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Chapter
10
Hypothesis Tests
Regarding a
Parameter
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Section
10.2
Hypothesis Tests
for a Population
Mean-Population
Standard Deviation
Known
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Objectives
1. Explain the logic of hypothesis testing
2. Test the hypotheses about a population mean with 
known using the classical approach
3. Test hypotheses about a population mean with 
known using P-values
4. Test hypotheses about a population mean with 
known using confidence intervals
5. Distinguish between statistical significance and
practical significance.
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Objective 1
• Explain the Logic of Hypothesis Testing
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To test hypotheses regarding the population mean
assuming the population standard deviation is
known, two requirements must be satisfied:
1. A simple random sample is obtained.
2. The population from which the sample is
drawn is normally distributed or the sample
size is large (n≥30).
If these requirements are met, the distribution of
x is normal with mean  and standard
deviation  .
n
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Recall the researcher who believes that the mean
length of a cell phone call has increased from its
March, 2006 mean of 3.25 minutes. Suppose we
take a simple random sample of 36 cell phone calls.
Assume the standard deviation of the phone call
lengths is known to be 0.78 minutes. What is the
sampling distribution of the sample mean?
x
is normally distributed with mean 3.25
and standard deviation 0.78 36  0.13.
Answer:
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Suppose the sample of 36 calls resulted in a sample
mean of 3.56 minutes. Do the results of this
sample suggest that the researcher is correct? In
other words, would it be unusual to obtain a
sample mean of 3.56 minutes from a population
whose mean is 3.25 minutes? What is
convincing or statistically significant evidence?
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When observed results are unlikely under
the assumption that the null hypothesis is
true, we say the result is statistically
significant. When results are found to be
statistically significant, we reject the null
hypothesis.
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The Logic of the Classical Approach
One criterion we may use for sufficient evidence
for rejecting the null hypothesis is if the sample
mean is too many standard deviations from the
assumed (or status quo) population mean. For
example, we may choose to reject the null
hypothesis if our sample mean is more than 2
standard deviations above the population mean
of 3.25 minutes.
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Recall that our simple random sample of 36 calls
resulted in a sample mean of 3.56 minutes with
standard deviation of 0.13. Thus, the sample
mean is
3.56  3.25
z
 2.38
0.13
standard deviations above the hypothesized mean
of 3.25 minutes.

Therefore, using our criterion, we would reject the
null hypothesis and conclude that the mean
cellular call length is greater than 3.25 minutes.
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Why does it make sense to reject the null
hypothesis if the sample mean is more than 2
standard deviations above the hypothesized mean?
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If the null hypothesis were true, then 10.0228=0.9772=97.72% of all sample
means will be less than
3.25+2(0.13)=3.51.
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Because sample means greater than 3.51 are unusual if
the population mean is 3.25, we are inclined to believe
the population mean is greater than 3.25.
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The Logic of the P-Value Approach
A second criterion we may use for sufficient
evidence to support the alternative hypothesis is
to compute how likely it is to obtain a sample
mean at least as extreme as that observed from a
population whose mean is equal to the value
assumed by the null hypothesis.
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We can compute the probability of obtaining a
sample mean of 3.56 or more using the normal
model.
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Recall
3.56  3.25
z
 2.38
0.13
So, we compute
P x  3.56  P(Z  2.38)  0.0087.
The 
probability of obtaining a sample mean of 3.56
minutes or more from a population whose mean is
3.25 minutes is 0.0087. This means that fewer than

1 sample in 100 will give us a mean as high or
higher than 3.56 if the population mean really is
3.25 minutes. Since this outcome is so unusual, we
take this as evidence against the null hypothesis.
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Premise of Testing a Hypothesis
Using the P-value Approach
Assuming that H0 is true, if the probability of
getting a sample mean as extreme or more
extreme than the one obtained is small, we
reject the null hypothesis.
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Objective 2
• Test Hypotheses about a Population Mean with
 Known Using the Classical Approach
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Testing Hypotheses Regarding the
Population Mean with σ Known Using the
Classical Approach
To test hypotheses regarding the population
mean with  known, we can use the steps
that follow, provided that two requirements
are satisfied:
1. The sample is obtained using simple random
sampling.
2. The sample has no outliers, and the
population from which the sample is drawn
is normally distributed or the sample size is
large (n ≥ 30).
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Step 1: Determine the null and alternative
hypotheses. Again, the hypotheses can
be structured in one of three ways:
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Step 2: Select a level of significance, , based
on the seriousness of making a
Type I error.
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Step 3: Provided that the population from which the
sample is drawn is normal or the sample size
is large, and the population standard
deviation, , is known, the distribution of the
sample mean, x , 0 is normal with mean
and standard deviation  .
n
Therefore,
x 
0
z0 

 
n 
represents the number of standard deviations that
the sample mean is from the assumed mean. This
value is called the test statistic.

