’s
mean bill payment time by more than 50 percent. This
conclusion will be used to help demonstrate the benefits of the
new billing system both to the Hamilton company and to other
trucking companies that are considering using such a system.
EXAMPLE 9.3: The Valentine’s Day Chocolate Case 2
A candy company annually markets a special 18 ounce box of
assorted chocolates to large retail stores for Valentine’s Day.
This year the candy company has designed an extremely
attractive new valentine box and will fill the box with an
especially appealing assortment or chocolates. For this reason,

the candy company subjectively projects—based on past
experience and knowledge of the candy market—that sales of its
valentine box will be 10 percent higher than last year. However,
since the candy company must decide how many valentine boxes
to produce, the company needs to assess whether it is
reasonable to plan for a 10 percent increase in sales.
Before the beginning of each Valentine’s Day sales season, the
candy company sends large retail stores information about its
newest valentine box of assorted chocolates. This information
includes a description of the box of chocolates, as well as a
preview of advertising displays that the candy company will
provide to help retail stores sell the chocolates. Each retail store
then places a single (nonreturnable) order of valentine boxes to
satisfy its anticipated customer demand for the Valentine’s Day
sales season. Last year the mean order quantity of large retail
stores was 300 boxes per store. If the projected 10 percent sales
increase will occur, the mean order quantity, μ, of large retail
stores this year will be 330 boxes per store. Therefore, the
candy company wishes to test the null hypothesis H0: μ = 330
versus the alternative hypothesis Ha: μ ≠ 330.
To perform the hypothesis test, the candy company will
randomly select a sample of n large retail stores and will make
an early mailing to these stores promoting this year’s valentine
box. The candy company will then ask each retail store to report
how many valentine boxes it anticipates ordering. If the sample
data do not provide sufficient evidence to reject H0: μ = 330 in
favor of Ha: μ ≠ 330, the candy company will base its
production on the projected 10 percent sales increase. On the
other hand, if there is sufficient evidence to reject H0: μ = 330,
the candy company will change its production plans.
We next summarize the sets of null and alternative hypotheses
that we have thus far considered.
The alternative hypothesis Ha: μ > 50 is called a one-sided,
greater than alternative hypothesis, whereas Ha: μ < 19.5 is

called a one-sided, less than alternative hypothesis, and Ha:
μ ≠ 330 is called a two-sided, not equal to alternative
hypothesis. Many of the alternative hypotheses we consider in
this book are one of these three types. Also, note that each null
hypothesis we have considered involves an equality. For
example, the null hypothesis H0: μ ≤ 50 says that μ is either
less than or equal to 50. We will see that, in general, the
approach we use to test a null hypothesis versus an alternative
hypothesis requires that the null hypothesis involve an equality.
The idea of a test statistic
Suppose that in the trash bag case the manufacturer randomly
selects a sample of n = 40 new trash bags. Each of these bags is
tested for breaking strength, and the sample mean of the 40
breaking strengths is calculated. In order to test H0: μ ≤ 50
versus Ha: μ > 50, we utilize the test statistic
The test statistic z measures the distance between and 50. The
division by says that this distance is measured in units of the
standard deviation of all possible sample means. For example, a
value of z equal to, say, 2.4 would tell us that is 2.4 such
standard deviations above 50. In general, a value of the test
statistic that is less than or equal to zero results when is less
than or equal to 50. This provides no evidence to support
rejecting H0 in favor of Ha because the point estimate indicates
that μ is probably less than or equal to 50. However, a value of
the test statistic that is greater than zero results when is greater
than 50. This provides evidence to support rejecting H0 in favor
of Ha because the point estimate indicates that μ might be
greater than 50. Furthermore, the farther the value of the test
statistic is above 0 (the farther is above 50), the stronger is the
evidence to support rejecting H0 in favor of Ha.
Hypothesis testing and the legal system
If the value of the test statistic z is far enough above 0, we
reject H0 in favor of Ha. To see how large z must be in order to
reject H0, we must understand that a hypothesis test rejects a
null hypothesis H0 only if there is strong statistical evidence

against H0. This is similar to our legal system, which rejects the
innocence of the accused only if evidence of guilt is beyond a
reasonable doubt. For instance, the network will reject H0: μ ≤
50 and run the trash bag commercial only if the test statistic z is
far enough above 0 to show beyond a reasonable doubt that H0:
μ ≤ 50 is false and Ha: μ > 50 is true. A test statistic that is only
slightly greater than 0 might not be convincing enough.
However, because such a test statistic would result from a
sample mean that is slightly greater than 50, it would provide
some evidence to support rejecting H0: μ ≤ 50, and it certainly
would not provide strong evidence sup porting H0: μ ≤ 50.
Therefore, if the value of the test statistic is not large enough to
convince us to reject H0, we do not say that we accept H0.
Rather we say that we do not reject H0 because the evidence
against H0 is not strong enough. Again, this is similar to our
legal system, where the lack of evidence of guilt beyond a
reasonable doubt results in a verdict of not guilty, but does not
prove that the accused is innocent.
Type I and Type II errors and their probabilities
To determine exactly how much statistical evidence is required
to reject H0, we consider the errors and the correct decisions
that can be made in hypothesis testing. These errors and correct
decisions, as well as their implications in the trash bag
advertising example, are summarized in Tables 9.1 and 9.2.
Across the top of each table are listed the two possible “states
of nature.” Either H0: μ ≤ 50 is true, which says the
manufacturer’s claim that μ is greater than 50 is false, or H0 is
false, which says the claim is true. Down the left side of each
table are listed the two possible decisions we can make in the
hypothesis test. Using the sample data, we will either reject H0:
μ ≤ 50, which implies that the claim will be advertised, or we
will not reject H0, which implies that the claim will not be
advertised.
Table 9.1: Type I and Type II Errors
Table 9.2: The Implications of Type I and Type II Errors in the

Trash Bag Example
In general, the two types of errors that can be made in
hypothesis testing are defined here:
Type I and Type II Errors
If we reject H0 when it is true, this is a Type I error.
If we do not reject H0 when it is false, this is a Type II error.
As can be seen by comparing Tables 9.1 and 9.2, if we commit a
Type I error, we will advertise a false claim. If we commit a
Type II error, we will fail to advertise a true claim.
We now let the symbol α (pronounced alpha) denote the
probability of a Type I error, and we let β (pronounced beta)
denote the probability of a Type II error. Obviously, we would
like both α and β to be small. A common (but not the only)
procedure is to base a hypothesis test on taking a sample of a
fixed size (for example, n = 40 trash bags) and on setting α
equal to a small prespecified value. Setting α low means there is
only a small chance of rejecting H0 when it is true. This implies
that we are requiring strong evidence against H0 before we
reject it.
We sometimes choose α as high as .10, but we usually choose α
between .05 and .01. A frequent choice for α is .05. In fact, our
former student tells us that the network often tests advertising
claims by setting the probability of a Type I error equal to .05.
That is, the network will run a commercial making a claim if the
sample evidence allows it to reject a null hypothesis that says
the claim is not valid in favor of an alternative hypothesis that
says the claim is valid with α set equal to .05. Since a Type I
error is deciding that the claim is valid when it is not, the policy
of setting α equal to .05 says that, in the long run, the network
will advertise only 5 percent of all invalid claims made by
advertisers.
One might wonder why the network does not set α lower—say at
.01. One reason is that it can be shown that, for a fixed sample
size, the lower we set α, the higher is β, and the higher we set α,
the lower is β. Setting α at .05 means that β, the probability of

failing to advertise a true claim (a Type II error), will be
smaller than it would be if α were set at .01. As long as (1) the
claim to be advertised is plausible and (2) the consequences of
advertising the claim even if it is false are not terribly serious,
then it is reasonable to set α equal to .05. However, if either (1)
or (2) is not true, then we might set α lower than .05. For
example, suppose a pharmaceutical company wishes to advertise
that it has developed an effective treatment for a disease that
has formerly been very resistant to treatment. Such a claim is
(perhaps) difficult to believe. Moreover, if the claim is false,
patients suffering from the disease would be subjected to false
hope and needless expense. In such a case, it might be
reasonable for the network to set α at .01 because this would
lower the chance of advertising the claim if it is false. We
usually do not set α lower than .01 because doing so often leads
to an unacceptably large value of β. We explain some methods
for computing the probability of a Type II error in optional
Section 9.6. However, β can be difficult or impossible to
calculate in many situations, and we often must rely on our
intuition when deciding how to set α.
Exercises for Section 9.1
CONCEPTS
9.1Which hypothesis (the null hypothesis, H0, or the alternative
hypothesis, Ha) is the “status quo” hypothesis (that is, the
hypothesis that states that things are remaining “as is”)? Which
hypothesis is the hypothesis that says that a “hoped for” or
“suspected” condition exists?
9.2Which hypothesis (H0 or Ha) is not rejected unless there is
convincing sample evidence that it is false? Which hypothesis
(H0 or Ha) will be accepted only if there is convincing sample
evidence that it is true?
9.3Define each of the following:
aType I error
bType II error
cα

