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Section
10.1
The Language of
Hypothesis
Testing
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Objectives
1. Determine the null and alternative
hypotheses
2. Explain Type I and Type II errors
3. State conclusions to hypothesis tests
10-2
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Objective 1
• Determine the Null and Alternative
Hypotheses
10-3
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A hypothesis is a statement regarding a
characteristic of one or more populations.
In this chapter, we look at hypotheses
regarding a single population parameter.
10-4
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Examples of Claims Regarding a
Characteristic of a Single Population
• In 2008, 62% of American adults regularly volunteered their
time for charity work. A researcher believes that this percentage
is different today.
Source: ReadersDigest.com poll created on 2008/05/02
10-5
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Examples of Claims Regarding a
Characteristic of a Single Population
• In 2008, 62% of American adults regularly volunteered their time
for charity work. A researcher believes that this percentage is
different today.
• According to a study published in March, 2006 the mean length
of a phone call on a cellular telephone was 3.25 minutes. A
researcher believes that the mean length of a call has increased
since then.
10-6
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Examples of Claims Regarding a
Characteristic of a Single Population
• In 2008, 62% of American adults regularly volunteered their time
for charity work. A researcher believes that this percentage is
different today.
• According to a study published in March, 2006 the mean length
of a phone call on a cellular telephone was 3.25 minutes. A
researcher believes that the mean length of a call has increased
since then.
• Using an old manufacturing process, the standard deviation of the
amount of wine put in a bottle was 0.23 ounces. With new
equipment, the quality control manager believes the standard
deviation has decreased.
10-7
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CAUTION!
We test these types of statements using sample
data because it is usually impossible or
impractical to gain access to the entire
population. If population data are available,
there is no need for inferential statistics.
10-8
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Hypothesis testing is a procedure, based
on sample evidence and probability, used to
test statements regarding a characteristic of
one or more populations.
10-9
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Steps in Hypothesis Testing
1. Make a statement regarding the nature of the
population.
2. Collect evidence (sample data) to test the
statement.
3. Analyze the data to assess the plausibility of the
statement.
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The null hypothesis, denoted H0, is a
statement to be tested. The null hypothesis
is a statement of no change, no effect or no
difference and is assumed true until
evidence indicates otherwise.
10-11
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The alternative hypothesis, denoted H1, is
a statement that we are trying to find
evidence to support.
10-12
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In this chapter, there are three ways to set up the
null and alternative hypotheses:
1.Equal versus not equal hypothesis (two-tailed test)
H0: parameter = some value
H1: parameter ≠ some value
10-13
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In this chapter, there are three ways to set up the
null and alternative hypotheses:
1. Equal versus not equal hypothesis (two-tailed test)
H0: parameter = some value
H1: parameter ≠ some value
2. Equal versus less than (left-tailed test)
H0: parameter = some value
H1: parameter < some value
10-14
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In this chapter, there are three ways to set up the
null and alternative hypotheses:
1. Equal versus not equal hypothesis (two-tailed test)
H0: parameter = some value
H1: parameter ≠ some value
2. Equal versus less than (left-tailed test)
H0: parameter = some value
H1: parameter < some value
3. Equal versus greater than (right-tailed test)
H0: parameter = some value
H1: parameter > some value
10-15
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“In Other Words”
The null hypothesis is a statement of status
quo or no difference and always contains a
statement of equality. The null hypothesis is
assumed to be true until we have evidence to
the contrary. We seek evidence that supports
the statement in the alternative hypothesis.
10-16
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Parallel Example 2: Forming Hypotheses
For each of the following claims, determine the null and alternative
hypotheses. State whether the test is two-tailed, left-tailed or righttailed.
a)
In 2008, 62% of American adults regularly volunteered their
time for charity work. A researcher believes that this
percentage is different today.
b)
According to a study published in March, 2006 the mean length
of a phone call on a cellular telephone was 3.25 minutes. A
researcher believes that the mean length of a call has increased
since then.
c)
Using an old manufacturing process, the standard deviation of
the amount of wine put in a bottle was 0.23 ounces. With new
equipment, the quality control manager believes the standard
deviation has decreased.
10-17
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Solution
a)
In 2008, 62% of American adults regularly volunteered
their time for charity work. A researcher believes that this
percentage is different today.
The hypothesis deals with a population proportion, p. If
the percentage participating in charity work is no different
than in 2008, it will be 0.62 so the null hypothesis is H0:
p=0.62.