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Step 4: The level of significance is used to
determine the critical value. The critical
region represents the maximum number
of standard deviations that the sample
mean can be from 0 before the null
hypothesis is rejected. The critical
region or rejection region is the set of all
values such that the null hypothesis is
rejected.
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Two-Tailed
(critical value)
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Left-Tailed
(critical value)
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Right-Tailed
(critical value)
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Step 5: Compare the critical value with the test
statistic:
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Step 6: State the conclusion.
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The procedure is robust, which means that
minor departures from normality will not
adversely affect the results of the test.
However, for small samples, if the data have
outliers, the procedure should not be used.
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Parallel Example 2: The Classical Approach to Hypothesis
Testing
A can of 7-Up states that the contents of the can are 355 ml.
A quality control engineer is worried that the filling machine
is miscalibrated. In other words, she wants to make sure the
machine is not under- or over-filling the cans. She randomly
selects 9 cans of 7-Up and measures the contents. She
obtains the following data.
351
358
360 358 356 359
355 361 352
Is there evidence at the =0.05 level of significance to
support the quality control engineer’s claim? Prior
experience indicates that =3.2ml.
Source: Michael McCraith, Joliet Junior College
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Solution
The quality control engineer wants to know if the mean
content is different from 355 ml. Since the sample size
is small, we must verify that the data come from a
population that is approximately normal with no
outliers.
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Normal Probability Plot for Contents (ml)
Assumption of normality appears reasonable.
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No outliers.
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Solution
Step 1: H0: =355
versus
H1: ≠355
Step 2: The level of significance is =0.05.
Step 3: The sample mean is calculated to be 356.667.
The test statistic is then
356.667  355
z0 
1.56
3.2 9
The sample mean of 356.667 is 1.56 standard deviations
above the assumed mean of 355 ml.
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Solution
Step 4: Since this is a two-tailed test, we determine the
critical values at the =0.05 level of
significance to be -z0.025= -1.96 and z0.025=1.96
Step 5: Since the test statistic, z0=1.56, is less than the
critical value 1.96, we fail to reject the null
hypothesis.
Step 6: There is insufficient evidence at the =0.05
level of significance to conclude that the mean
content differs from 355 ml.
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Objective 3
• Test Hypotheses about a Population Mean with
 Known Using P-values.
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A P-value is the probability of observing a
sample statistic as extreme or more extreme
than the one observed under the assumption
that the null hypothesis is true.
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Testing Hypotheses Regarding the
Population Mean with σ Known Using
P-values
To test hypotheses regarding the population mean
with  known, we can use the steps that follow
to compute the P-value, provided that two
requirements are satisfied:
1. The sample is obtained using simple random
sampling.
2. The sample has no outliers, and the population
from which the sample is drawn is normally
distributed or the sample size is large (n ≥ 30).
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Step 1: A claim is made regarding the population
mean. The claim is used to determine the
null and alternative hypotheses. Again,
the hypothesis can be structured in one of
three ways:
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Step 2: Select a level of significance, , based
on the seriousness of making a
Type I error.
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Step 3: Compute the test statistic,
z0 
x  0

n

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Step 4: Determine the P-value
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Step 5: Reject the null hypothesis if the
P-value is less than the level of
significance, . The comparison of
the P-value and the level of
significance is called the
decision rule.
Step 6: State the conclusion.
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Parallel Example 3: The P-Value Approach to Hypothesis
Testing: Left-Tailed, Large Sample
The volume of a stock is the number of shares traded in
the stock in a day. The mean volume of Apple stock in
2007 was 35.14 million shares with a standard deviation
of 15.07 million shares. A stock analyst believes that the
volume of Apple stock has increased since then. He
randomly selects 40 trading days in 2008 and determines
the sample mean volume to be 41.06 million shares.
Test the analyst’s claim at the =0.10 level of
significance using P-values.
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Solution
Step 1: The analyst wants to know if the stock volume
has increased. This is a right-tailed test with
H0: =35.14
versus
H1: >35.14.
We want to know the probability of obtaining a
sample mean of 41.06 or more from a
population where the mean is assumed to be
35.14.
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Solution
Step 2: The level of significance is =0.10.
Step 3: The test statistic is
41.06  35.14
z0 
 2.48
15.07 40