dβ
9.4For each of the following situations, indicate whether an
error has occurred and, if so, indicate what kind of error (Type I
or Type II) has occurred.
aWe do not reject H0 and H0 is true.
bWe reject H0 and H0 is true.
cWe do not reject H0 and H0 is false.
dWe reject H0 and H0 is false.
9.5If we reject H0, what is the only type of error that we could
be making? Explain.
9.6If we do not reject H0, what is the only type of error that we
could be making? Explain.
9.7When testing a hypothesis, why don’t we set the probability
of a Type I error to be extremely small? Explain.
METHODS AND APPLICATIONS
9.8THE VIDEO GAME SATISFACTION RATING CASE
VideoGame
Recall that “very satisfied” customers give the XYZ-Box video
game system a rating that is at least 42. Suppose that the
manufacturer of the XYZ-Box wishes to use the random sample
of 65 satisfaction ratings to provide evidence supporting the
claim that the mean composite satisfaction rating for the XYZ-
Box exceeds 42.
aLetting μ represent the mean composite satisfaction rating for
the XYZ-Box, set up the null and alternative hypotheses needed
if we wish to attempt to provide evidence supporting the claim
that μ exceeds 42.
bIn the context of this situation, interpret making a Type I
error; interpret making a Type II error.
9.9THE BANK CUSTOMER WAITING TIME CASE WaitTime
Recall that a bank manager has developed a new system to
reduce the time customers spend waiting for teller service
during peak hours. The manager hopes the new system will
reduce waiting times from the current 9 to 10 minutes to less
than 6 minutes.
Suppose the manager wishes to use the random sample of 100

waiting times to support the claim that the mean waiting time
under the new system is shorter than six minutes.
aLetting μ represent the mean waiting time under the new
system, set up the null and alternative hypotheses needed if we
wish to attempt to provide evidence supporting the claim that μ
is shorter than six minutes.
9.10An automobile parts supplier owns a machine that produces
a cylindrical engine part. This part is supposed to have an
outside diameter of three inches. Parts with diameters that are
too small or too large do not meet customer requirements and
must be rejected. Lately, the company has experienced problems
meeting customer requirements. The technical staff feels that
the mean diameter produced by the machine is off target. In
order to verify this, a special study will randomly sample 100
parts produced by the machine. The 100 sampled parts will be
measured, and if the results obtained cast a substantial amount
of doubt on the hypothesis that the mean diameter equals the
target value of three inches, the company will assign a problem-
solving team to intensively search for the causes of the problem.
aThe parts supplier wishes to set up a hypothesis test so that the
problem-solving team will be assigned when the null hypothesis
is rejected. Set up the null and alternative hypotheses for this
situation.
cSuppose it costs the company $3,000 a day to assign the
problem-solving team to a project. Is this $3,000 figure the
daily cost of a Type I error or a Type II error? Explain.
9.11The Crown Bottling Company has just installed a new
bottling process that will fill 16-ounce bottles of the popular
Crown Classic Cola soft drink. Both overfilling and underfilling
bottles are undesirable: Underfilling leads to customer
complaints and overfilling costs the company considerable
money. In order to verify that the filler is set up correctly, the

company wishes to see whether the mean bottle fill, μ, is close
to the target fill of 16 ounces. To this end, a random sample of
36 filled bottles is selected from the output of a test filler run.
If the sample results cast a substantial amount of doubt on the
hypothesis that the mean bottle fill is the desired 16 ounces,
then the filler’s initial setup will be readjusted.
aThe bottling company wants to set up a hypothesis test so that
the filler will be readjusted if the null hypothesis is rejected.
Set up the null and alternative hypotheses for this hypothesis
test.
9.12Consolidated Power, a large electric power utility, has just
built a modern nuclear power plant. This plant discharges waste
water that is allowed to flow into the Atlantic Ocean. The
Environmental Protection Agency (EPA) has ordered that the
waste water may not be excessively warm so that thermal
pollution of the marine environment near the plant can be
avoided. Because of this order, the waste water is allowed to
cool in specially constructed ponds and is then released into the
ocean. This cooling system works properly if the mean
temperature of waste water discharged is 60°F or cooler.
Consolidated Power is required to monitor the temperature of
the waste water. A sample of 100 temperature readings will be
obtained each day, and if the sample results cast a substantial
amount of doubt on the hypothesis that the cooling system is
working properly (the mean temperature of waste water
discharged is 60°F or cooler), then the plant must be shut down
and appropriate actions must be taken to correct the problem.
aConsolidated Power wishes to set up a hypothesis test so that
the power plant will be shut down when the null hypothesis is
rejected. Set up the null and alternative hypotheses that should
be used.
cThe EPA periodically conducts spot checks to determine

whether the waste water being discharged is too warm. Suppose
the EPA has the power to impose very severe penalties (for
example, very heavy fines) when the waste water is excessively
warm. Other things being equal, should Consolidated Power set
the probability of a Type I error equal to α = .01 or α = .05?
Explain.
9.13Consider Exercise 9.12, and suppose that Consolidated
Power has been experiencing technical problems with the
cooling system. Because the system has been unreliable, the
company feels it must take precautions to avoid failing to shut
down the plant when its waste water is too warm. Other things
being equal, should Consolidated Power set the probability of a
Type I error equal to α = .01 or α = .05? Explain.
9.2: z Tests about a Population Mean: σ Known
In this section we discuss hypothesis tests about a population
mean that are based on the normal distribution. These tests are
called z tests, and they require that the true value of the
population standard deviation σ is known. Of course, in most
real-world situations the true value of σ is not known. However,
the concepts and calculations of hypothesis testing are most
easily illustrated using the normal distribution. Therefore, in
this section we will assume that—through theory or history
related to the population under consideration—we know σ.
When σ is unknown, we test hypotheses about a population
mean by using the t distribution. In Section 9.3 we study t tests,
and we will revisit the examples of this section assuming that σ
is unknown.
Chapter 9
Testing a “greater than” alternative hypothesis by using a
critical value rule
In Section 9.1 we explained how to set up appropriate null and
alternative hypotheses. We also discussed how to specify a
value for α, the probability of a Type I error (also called the
level of significance) of the hypothesis test, and we introduced
the idea of a test statistic. We can use these concepts to begin

developing a seven step hypothesis testing procedure. We will
introduce these steps in the context of the trash bag case and
testing a “greater than” alternative hypothesis.
Step 1: State the null hypothesis H0 and the alternative
hypothesis Ha. In the trash bag case, we will test H0: μ ≤ 50
versus Ha: μ > 50. Here, μ is the mean breaking strength of the
new trash bag.
Step 2: Specify the level of significance α. The television
network will run the commercial stating that the new trash bag
is stronger than the former bag if we can reject H0: μ ≤ 50 in
favor of Ha: μ > 50 by setting α equal to .05.
Step 3: Select the test statistic. In order to test H0: μ ≤ 50
versus Ha: μ > 50, we will test the modified null hypothesis H0:
μ = 50 versus Ha: μ > 50. The idea here is that if there is
sufficient evidence to reject the hypothesis that μ equals 50 in
favor of μ > 50, then there is certainly also sufficient evidence
to reject the hypothesis that μ is less than or equal to 50. In
order to test H0: μ = 50 versus Ha: μ > 50, we will randomly
select a sample of n = 40 new trash bags and calculate the mean
of the breaking strengths of these bags. We will then utilize the
test statistic
A positive value of this test statistic results from an that is
greater than 50 and thus provides evidence against H0: μ = 50
and in favor of Ha: μ > 50.
Step 4: Determine the critical value rule for deciding whether to
reject H0. To decide how large the test statistic z must be to
reject H0 in favor of Ha by setting the probability of a Type I
error equal to α, we note that different samples would give
different sample means and thus different values of z. Because
the sample size n = 40 is large, the Central Limit Theorem tells
us that the sampling distribution of z is (approximately) a
standard normal distribution if the null hypothesis H0: μ = 50 is
true. Therefore, we do the following:
Place the probability of a Type I error, α, in the right-hand tail
of the standard normal curve and use the normal table (see

Table A.3, page 863) to find the normal point zα. Here zα,
which we call a critical value, is the point on the horizontal axis
under the standard normal curve that gives a right-hand tail area
equal to α.
Reject H0: μ = 50 in favor of Ha: μ > 50 if and only if the test
statistic z is greater than the critical value zα (This is the
critical value rule.)
Figure 9.1 illustrates that since we have set α equal to .05, we
should use the critical value zα = z.05 = 1.645 (see Table A.3).
This says that we should reject H0 if z > 1.645 and we should
not reject H0 if z ≤ 1.645.
Figure 9.1: The Critical Value for Testing H0: μ = 50 versus
Ha: μ > 50 by Setting α = .05
To better understand the critical value rule, consider the
standard normal curve in Figure 9.1. The area of .05 in the
right-hand tail of this curve implies that values of the test
statistic z that are greater than 1.645 are unlikely to occur if the
null hypothesis H0: μ = 50 is true. There is a 5 percent chance
of observing one of these values—and thus wrongly rejecting
H0—if H0 is true. However, we are more likely to observe a
value of z greater than 1.645—and thus correctly reject H0—if
H0 is false. Therefore, it is intuitively reasonable to reject H0 if
the value of the test statistic z is greater than 1.645.
Step 5: Collect the sample data and compute the value of the
test statistic. When the sample of n = 40 new trash bags is
randomly selected, the mean of the breaking strengths is
calculated to be . Assuming that σ is known to equal 1.65, the
value of the test statistic is
Step 6: Decide whether to reject H0 by using the test statistic
value and the critical value rule. Since the test statistic value z
= 2.20 is greater than the critical value z.05 = 1.645, we can
reject H0: μ = 50 in favor of Ha: μ > 50 by setting α equal to
.05. Furthermore, we can be intuitively confident that H0: μ =
50 is false and Ha: μ > 50 is true. This is because, since we

have rejected H0 by setting α equal to .05, we have rejected H0
by using a test that allows only a 5 percent chance of wrongly
rejecting H0. In general, if we can reject a null hypothesis in
favor of an alternative hypothesis by setting the probability of a
Type I error equal to α, we say that we have statistical
significance at the α level.
Step 7: Interpret the statistical results in managerial (real-
world) terms and assess their practical importance. Since we
have rejected H0: μ = 50 in favor of Ha: μ > 50 by setting α
equal to .05, we conclude (at an α of .05) that the mean
breaking strength of the new trash bag exceeds 50 pounds.
Furthermore, this conclusion has practical importance to the
trash bag manufacturer because it means that the television
network will approve running commercials claiming that the
new trash bag is stronger than the former bag. Note, however,
that the point estimate of μ, , indicates that μ is not much larger
than 50. Therefore, the trash bag manufacturer can claim only
that its new bag is slightly stronger than its former bag. Of
course, this might be practically important to consumers who
feel that, because the new bag is 25 percent less expensive and
is more environmentally sound, it is definitely worth purchasing
if it has any strength advantage. However, to customers who are
looking only for a substantial increase in bag strength, the
statistical results would not be practically important. This
illustrates that, in general, a finding of statistical significance
(that is, concluding that the alternative hypothesis is true) can
be practically important to some people but not to others.
Notice that the point estimate of the parameter involved in a
hypothesis test can help us to assess practical importance. We
can also use confidence intervals to help assess practical
importance.
Considerations in setting α
We have reasoned in Section 9.1 that the television network has
set α equal to .05 rather than .01 because doing so means that β,
the probability of failing to advertise a true claim (a Type II

error), will be smaller than it would be if α were set at .01. It is
informative, however, to see what would have happened if the
network had set α equal to .01. Figure 9.2 illustrates that as we
decrease α from .05 to .01, the critical value zα increases from
z.05 = 1.645 to z.01 = 2.33. Because the test statistic value z =
2.20 is less than z.01 = 2.33, we cannot reject H0: μ = 50 in
favor of Ha: μ > 50 by setting α equal to .01. This illustrates the
point that, the smaller we set α, the larger is the critical value,
and thus the stronger is the statistical evidence that we are
requiring to reject the null hypothesis H0. Some statisticians
have concluded (somewhat subjectively) that (1) if we set α
equal to .05, then we are requiring strong evidence to reject H0;
and (2) if we set α equal to .01, then we are requiring very
strong evidence to reject H0.
Figure 9.2: The Critical Values for Testing H0: μ = 50 versus
Ha: μ > 50 by Setting α = .05 and .01
A p-value for testing a “greater than” alternative hypothesis
To decide whether to reject the null hypothesis H0 at level of
significance α, steps 4, 5, and 6 of the seven-step hypoth esis
testing procedure compare the test statistic value with a critical
value. Another way to make this decision is to calculate a p-
value, which measures the likelihood of the sample results if the
null hypothesis H0 is true. Sample results that are not likely if
H0 is true are evidence that H0 is not true. To test H0 by using
a p-value, we use the following steps 4, 5, and 6:
Step 4: Collect the sample data and compute the value of the
test statistic. In the trash bag case, we have computed the value
of the test statistic to be z = 2.20.
Step 5: Calculate the p-value by using the test statistic value.
The p-value for testing H0: μ = 50 versus Ha: μ > 50 in the
trash bag case is the area under the standard normal curve to the
right of the test statistic value z = 2.20. As illustrated in Figure
9.3(b), this area is 1 − .9861 = .0139. The p-value is the
probability, computed assuming that H0: μ = 50 is true, of
observing a value of the test statistic that is greater than or

equal to the value z = 2.20 that we have actually computed from
the sample data. The p-value of .0139 says that, if H0: μ = 50 is
true, then only 139 in 10,000 of all possible test statistic values
are at least as large, or extreme, as the value z = 2.20. That is,
if we are to believe that H0 is true, we must believe that we
have observed a test statistic value that can be described as a
139 in 10,000 chance. Because it is difficult to believe that we
have observed a 139 in 10,000 chance, we intuitively have
strong evidence that H0: μ = 50 is false and Ha: μ > 50 is true.
Figure 9.3: Testing H0: μ = 50 versus Ha: μ > 50 by Using
Critical Values and the p-Value
Step 6: Reject H0 if the p-value is less than α. Recall that the
television network has set α equal to .05. The p-value of .0139
is less than the α of .05. Comparing the two normal curves in
Figures 9.3(a) and (b), we see that this implies that the test
statistic value z = 2.20 is greater than the critical value z.05 =
1.645. Therefore, we can reject H0 by setting α equal to .05. As
another example, suppose that the television network had set α
equal to .01. The p-value of .0139 is greater than the α of .01.
Comparing the two normal curves in Figures 9.3(b) and (c), we
see that this implies that the test statistic value z = 2.20 is less
than the critical value z.01 = 2.33. Therefore, we cannot reject
H0 by setting α equal to .01. Generalizing these examples, we
conclude that the value of the test statistic z will be greater than
the critical value zα if and only if the p-value is less than α.
That is, we can reject H0 in favor of Ha at level of significance
α if and only if the p-value is less than α.
© NBC, Inc. Used with permission.
Note: This logo appears on an NBC advertising standards
booklet. This booklet, along with other information provided by
NBC and CBS, forms the basis for much of the discussion in the
paragraph to the right.
Comparing the critical value and p-value methods
Thus far we have considered two methods for testing H0: μ = 50

versus Ha: μ > 50 at the .05 and .01 values of α. Using the first
method, we determine if the test statistic value z = 2.20 is
greater than the critical values z.05 = 1.645 and z.01 = 2.33.
Using the second method, we determine if the p-value of .0139
is less than .05 and .01. Whereas the critical value method
requires that we look up a different critical value for each
different α value, the p-value method requires only that we
calculate a single p-value and compare it directly with the
different α values. It follows that the p-value method is the most
efficient way to test a hypothesis at different α values. This can
be useful when there are different decision makers who might
use different α values. For example, television networks do not
always evaluate advertising claims by setting α equal to .05.
The reason is that the consequences of a Type I error
(advertising a false claim) are more serious for some claims
than for others. For example, the consequences of a Type I error
would be fairly serious for a claim about the effectiveness of a
drug or for the superiority of one product over another.
However, these consequences might not be as serious for a
noncomparative claim about an inexpensive and safe product,
such as a cosmetic. Networks sometimes use α values between
.01 and .04 for claims having more serious Type I error
consequences, and they sometimes use α values between .06 and
.10 for claims having less serious Type I error consequences.
Furthermore, one network’s policies for setting α can differ
somewhat from those of another. As a result, reporting an
advertising claim’s p-value to each network is the most efficient
way to tell the network whether to allow the claim to be
advertised. For example, most networks would evaluate the
trash bag claim by choosing an α value between .025 and .10.
Since the p-value of .0139 is less than all these α values, most
networks would allow the trash bag claim to be advertised.
A summary of the seven steps of hypothesis testing
For almost every hypothesis test discussed in this book,
statisticians have developed both a critical value rule and a p-
value that can be used to perform the hypothesis test.

Furthermore, it can be shown that for each hypothesis test the p-
value has been defined so that we can reject the null hypothesis
at level of significance α if and only if the p-value is less than
α. We now summarize a seven-step procedure for performing a
hypothesis test.
The Seven Steps of Hypothesis Testing
1State the null hypothesis H0 and the alternative hypothesis Ha.
2Specify the level of significance α.
3Select the test statistic.
Using a critical value rule:
4Determine the critical value rule for deciding whether to reject
H0. Use the specified value of α to find the critical value in the
critical value rule.
5Collect the sample data and compute the value of the test
statistic.
6Decide whether to reject H0 by using the test statistic value
and the critical value rule.
Using a p-value:
4Collect the sample data and compute the value of the test
statistic.
5Calculate the p-value by using the test statistic value.
6Reject H0 at level of significance α if the p-value is less than
α.
7Interpret your statistical results in managerial (real-world)
terms and assess their practical importance.
In the real world both critical value rules and p-values are used
to carry out hypothesis tests. For example, NBC uses critical
value rules, whereas CBS uses p-values, to statistically verify
the validity of advertising claims. Throughout this book we will
continue to present both the critical value and the p-value
approaches to hypothesis testing.
Testing a “less than” alternative hypothesis
We next consider the payment time case and testing a “less
than” alternative hypothesis:
Step 1: In order to study whether the new electronic billing
system reduces the mean bill payment time by more than 50