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Solution
Step 4: P(Z > z0)=P(Z > 2.48)=0.0066.
The probability of obtaining a sample mean of
41.06 or more from a population whose mean is
35.14 is 0.0066.
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Solution
Step 5: Since the P-value= 0.0066 is less than the level
of significance, 0.10, we reject the null
hypothesis.
Step 6: There is sufficient evidence to reject the null
hypothesis and to conclude that the mean
volume of Apple stock is greater than
35.14 million shares.
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Parallel Example 4: The P-Value Approach of Hypothesis
Testing: Two-Tailed, Small Sample
A can of 7-Up states that the contents of the can are 355 ml.
A quality control engineer is worried that the filling machine
is miscalibrated. In other words, she wants to make sure the
machine is not under- or over-filling the cans. She randomly
selects 9 cans of 7-Up and measures the contents. She
obtains the following data.
351
360
358
356
359 358 355 361 352
Use the P-value approach to determine if there is evidence
at the =0.05 level of significance to support the quality
control engineer’s claim. Prior experience indicates that
=3.2ml.
Source: Michael McCraith, Joliet Junior College
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Solution
The quality control engineer wants to know if the mean
content is different from 355 ml. Since we have already
verified that the data come from a population that is
approximately normal with no outliers, we will continue
with step 1.
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Solution
Step 1: The quality control engineer wants to know if
the content has changed. This is a two-tailed
test with
H0: =355
versus
H1: ≠355.
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Solution
Step 2: The level of significance is =0.05.
Step 3: Recall that the sample mean is 356.667. The
test statistic is then
356.667  355
z0 
1.56
3.2 9

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Solution
Step 4: Since this is a two-tailed test,
P-value = P(Z < -1.56 or Z > 1.56) =
2*(0.0594)=0.1188.
The probability of obtaining a sample mean that
is more than 1.56 standard deviations away from
the assumed mean of 355 ml is 0.1188.
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Solution
Step 5: Since the P-value= 0.1188 is greater than the
level of significance, 0.05, we fail to reject the
null hypothesis.
Step 6: There is insufficient evidence to conclude that
the mean content of 7-Up cans differs from
355 ml.
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One advantage of using P-values over the
classical approach in hypothesis testing is that
P-values provide information regarding the
strength of the evidence. Another is that Pvalues are interpreted the same way regardless
of the type of hypothesis test being performed.
the lower the P-value, the stronger the
evidence against the statement in the null
hypothesis.
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Objective 4
• Test Hypotheses about a Population Mean with
 Known Using Confidence Intervals
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When testing H0:  = 0 versus H1:  ≠ 0,
if a (1-)·100% confidence interval contains
0, we do not reject the null hypothesis.
However, if the confidence interval does not
contain 0, we have sufficient evidence that
supports the statement in the alternative
hypothesis and conclude that  ≠ 0 at the
level of significance, .
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Parallel Example 6: Testing Hypotheses about a Population
Mean Using a Confidence Interval
Test the hypotheses presented in Parallel Examples 2 and
4 at the =0.05 level of significance by constructing a
95% confidence interval about , the population mean can
content.
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Solution
3.2
 354.58
Lower bound: 356.667 1.96 
9
3.2
Upper bound: 356.667  1.96 
 358.76
9

We are 95% confident that the mean can content is
between
 354.6 ml and 358.8 ml. Since the mean stated in
the null hypothesis is in this interval, there is insufficient
evidence to reject the hypothesis that the mean can
content is 355 ml.
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Objective 5
• Distinguish between Statistical Significance
and Practical Significance
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When a large sample size is used in a hypothesis
test, the results could be statistically significant
even though the difference between the sample
statistic and mean stated in the null hypothesis
may have no practical significance.
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Practical significance refers to the idea that,
while small differences between the statistic
and parameter stated in the null hypothesis are
statistically significant, the difference may not
be large enough to cause concern or be
considered important.
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Parallel Example 7: Statistical versus Practical Significance
In 2003, the average age of a mother at the time of her
first childbirth was 25.2. To determine if the average age
has increased, a random sample of 1200 mothers is taken
and is found to have a sample mean age of 25.5.
Assuming a standard deviation of 4.8, determine whether
the mean age has increased using a significance level of
=0.05.
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Solution
Step 1: To determine whether the mean age has
increased, this is a right-tailed test with
H0: =25.2
versus
H1: >25.2.
Step 2: The level of significance is =0.05.
Step 3: Recall that the sample mean is 25.5. The
test statistic is then
25.5  25.2
z0 
 2.17
4.8 1200
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Solution
Step 4: Since this is a right-tailed test,
P-value = P(Z > 2.17) = 0.015.
The probability of obtaining a sample mean that
is more than 2.17 standard deviations above
the assumed mean of 25.2 is 0.015.
Step 5: Because the P-value is less than the level of
significance, 0.05, we reject the null
hypothesis.
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Solution
Step 6: There is sufficient evidence at the 0.05
significance level to conclude that the mean age
of a mother at the time of her first childbirth is
greater than 25.2.
Although we found the difference in age to be
significant, there is really no practical significance in the
age difference (25.2 versus 25.5).
Large sample sizes can lead to statistically significant
results while the difference between the statistic and
parameter is not enough to be considered practically
significant.
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