percent, the management consulting firm will test H0: μ ≥ 19.5
versus Ha: μ < 19.5.
Step 2: The management consulting firm wishes to make sure
that it truthfully describes the benefits of the new system both
to the Hamilton, Ohio, trucking company and to other
companies that are considering installing such a system.
Therefore, the firm will require very strong evidence to
conclude that μ is less than 19.5, which implies that it will test
H0: μ ≥ 19.5 versus Ha: μ < 19.5 by setting α equal to .01.
Step 3: In order to test H0: μ ≥ 19.5 versus Ha: μ < 19.5, we
will test the modified null hypothesis H0: μ = 19.5 versus Ha: μ
< 19.5. The idea here is that if there is sufficient evidence to
reject the hypothesis that μ equals 19.5 in favor of μ < 19.5,
then there is certainly also sufficient evidence to reject the
hypothesis that μ is greater than or equal to 19.5. In order to
test H0: μ = 19.5 versus Ha: μ < 19.5, we will randomly select a
sample of n = 65 invoices paid using the billing system and
calculate the mean of the payment times of these invoices.
Since the sample size is large, the Central Limit Theorem
applies, and we will utilize the test statistic
A value of the test statistic z that is less than zero results when
is less than 19.5. This provides evidence to support rejecting H0
in favor of Ha because the point estimate indicates that μ might
be less than 19.5.
Step 4: To decide how much less than zero the test statistic
must be to reject H0 in favor of Ha by setting the probability of
a Type I error equal to α, we do the following:
Place the probability of a Type I error, α, in the left-hand tail of
the standard normal curve and use the normal table to find the
critical value −zα. Here −zα is the negative of the normal point
zα. That is, −zα is the point on the horizontal axis under the
standard normal curve that gives a left-hand tail area equal to α.
Reject H0: μ = 19.5 in favor of Ha: μ < 19.5 if and only if the
test statistic z is less than the critical value −zα. Because α
equals .01, the critical value −zα is −z.01 = −2.33 [see Fig.

9.4(a)].
Figure 9.4: Testing H0: μ = 19.5 versus Ha: μ < 19.5 by Using
Step 5: When the sample of n = 65 invoices is randomly
selected, the mean of the payment times of these invoices is
calculated to be . Assuming that σ is known to equal 4.2, the
Step 6: Since the test statistic value z = −2.67 is less than the
critical value −z.01 = −2.33, we can reject H0: μ = 19.5 in favor
of Ha: μ < 19.5 by setting α equal to .01.
Step 7: We conclude (at an α of .01) that the mean payment time
for the new electronic billing system is less than 19.5 days.
This, along with the fact that the sample mean is slightly less
than 19.5, implies that it is reasonable for the management
consulting firm to conclude that the new electronic billing
system has reduced the mean payment time by slightly more
than 50 percent (a substantial improvement over the old
system).
A p-value for testing a “less than” alternative hypothesis
To test H0: μ = 19.5 versus Ha: μ < 19.5 in the payment time
case by using a p-value, we use the following steps 4, 5, and 6:
Step 4: We have computed the value of the test statistic in the
payment time case to be z = −2.67.
Step 5: The p-value for testing H0: μ = 19.5 versus Ha: μ < 19.5
is the area under the standard normal curve to the left of the test
statistic value z = −2.67. As illustrated in Figure 9.4(b), this
area is .0038. The p-value is the probability, computed
assuming that H0: μ = 19.5 is true, of observing a value of the
test statistic that is less than or equal to the value z = −2.67 that
we have actually computed from the sample data. The p-value
of .0038 says that, if H0: μ = 19.5 is true, then only 38 in
10,000 of all possible test statistic values are at least as
negative, or extreme, as the value z = −2.67. That is, if we are

to believe that H0 is true, we must believe that we have
observed a test statistic value that can be described as a 38 in
10,000 chance.
Step 6: The management consulting firm has set α equal to .01.
The p-value of .0038 is less than the α of .01. Therefore, we can
reject H0 by setting α equal to .01.
Testing a “not equal to” alternative hypothesis
We next consider the Valentine’s Day chocolate case and
testing a “not equal to” alternative hypothesis.
Step 1: To assess whether this year’s sales of its valentine box
of assorted chocolates will be ten percent higher than last
year’s, the candy company will test H0: μ = 330 versus Ha:
μ ≠ 330. Here, μ is the mean order quantity of this year’s
valentine box by large retail stores.
Step 2: If the candy company does not reject H0: μ = 330 and
H0: μ = 330 is false—a Type II error—the candy company will
base its production of valentine boxes on a 10 percent projected
sales increase that is not correct. Since the candy company
wishes to have a reasonably small probability of making this
Type II error, the company will set α equal to .05. Setting α
equal to .05 rather than .01 makes the probability of a Type II
error smaller than it would be if α were set at .01. Note that in
optional Section 9.5 we will verify that the probability of a
Type II error in this situation is reasonably small. Therefore, if
the candy company ends up not rejecting H0: μ = 330 and
therefore decides to base its production of valentine boxes on
the ten percent projected sales increase, the company can be
intuitively confident that it has made the right decision.
Step 3: The candy company will randomly select n = 100 large
retail stores and will make an early mailing to these stores
promoting this year’s valentine box of assorted chocolates. The
candy company will then ask each sampled retail store to report
its anticipated order quantity of valentine boxes and will
calculate the mean of the reported order quantities. Since the
sample size is large, the Central Limit Theorem applies, and we
will utilize the test statistic

A value of the test statistic that is greater than 0 results when
is greater than 330. This provides evidence to support rejecting
H0 in favor of Ha because the point estimate indicates that μ
might be greater than 330. Similarly, a value of the test statistic
that is less than 0 results when is less than 330. This also
provides evidence to support rejecting H0 in favor of Ha
because the point estimate indicates that μ might be less than
330.
Step 4: To decide how different from zero (positive or negative)
the test statistic must be in order to reject H0 in favor of Ha by
setting the probability of a Type I error equal to α, we do the
following:
Divide the probability of a Type I error, α, into two equal parts,
and place the area α/2 in the right-hand tail of the standard
normal curve and the area α/2 in the left-hand tail of the
standard normal curve. Then use the normal table to find the
critical values zα/2 and −zα/2. Here zα/2 is the point on the
horizontal axis under the standard normal curve that gives a
right-hand tail area equal to α/2, and −zα/2 is the point giving a
left-hand tail area equal to α/2.
Reject H0: μ = 330 in favor of Ha: μ ≠ 330 if and only if the
test statistic z is greater than the critical value zα/2 or less than
the critical value −zα/2. Note that this is equivalent to saying
that we should reject H0 if and only if the absolute value of the
test statistic, | z | is greater than the critical value zα/2.
Because α equals .05, the critical values are [see Figure 9.5(a)]
Figure 9.5: Testing H0: μ = 330 versus Ha: μ ≠ 330 by Using
Step 5: When the sample of n = 100 large retail stores is
randomly selected, the mean of their reported order quantities is
calculated to be . Assuming that σ is known to equal 40, the

Step 6: Since the test statistic value z = −1 is greater than −
z.025 = −1.96 (or, equivalently, since | z | = 1 is less than
z.025 = 1.96), we cannot reject H0: μ = 330 in favor of Ha:
μ ≠ 330 by setting α equal to .05.
Step 7: We cannot conclude (at an α of .05) that the mean order
quantity of this year’s valentine box by large retail stores will
differ from 330 boxes. Therefore, the candy company will base
its production of valentine boxes on the ten percent projected
sales increase.
A p-value for testing a “not equal to” alternative hypothesis
To test H0: μ = 330 versus Ha: μ ≠ 330 in the Valentine’s Day
chocolate case by using a p-value, we use the following steps 4,
5, and 6:
Step 4: We have computed the value of the test statistic in the
Valentine’s Day chocolate case to be z = −1.
Step 5: Note from Figure 9.5(b) that the area under the standard
normal curve to the right of | z | = 1 is .1587. Twice this
area—that is, 2(.1587) = .3174—is the p-value for testing H0: μ
= 330 versus Ha: μ ≠ 330. To interpret the p-value as a
probability, note that the symmetry of the standard normal
curve implies that twice the area under the curve to the right of
| z | = 1 equals the area under this curve to the right of 1 plus
the area under the curve to the left of −1 [see Figure 9.5(b)].
Also, note that since both positive and negative test statistic
values count against H0: μ = 330, a test statistic value that is
either greater than or equal to 1 or less than or equal to −1 is at
least as extreme as the observed test statistic value z = −1. It
follows that the p-value of .3174 says that, if H0: μ = 330 is
true, then 31.74 percent of all possible test statistic values are
at least as extreme as z = −1. That is, if we are to believe that
H0 is true, we must believe that we have observed a test
statistic value that can be described as a 31.74 percent chance.
Step 6: The candy company has set α equal to .05. The p-value
of .3174 is greater than the α of .05. Therefore, we cannot reject
H0 by setting α equal to .05.

A general procedure for testing a hypothesis about a population
mean
In the trash bag case we have tested H0: μ ≤ 50 versus Ha: μ >
50 by testing H0: μ = 50 versus Ha: μ > 50. In the payment time
case we have tested H0: μ ≥ 19.5 versus Ha: μ < 19.5 by testing
H0: μ = 19.5 versus Ha: μ < 19.5. In general, the usual
procedure for testing a “less than or equal to” null hypothesis or
a “greater than or equal to” null hypothesis is to change the null
hypothesis to an equality. We then test the “equal to” null
hypothesis versus the alternative hypothesis. Furthermore, the
critical value and p-value procedures for testing a null
hypothesis versus an alternative hypothesis depend upon
whether the alternative hypothesis is a “greater than,” a “less
than,” or a “not equal to” alternative hypothesis. The following
summary box gives the appropriate procedures. Specifically,
letting μ0 be a particular number, the summary box shows how
to test H0: μ = μ0 versus either Ha: μ > μ0, Ha: μ < μ0, or Ha: μ
≠ μ0:
Testing a Hypothesis about a Population Mean when σ Is
Known
Define the test statistic
and assume that the population sampled is normally distributed,
or that the sample size n is large. We can test H0: μ = μ0 versus
a particular alternative hypothesis at level of significance α by
using the appropriate critical value rule, or, equivalently, the
corresponding p-value.
Using confidence intervals to test hypotheses
Confidence intervals can be used to test hypotheses.
Specifically, it can be proven that we can reject H0: μ = μ0 in
favor of Ha: μ ≠ μ0 by setting the probability of a Type I error
equal to α if and only if the 100(1 − α) percent confidence
interval for μ does not contain μ0. For example, consider the
Valentine’s Day chocolate case and testing H0: μ = 330 versus
Ha: μ ≠ 330 by setting α equal to .05. To do this, we use the

mean of the sample of n = 100 reported order quantities to
calculate the 95 percent confidence interval for μ to be
Because this interval does contain 330, we cannot reject H0: μ =
330 in favor of Ha: μ ≠ 330 by setting α equal to .05.
Whereas we can use two-sided confidence intervals to test “not
equal to” alternative hypotheses, we must use one-sided
confidence intervals to test “greater than” or “less than”
alternative hypotheses. We will not study one-sided confidence
intervals in this book. However, it should be emphasized that
we do not need to use confidence intervals (one-sided or two-
sided) to test hypotheses. We can test hypotheses by using test
statistics and critical values or p-values, and these are the
approaches that we will feature throughout this book.
Measuring the weight of evidence against the null hypothesis
We have seen that in some situations the decision to take an
action is based solely on whether a null hypothesis can be
rejected in favor of an alternative hypothesis by setting α equal
to a single, prespecified value. For example, in the trash bag
case the television network decided to run the trash bag
commercial because H0: μ = 50 was rejected in favor of Ha: μ >
50 by setting α equal to .05. Also, in the payment time case the
management consulting firm decided to claim that the new
electronic billing system has reduced the Hamilton trucking
company’s mean payment time by more than 50 percent because
H0: μ = 19.5 was rejected in favor of Ha: μ < 19.5 by setting α
equal to .01. Furthermore, in the Valentine’s Day chocolate
case, the candy company decided to base its production of
valentine boxes on the ten percent projected sales increase
because H0: μ = 330 could not be rejected in favor of Ha:
μ ≠ 330 by setting α equal to .05.
Although hypothesis testing at a fixed α level is sometimes used
as the sale basis for deciding whether to take an action, this is
not always the case. For example, consider again the payment
time case. The reason that the management consulting firm
wishes to make the claim about the new electronic billing

system is to demonstrate the benefits of the new system both to
the Hamilton company and to other trucking companies that are
considering using such a system. Note, however, that a potential
user will decide whether to install the new system by
considering factors beyond the results of the hypothesis test.
For example, the cost of the new billing system and the
receptiveness of the company’s clients to using the new system
are among other factors that must be considered. In complex
business and industrial situations such as this, hypothesis
testing is used to accumulate knowledge about and understand
the problem at hand. The ultimate decision (such as whether to
adopt the new billing system) is made on the basis of
nonstatistical considerations, intuition, and the results of one or
more hypothesis tests. Therefore, it is important to know all the
information—called the weight of evidence—that a hypothesis
test provides against the null hypothesis and in favor of the
alternative hypothesis. Furthermore, even when hypothesis
testing at a fixed α level is used as the sole basis for deciding
whether to take an action, it is useful to evaluate the weight of
evidence. For example, the trash bag manufacturer would almost
certainly wish to know how much evidence there is that its new
bag is stronger than its former bag.
The most informative way to measure the weight of evidence is
to use the p-value. For every hypothesis test considered in this
book we can interpret the p-value to be the probability,
computed assuming that the null hypothesis H0 is true, of
observing a value of the test statistic that is at least as extreme
as the value actually computed from the sample data. The
smaller the p-value is, the less likely are the sample results if
the null hypothesis H0 is true. Therefore, the stronger is the
evidence that H0 is false and that the alternative hypothesis Ha
is true. Experience with hypothesis testing has resulted in
statisticians making the following (somewhat subjective)
conclusions:
Interpreting the Weight of Evidence against the Null Hypothesis
If the p-value for testing H0 is less than

•.10, we have some evidence that H0 is false.
•.05, we have strong evidence that H0 is false.
•.01, we have very strong evidence that H0 is false.
•.001, we have extremely strong evidence that H0 is false.
We will frequently use these conclusions in future examples.
Understand, however, that there are really no sharp borders
between different weights of evidence. Rather, there is really
only increasingly strong evidence against the null hypothesis as
the p-value decreases.
For example, recall that the p-value for testing H0: μ = 50
versus Ha: μ > 50 in the trash bag case is .0139. This p-value is
less than .05 but not less than .01. Therefore, we have strong
evidence, but not very strong evidence, that H0: μ = 50 is false
and Ha: μ > 50 is true. That is, we have strong evidence that the
mean breaking strength of the new trash bag exceeds 50 pounds.
As another example, the p-value for testing H0: μ = 19.5 versus
Ha: μ < 19.5 in the payment time case is .0038. This p-value is
less than .01 but not less than .001. Therefore, we have very
strong evidence, but not extremely strong evidence, that H0: μ =
19.5 is false and Ha: μ < 19.5 is true. That is, we have very
strong evidence that the new billing system has reduced the
mean payment time by more than 50 percent. Finally, the p-
value for testing H0: μ = 330 versus Ha: μ ≠ 330 in the
Valentine’s Day chocolate case is .3174. This p-value is greater
than .10. Therefore, we have little evidence that H0: μ = 330 is
false and Ha: μ ≠ 330 is true. That is, we have little evidence
that the increase in the mean order quantity of the valentine box
by large retail stores will differ from ten percent.
CONCEPTS
9.14Explain what a critical value is, and explain how it is used
to test a hypothesis.
9.15Explain what a p-value is, and explain how it is used to test
a hypothesis.

In Exercises 9.16 through 9.22 we consider using a random
sample of 100 measurements to test H0: μ = 80 versus Ha: μ >
80. If and σ = 20:
9.16Calculate the value of the test statistic z.
9.17Use a critical value to test H0 versus Ha by setting α equal
to .10.
to .05.
to .01.
to .001.
9.21Calculate the p-value and use it to test H0 versus Ha at each
of α = .10, .05, .01, and .001.
9.22How much evidence is there that H0: μ = 80 is false and
Ha: μ > 80 is true?
sample of 49 measurements to test H0: μ = 20 versus Ha: μ <
20. If and σ = 7:
to .10.
to .05.
to .01.
to .001.
of α = .10, .05, .01, and .001.
Ha: μ < 20 is true?
sample of n = 81 measurements to test H0: μ = 40 versus Ha:
μ ≠ 40. If and σ = 18:

9.31Use critical values to test H0 versus Ha by setting α equal
to .10.
to .05.
to .01.
to .001.
of α = .10, .05, .01, and .001.
Ha: μ ≠ 40 is true?
VideoGame
Recall that “very satisfied” customers give the XYZ-Box video
game system a rating that is at least 42. Suppose that the
manufacturer of the XYZ-Box wishes to use the random sample
of 65 satisfaction ratings to provide evidence supporting the
claim that the mean composite satisfaction rating for the XYZ-
Box exceeds 42.
aLetting μ represent the mean composite satisfaction rating for
the XYZ-Box, set up the null hypothesis H0 and the alternative
hypothesis Ha needed if we wish to attempt to provide evidence
supporting the claim that μ exceeds 42.
bThe random sample of 65 satisfaction ratings yields a sample
mean of . Assuming that σ equals 2.64, use critical values to
test H0 versus Ha at each of α = .10, .05, .01, and .001.
cUsing the information in part (b), calculate the p-value and use
it to test H0 versus Ha at each of α = .10, .05, .01, and .001.
dHow much evidence is there that the mean composite
satisfaction rating exceeds 42?
9.38THE BANK CUSTOMER WAITING TIME CASE
WaitTime
Letting μ be the mean waiting time under the new system, we
found in Exercise 9.9 that we should test H0: μ ≥ 6 versus Ha: μ
< 6 in order to attempt to provide evidence that μ is less than

six minutes. The random sample of 100 waiting times yields a
sample mean of minutes. Moreover, Figure 9.6 gives the
MINITAB output obtained when we use the waiting time data to
test H0: μ = 6 versus Ha: μ < 6. On this output the label “SE
Mean,” which stands for “the standard error of the mean,”
denotes the quantity , and the label “Z” denotes the calculated
test statistic. Assuming that σ equals 2.47:
aUse critical values to test H0 versus Ha at each of α = .10, .05,
.01, and .001.
bCalculate the p-value and verify that it equals .014, as shown
on the MINITAB output. Use the p-value to test H0 versus Ha at
each of α = .10, .05, .01, and .001.
cHow much evidence is there that the new system has reduced
the mean waiting time to below six minutes?
Figure 9.6: MINITAB Output of the Test of H0: μ = 6 versus
Ha: μ < 6 in the Bank Customer Waiting Time Case
Note: Because the test statistic z has a denominator that uses
the population standard deviation σ, MINITAB makes the user
specify an assumed value for σ.
9.39Again consider the audit delay situation of Exercise 8.11.
Letting μ be the mean audit delay for all public owner-
controlled companies in New Zealand, formulate the null
hypothesis H0 and the alternative hypothesis Ha that would be
used to attempt to provide evidence supporting the claim that μ
is less than 90 days. Suppose that a random sample of 100
public owner-controlled companies in New Zealand is found to
give a mean audit delay of days. Assuming that σ equals 32.83,
calculate the p-value for testing H0 versus Ha and determine
how much evidence there is that the mean audit delay for all
public owner-controlled companies in New Zealand is less than
90 days.
9.40Consolidated Power, a large electric power utility, has just
built a modern nuclear power plant. This plant discharges waste
water that is allowed to flow into the Atlantic Ocean. The
Environmental Protection Agency (EPA) has ordered that the

waste water may not be excessively warm so that thermal
pollution of the marine environment near the plant can be
avoided. Because of this order, the waste water is allowed to
cool in specially constructed ponds and is then released into the
ocean. This cooling system works properly if the mean
temperature of waste water discharged is 60°F or cooler.
Consolidated Power is required to monitor the temperature of
the waste water. A sample of 100 temperature readings will be
obtained each day, and if the sample results cast a substantial
amount of doubt on the hypothesis that the cooling system is
working properly (the mean temperature of waste water
discharged is 60°F or cooler), then the plant must be shut down
and appropriate actions must be taken to correct the problem.
aConsolidated Power wishes to set up a hypothesis test so that
the power plant will be shut down when the null hypothesis is
rejected. Set up the null hypothesis H0 and the alternative
hypothesis Ha that should be used.
bSuppose that Consolidated Power decides to use a level of
significance of α = .05, and suppose a random sample of 100
temperature readings is obtained. If the sample mean of the 100
temperature readings is , test H0 versus Ha and determine
whether the power plant should be shut down and the cooling
system repaired. Perform the hypothesis test by using a critical
value and a p-value. Assume σ = 2.
9.41Do part (b) of Exercise 9.40 if .
9.42Do part (b) of Exercise 9.40 if .
9.43An automobile parts supplier owns a machine that produces
a cylindrical engine part. This part is supposed to have an
outside diameter of three inches. Parts with diameters that are
too small or too large do not meet customer requirements and
must be rejected. Lately, the company has experienced problems
meeting customer requirements. The technical staff feels that
the mean diameter produced by the machine is off target. In
order to verify this, a special study will randomly sample 100
parts produced by the machine. The 100 sampled parts will be
measured, and if the results obtained cast a substantial amount

of doubt on the hypothesis that the mean diameter equals the
target value of three inches, the company will assign a problem-
solving team to intensively search for the causes of the problem.
aThe parts supplier wishes to set up a hypothesis test so that the
problem-solving team will be assigned when the null hypothesis
is rejected. Set up the null and alternative hypotheses for this
situation.
bA sample of 40 parts yields a sample mean diameter of inches.
Assuming σ equals .016, use a critical value and a p-value to
test H0 versus Ha by setting α equal to .05. Should the problem-
solving team be assigned?
9.44The Crown Bottling Company has just installed a new
bottling process that will fill 16-ounce bottles of the popular
Crown Classic Cola soft drink. Both overfilling and underfilling
bottles are undesirable: Underfilling leads to customer
complaints and overfilling costs the company considerable
money. In order to verify that the filler is set up correctly, the
company wishes to see whether the mean bottle fill, μ, is close
to the target fill of 16 ounces. To this end, a random sample of
36 filled bottles is selected from the output of a test filler run.
If the sample results cast a substantial amount of doubt on the
hypothesis that the mean bottle fill is the desired 16 ounces,
then the filler’s initial setup will be readjusted.
aThe bottling company wants to set up a hypothesis test so that
the filler will be readjusted if the null hypothesis is rejected.
Set up the null and alternative hypotheses for this hypothesis
test.
bSuppose that Crown Bottling Company decides to use a level
of significance of α = .01, and suppose a random sample of 36
bottle fills is obtained from a test run of the filler. For each of
the following three sample means, determine whether the
filler’s initial setup should be readjusted. In each case, use a
critical value and a p-value, and assume that σ equals .1.
9.45Use the first sample mean in Exercise 9.44 and a confidence
interval to perform the hypothesis test by setting α equal to .05.

What considerations would help you to decide whether the
result has practical importance?
9.46THE DISK BRAKE CASE
National Motors has equipped the ZX-900 with a new disk brake
system. We define the stopping distance for a ZX-900 as the
distance (in feet) required to bring the automobile to a complete
stop from a speed of 35 mph under normal driving conditions
using this new brake system. In addition, we define μ to be the
mean stopping distance of all ZX-900s. One of the ZX-900’s
major competitors is advertised to achieve a mean stopping
distance of 60 ft. National Motors would like to claim in a new
television commercial that the ZX-900 achieves a shorter mean
stopping distance.
aSet up the null hypothesis H0 and the alternative hypothesis
Ha that would be used to attempt to provide evidence supporting
the claim that μ is less than 60.
bA television network will permit National Motors to claim that
the ZX-900 achieves a shorter mean stopping distance than the
competitor if H0 can be rejected in favor of Ha by setting α
equal to .05. If the stopping distances of a random sample of n =
81 ZX-900s have a mean of , will National Motors be allowed to
run the commercial? Perform the hypothesis test by using a
critical value and a p-value. Assume here that σ = 6.02.
9.47Consider part (b) of Exercise 9.46, and calculate a 95
percent confidence interval for μ. Do the point estimate of μ and
confidence interval for μ indicate that μ might be far enough
below 60 feet to suggest that we have a practically important
result?
9.48Recall from Exercise 8.12 that Bayus (1991) studied the
mean numbers of auto dealers visited by early and late
replacement buyers.
aLetting μ be the mean number of dealers visited by early
replacement buyers, suppose that we wish to test H0: μ = 4
versus Ha: μ ≠ 4. A random sample of 800 early replacement
buyers yields a mean number of dealers visited of . Assuming σ
equals .71, calculate the p-value and test H0 versus Ha. Do we

estimate that μ is less than 4 or greater than 4?
bLetting μ be the mean number of dealers visited by late
replacement buyers, suppose that we wish to test H0: μ = 4
versus Ha: μ ≠ 4. A random sample of 500 late replacement
buyers yields a mean number of dealers visited of . Assuming σ
equals .66, calculate the p-value and test H0 versus Ha. Do we
estimate that μ is less than 4 or greater than 4?
9.3: t Tests about a Population Mean: σ Unknown
If we do not know σ (which is usually the case), we can base a
hypothesis test about μ on the sampling distribution of
If the sampled population is normally distributed, then this
sampling distribution is a t distribution having n − 1 degrees of
freedom. This leads to the following results:
A t Tests about a Population Mean: σ UnKnown
and assume that the population sampled is normally distributed.
We can test H0: μ = μ0 versus a particular alternative
hypothesis at level of significance α by using the appropriate
critical value rule, or, equivalently, the corresponding p-value.
Here tα, tα/2, and the p-values are based on n − 1 degrees of
freedom.
In the rest of this chapter and in Chapter 10 we will present
most of the hypothesis testing examples by using hypothesis
testing summary boxes and the seven hypothesis testing steps
given in the previous section. However, to be concise, we will
not formally number each hypothesis testing step. Rather, for
each of the first six steps, we will set out in boldface font a key
phrase that indicates that the step is being carried out. Then, we
will highlight the seventh step—the business improvement
conclusion—as we highlight all business improvement
conclusions in this book. After Chapter 10, we will continue to
use hypothesis testing summary boxes, and we will more
informally use the seven steps.

As illustrated in the following example, we will often first use a
critical value rule to test the hypotheses under consideration at
a fixed value of α and then use a p-value to assess the weight of
evidence against the null hypothesis.
EXAMPLE 9.4
In 1991 the average interest rate charged by U.S. credit card
issuers was 18.8 percent. Since that time, there has been a
proliferation of new credit cards affiliated with retail stores, oil
companies, alumni associations, professional sports teams, and
so on. A financial officer wishes to study whether the increased
competition in the credit card business has reduced interest
rates. To do this, the officer will test a hypothesis about the
current mean interest rate, μ, charged by U.S. credit card
issuers. The null hypothesis to be tested is H0: μ = 18.8%, and
the alternative hypothesis is Ha: μ < 18.8%. If H0 can be
rejected in favor of Ha at the .05 level of significance, the
officer will conclude that the current mean interest rate is less
than the 18.8% mean interest rate charged in 1991. To perform
the hypothesis test, suppose that we randomly select n = 15
credit cards and determine their current interest rates. The
interest rates for the 15 sampled cards are given in Table 9.3. A
stem-and-leaf display and MINITAB box plot are given in
Figure 9.7. The stem-and-leaf display looks reasonably mound-
shaped, and both the stem-and-leaf display and the box plot
look reasonably symmetrical. It follows that it is appropriate to
calculate the value of the test statistic t in the summary box.
Furthermore, since Ha: μ < 18.8% is of the form Ha: μ < μ0, we
should reject H0: μ = 18.8% if the value of t is less than the
critical value −tα = −t.05 = −1.761. Here, −t.05 = −1.761 is
based on n − 1 = 15 −1 = 14 degrees of freedom and this critical
value is illustrated in Figure 9.8(a). The mean and the standard
deviation of the n = 15 interest rates in Table 9.3 are and s =
1.538. This implies that the value of the test statistic is
Figure 9.7: Stem-and-Leaf Display and Box Plot of the Interest
Rates

Figure 9.8: Testing H0: μ = 18.8% versus Ha: μ < 18.8% by
Using a Critical Value and a p-Value
Table 9.3: Interest Rates Charged by 15 Randomly Selected
Credit Cards CreditCd
Since t = −4.97 is less than −t.05 = −1.761, we reject H0: μ =
18.8% in favor of Ha: μ < 18.8%. That is, we conclude (at an α
of .05) that the current mean credit card interest rate is lower
than 18.8%, the mean interest rate in 1991. Furthermore, the
sample mean says that we estimate the mean interest rate is
18.8% − 16.827% = 1.973% lower than it was in 1991.
The p-value for testing H0: μ = 18.8% versus Ha: μ < 18.8% is
the area under the curve of the t distribution having 14 degrees
of freedom to the left of t = −4.97. Tables of t points (such as
Table A.4, page 864) are not complete enough to give such
areas for most t statistic values, so we use computer software
packages to calculate p-values that are based on the t
distribution. For example, the MINITAB output in Figure 9.9(a)
and the MegaStat output in Figure 9.10 tell us that the p-value
for testing H0: μ = 18.8% versus Ha: μ < 18.8% is .0001. Notice
that both MINITAB and MegaStat round p-values to three or
four decimal places. The Excel output in Figure 9.9(b) gives the
slightly more accurate value of 0.000103 for the p-value.
Because this p-value is less than .05, .01, and .001, we can
reject H0 at the .05, .01, and .001 levels of significance. Also
note that the p-value of .0001 on the MegaStat output is shaded
dark yellow. This indicates that we can reject H0 at the .01
level of significance (light yellow shading would indicate
significance at the .05 level, but not at the .01 level). As a
probability, the p-value of .0001 says that if we are to believe
that H0: μ = 18.8% is true, we must believe that we have
observed a t statistic value (t = −4.97) that can be described as a
1 in 10,000 chance. In summary, we have extremely strong

evidence that H0: μ = 18.8% is false and Ha: μ < 18.8% is true.
That is, we have extremely strong evidence that the current
mean credit card interest rate is less than 18.8%.
Figure 9.9: The MINITAB and Excel Outputs for Testing H0: μ
= 18.8% versus Ha: μ < 18.8%
Figure 9.10: The MegaStat Output for Testing H0: μ = 18.8%
versus Ha: μ < 18.8%
Recall that in three cases discussed in Section 9.2 we tested
hypotheses by assuming that the population standard deviation σ
is known and by using z tests. If σ is actually not known in
these cases (which would probably be true), we should test the
hypotheses under consideration by using t tests. Furthermore,
recall that in each case the sample size is large (at least 30). In
general, it can be shown that if the sample size is large, the t
test is approximately valid even if the sampled population is not
normally distributed (or mound shaped). Therefore, consider the
Valentine’s Day chocolate case and testing H0: μ = 330 versus
Ha: μ ≠ 330 at the .05 level of significance. To perform the
hypothesis test, assume that we will randomly select n = 100
large retail stores and use their anticipated order quantities to
calculate the value of the test statistic t in the summary box.
Then, since the alternative hypothesis Ha: μ ≠ 330 is of the
form Ha: μ ≠ μ0, we will reject H0: μ = 330 if the absolute
value of t is greater than tα/2 = t.025 = 1.984 (based on n − 1 =
99 degrees of freedom). Suppose that when the sample is
randomly selected, the mean and the standard deviation of the n
= 100 reported order quantities are calculated to be and s =
39.1. The value of the test statistic is
Since | t | = 1.023 is less than t.025 = 1.984, we cannot reject
H0: μ = 330 by setting α equal to .05. It follows that we cannot
conclude (at an α of .05) that this year’s mean order quantity of
the valentine box by large retail stores will differ from 330

boxes. Therefore, the candy company will base its production of
valentine boxes on the ten percent projected sales increase. The
p-value for the hypothesis test is twice the area under the t
distribution curve having 99 degrees of freedom to the right of |
t | = 1.023. Using a computer, we find that this p-value is .3088,
which provides little evidence against H0: μ = 330 and in favor
of Ha: μ ≠ 330.
As another example, consider the trash bag case and note that
the sample of n = 40 trash bag breaking strengths has mean and
standard deviation s = 1.6438. The p-value for testing H0: μ =
50 versus Ha: μ > 50 is the area under the t distribution curve
having n − 1 = 39 degrees of freedom to the right of
Using a computer, we find that this p-value is .0164, which
provides strong evidence against H0: μ = 50 and in favor of Ha:
μ > 50. In particular, recall that most television networks would
evaluate the claim that the new trash bag has a mean breaking
strength that exceeds 50 pounds by choosing on α value between
.025 and .10. It follows, since the p-value of .0164 is less than
all these α values, that most networks would allow the trash bag
claim to be advertised.
As a third example, consider the payment time case and note
that the sample of n = 65 payment times has mean and standard
deviation s = 3.9612. The p-value for testing H0: μ = 19.5
versus Ha: μ < 19.5 is the area under the t distribution curve
having n − 1 = 64 degrees of freedom to the left of
Using a computer, we find that this p-value is .0031, which is
less than the management consulting firm’s α value of .01. It
follows that the consulting firm will claim that the new
electronic billing system has reduced the Hamilton, Ohio,
trucking company’s mean bill payment time by more than 50
percent.
To conclude this section, note that if the sample size is small

(<30) and the sampled population is not mound-shaped, or if the
sampled population is highly skewed, then it might be
appropriate to use a nonparametric test about the population
median. Such a test is discussed in Chapter 18.
CONCEPTS
9.49What assumptions must be met in order to carry out the test
about a population mean based on the t distribution?
9.50How do we decide whether to use a z test or a t test when
testing a hypothesis about a population mean?
9.51Suppose that a random sample of 16 measurements from a
normally distributed population gives a sample mean of and a
sample standard deviation of s = 6. Use critical values to test
H0: μ ≤ 10 versus Ha: μ > 10 using levels of significance α =
.10, α = .05, α = .01, and α = .001. What do you conclude at
each value of α?
9.52Suppose that a random sample of nine measurements from a
normally distributed population gives a sample mean of and a
sample standard deviation of s = .3. Use critical values to test
H0: μ = 3 versus Ha: μ ≠ 3 using levels of significance α = .10,
α = .05, α = .01, and α = .001. What do you conclude at each
value of α?
9.53THE AIR TRAFFIC CONTROL CASE AlertTime
Recall that it is hoped that the mean alert time, μ, using the new
display panel is less than eight seconds. Formulate the null
hypothesis H0 and the alternative hypothesis Ha that would be
used to attempt to provide evidence that μ is less than eight
seconds. The mean and the standard deviation of the sample of n
= 15 alert times are and s = 1.0261. Perform a t test of H0
versus Ha by setting α equal to .05 and using a critical value.
Interpret the results of the test.
9.54THE AIR TRAFFIC CONTROL CASE AlertTime
The p-value for the hypothesis test of Exercise 9.53 can be
computer calculated to be .0200. How much evidence is there

that μ is less than eight seconds?
9.55The bad debt ratio for a financial institution is defined to be
the dollar value of loans defaulted divided by the total dollar
value of all loans made. Suppose that a random sample of seven
Ohio banks is selected and that the bad debt ratios (written as
percentages) for these banks are 7%, 4%, 6%, 7%, 5%, 4%, and
9%. BadDebt
aBanking officials claim that the mean bad debt ratio for all
Midwestern banks is 3.5 percent and that the mean bad debt
ratio for Ohio banks is higher. Set up the null and alternative
hypotheses needed to attempt to provide evidence supporting
the claim that the mean bad debt ratio for Ohio banks exceeds
3.5 percent.
bAssuming that bad debt ratios for Ohio banks are
approximately normally distributed, use critical values and the
given sample information to test the hypotheses you set up in
part a by setting α equal to .10, .05, .01, and .001. How much
evidence is there that the mean bad debt ratio for Ohio banks
exceeds 3.5 percent? What does this say about the banking
official’s claim?
cAre you qualified to decide whether we have a practically
important result? Who would be? How might practical
importance be defined in this situation?
dThe p-value for the hypothesis test of part (b) can be computer
calculated to be .006. What does this p-value say about whether
the mean bad debt ratio for Ohio banks exceeds 3.5 percent?
9.56In the book Business Research Methods, Donald R. Cooper
and C. William Emory (1995) discuss using hypothesis testing
to study receivables outstanding. To quote Cooper and Emory:
…the controller of a large retail chain may be concerned about a
possible slowdown in payments by the company’s customers.
She measures the rate of payment in terms of the average
number of days receivables outstanding. Generally, the
company has maintained an average of about 50 days with a
standard deviation of 10 days. Since it would be too expensive
to analyze all of a company’s receivables frequently, we

normally resort to sampling.
aSet up the null and alternative hypotheses needed to attempt to
show that there has been a slowdown in payments by the
company’s customers (there has been a slowdown if the average
days outstanding exceeds 50).
bAssume approximate normality and suppose that a random
sample of 25 accounts gives an average days outstanding of
with a standard deviation of s = 8. Use critical values to test the
hypotheses you set up in part a at levels of significance α = .10,
α = .05, α = .01, and α = .001. How much evidence is there of a
slowdown in payments?
cAre you qualified to decide whether this result has practical
importance? Who would be?
9.57Consider a chemical company that wishes to determine
whether a new catalyst, catalyst XA-100, changes the mean
hourly yield of its chemical process from the historical process
mean of 750 pounds per hour. When five trial runs are made
using the new catalyst, the following yields (in pounds per
hour) are recorded: 801, 814, 784, 836, and 820. ChemYield
aLet μ be the mean of all possible yields using the new catalyst.
Assuming that chemical yields are approximately normally
distributed, the MegaStat output of the test statistic and p-value,
and the Excel output of the p-value, for testing H0: μ = 750
versus Ha: μ ≠ 750 are as follows:
(Here we had Excel calculate twice the area under the t
distribution curve having 4 degrees of freedom to the right of
6.942585.) Use the sample data to verify that the values of , s,
and t given on the output are correct.
bUse the test statistic and critical values to test H0 versus Ha by
setting α equal to .10, .05, .01, and .001.
9.58Consider Exercise 9.57. Use the p-value to test H0: μ = 750
versus Ha: μ ≠ 750 by setting α equal to .10, .05, .01, and .001.
How much evidence is there that the new catalyst changes the
mean hourly yield?
9.59Whole Foods is an all-natural grocery chain that has 50,000

square foot stores, up from the industry average of 34,000
square feet. Sales per square foot of supermarkets average just
under $400 per square foot, as reported by USA Today in an
article on “A whole new ballgame in grocery shopping.”
Suppose that sales per square foot in the most recent fiscal year
are recorded for a random sample of 10 Whole Foods
supermarkets. The data (sales dollars per square foot) are as
follows: 854, 858, 801, 892, 849, 807, 894, 863, 829, 815. Let μ
denote the mean sales dollars per square foot for all Whole
Foods supermarkets during the most recent fiscal year, and note
that the historical mean sales dollars per square foot for Whole
Foods supermarkets in previous years has been $800. Below we
present the MINITAB output obtained by using the sample data
to test H0: μ = 800 versus Ha: μ > 800. WholeFoods
aUse the p-value to test H0 versus Ha by setting α equal to .10,
.05, and .01.
bHow much evidence is there that μ exceeds $800?
9.60Consider Exercise 9.59. Do you think that the difference
between the sample mean of $846.20 and the historical average
of $800 has practical importance?
VideoGame
The mean and the standard deviation of the sample of n = 65
customer satisfaction ratings are and s = 2.6424. Let μ denote
the mean of all possible customer satisfaction ratings for the
XYZ-Box video game system, and consider testing H0: μ = 42
versus Ha: μ > 42 Perform a t test of these hypotheses by setting
α equal to .05 and using a critical value. Also, interpret the p-
value of .0025 for the hypothesis test.
9.62THE BANK CUSTOMER WAITING TIME CASE
WaitTime
The mean and the standard deviation of the sample of 100 bank
customer waiting times are and s = 2.475. Let μ denote the
mean of all possible bank customer waiting times using the new
system and consider testing H0: μ = 6 versus Ha: μ < 6. Perform

a t test of these hypotheses by setting α equal to .05 and using a
critical value. Also, interpret the p-value of .0158 for the
hypothesis test.
9.4: z Tests about a Population Proportion
In this section we study a large sample hypothesis test about a
population proportion (that is, about the fraction of population
units that possess some characteristic). We begin with an
example.
EXAMPLE 9.5: The Cheese Spread Case
Recall that the soft cheese spread producer has decided that
replacing the current spout with the new spout is profitable only
if p, the true proportion of all current purchasers who would
stop buying the cheese spread if the new spout were used, is
less than .10. The producer feels that it is unwise to change the
spout unless it has very strong evidence that p is less than .10.
Therefore, the spout will be changed if and only if the null
hypothesis H0: p = .10 can be rejected in favor of the
alternative hypothesis Ha: p < .10 at the .01 level of
significance.
In order to see how to test this kind of hypothesis, remember
that when n is large, the sampling distribution of
is approximately a standard normal distribution. Let p0 denote a
specified value between 0 and 1 (its exact value will depend on
the problem), and consider testing the null hypothesis H0: p =
p0. We then have the following result:
A Large Sample Test about a Population Proportion
If the sample size n is large, we can test H0: p = p0 versus a
particular alternative hypothesis at level of significance α by
using the appropriate critical value rule, or, equivalently, the
corresponding p-value.
Here n should be considered large if both np0 and n(1 − p0) are
at least 5.3

EXAMPLE 9.6: The Cheese Spread Case
We have seen that the cheese spread producer wishes to test H0:
p = .10 versus Ha: p < .10, where p is the proportion of all
current purchasers who would stop buying the cheese spread if
the new spout were used. The producer will use the new spout if
H0 can be rejected in favor of Ha at the .01 level of
significance. To perform the hypothesis test, we will randomly
select n = 1,000 current purchasers of the cheese spread, find
the proportion of these purchasers who would stop buying the
cheese spread if the new spout were used, and calculate the
value of the test statistic z in the summary box. Then, since the
alternative hypothesis Ha: p < .10 is of the form Ha: p < p0, we
will reject H0: p = .10 if the value of z is less than −zα =
−z.01= −2.33. (Note that using this procedure is valid because
np0 = 1,000(.10) = 100 and n(1 − p0) = 1,000(1 − .10) = 900 are
both at least 5.) Suppose that when the sample is randomly
selected, we find that 63 of the 1,000 current purchasers say
they would stop buying the cheese spread if the new spout were
used. Since , the value of the test statistic is
Because z = −3.90 is less than −z.01 = −2.33, we reject H0: p =
.10 in favor of Ha: p < .10. That is, we conclude (at an α of .01)
that the proportion of current purchasers who would stop buying
the cheese spread if the new spout were used is less than .10. It
follows that the company will use the new spout. Furthermore,
the point estimate says we estimate that 6.3 percent of all
current customers would stop buying the cheese spread if the
new spout were used.
Although the cheese spread producer has made its decision by
setting α equal to a single, prechosen value (.01), it would
probably also wish to know the weight of evidence against H0
and in favor of Ha. The p-value is the area under the standard
normal curve to the left of z = − 3.90. Table A.3 (page 862)
tells us that this area is .00005. Because this p-value is less than
.001, we have extremely strong evidence that Ha: p < .10 is

true. That is, we have extremely strong evidence that fewer than
10 percent of current purchasers would stop buying the cheese
spread if the new spout were used.
EXAMPLE 9.7
Recent medical research has sought to develop drugs that lessen
the severity and duration of viral infections. Virol, a relatively
new drug, has been shown to provide relief for 70 percent of all
patients suffering from viral upper respiratory infections. A
major drug company is developing a competing drug called
Phantol. The drug company wishes to investigate whether
Phantol is more effective than Virol. To do this, the drug
company will test a hypothesis about the true proportion, p, of
all patients whose symptoms would be relieved by Phantol. The
null hypothesis to be tested is H0: p = .70, and the alternative
hypothesis is Ha: p > .70. If H0 can be rejected in favor of Ha
at the .05 level of significance, the drug company will conclude
that Phantol helps more than the 70 percent of patients helped
by Virol. To perform the hypothesis test, we will randomly
select n = 300 patients having viral upper respiratory infections,
find the proportion of these patients whose symptoms are
relieved by Phantol and calculate the value of the test statistic z
in the summary box. Then, since the alternative hypothesis Ha:
p > .70 is of the form Ha: p> p0, we will reject H0: p = .70 if
the value of z is greater than zα = z.05 = 1.645. (Note that using
this procedure is valid because np0 = 300(.70) = 210 and n(1 −
p0) = 300(1 − .70) = 90 are both at least 5.) Suppose that when
the sample is randomly selected, we find that Phantol provides
relief for 231 of the 300 patients. Since the value of the test
statistic is
Because z = 2.65 is greater than z.05 = 1.645, we reject H0: p =
.70 in favor of Ha: p > .70. That is, we conclude (at an α of .05)
that Phantol will provide relief for more than 70 percent of all
patients suffering from viral upper respiratory infections. More
specifically, the point estimate of p says that we estimate that

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