Since the researcher believes that the percentage is
different today, the alternative hypothesis is a two-tailed
hypothesis: H1: p≠0.62.
10-18
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Solution
b)
According to a study published in March, 2006 the mean
length of a phone call on a cellular telephone was 3.25
minutes. A researcher believes that the mean length of a
call has increased since then.
The hypothesis deals with a population mean, μ. If the
mean call length on a cellular phone is no different than in
2006, it will be 3.25 minutes so the null hypothesis is
H0: μ = 3.25.
Since the researcher believes that the mean call length has
increased, the alternative hypothesis is:
H1: μ > 3.25, a right-tailed test.
10-19
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Solution
c)
Using an old manufacturing process, the standard
deviation of the amount of wine put in a bottle was 0.23
ounces. With new equipment, the quality control manager
believes the standard deviation has decreased.
The hypothesis deals with a population standard deviation,
σ. If the standard deviation with the new equipment has
not changed, it will be 0.23 ounces so the null hypothesis
is H0: σ = 0.23.
Since the quality control manager believes that the
standard deviation has decreased, the alternative
hypothesis is: H1: σ < 0.23, a left-tailed test.
10-20
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Objective 2
• Explain Type I and Type II Errors
10-21
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Four Outcomes from Hypothesis Testing
1. Reject the null hypothesis when the alternative
hypothesis is true. This decision would be correct.
10-22
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Four Outcomes from Hypothesis Testing
1. Reject the null hypothesis when the alternative
hypothesis is true. This decision would be correct.
2. Do not reject the null hypothesis when the null
hypothesis is true. This decision would be correct.
10-23
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Four Outcomes from Hypothesis Testing
1. Reject the null hypothesis when the alternative
hypothesis is true. This decision would be correct.
2. Do not reject the null hypothesis when the null
hypothesis is true. This decision would be correct.
3. Reject the null hypothesis when the null hypothesis is
true. This decision would be incorrect. This type of
error is called a Type I error.
10-24
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Four Outcomes from Hypothesis Testing
1. Reject the null hypothesis when the alternative
hypothesis is true. This decision would be correct.
2. Do not reject the null hypothesis when the null
hypothesis is true. This decision would be correct.
3. Reject the null hypothesis when the null hypothesis is
true. This decision would be incorrect. This type of
error is called a Type I error.
4. Do not reject the null hypothesis when the alternative
hypothesis is true. This decision would be incorrect.
This type of error is called a Type II error.
10-25
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Parallel Example 3: Type I and Type II Errors
For each of the following claims, explain what it would
mean to make a Type I error. What would it mean to make
a Type II error?
a) In 2008, 62% of American adults regularly
volunteered their time for charity work. A researcher
believes that this percentage is different today.
b) According to a study published in March, 2006 the
mean length of a phone call on a cellular telephone
was 3.25 minutes. A researcher believes that the mean
length of a call has increased since then.
10-26
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Solution
a)
In 2008, 62% of American adults regularly volunteered
their time for charity work. A researcher believes that this
percentage is different today.
A Type I error is made if the researcher concludes that
p ≠ 0.62 when the true proportion of Americans 18 years
or older who participated in some form of charity work is
currently 62%.
A Type II error is made if the sample evidence leads the
researcher to believe that the current percentage of
Americans 18 years or older who participated in some
form of charity work is still 62% when, in fact, this
percentage differs from 62%.
10-27
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Solution
b)
According to a study published in March, 2006 the mean
length of a phone call on a cellular telephone was 3.25
minutes. A researcher believes that the mean length of a
call has increased since then.
A Type I error occurs if the sample evidence leads the
researcher to conclude that μ > 3.25 when, in fact, the
actual mean call length on a cellular phone is still 3.25
minutes.
A Type II error occurs if the researcher fails to reject the
hypothesis that the mean length of a phone call on a
cellular phone is 3.25 minutes when, in fact, it is longer
than 3.25 minutes.
10-28
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α = P(Type I Error)
= P(rejecting H0 when H0 is true)
β = P(Type II Error)
= P(not rejecting H0 when H1 is true)
10-29
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The probability of making a Type I error, α,
is chosen by the researcher before the
sample data is collected.
The level of significance, α, is the
probability of making a Type I error.
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“In Other Words”
As the probability of a Type I error
increases, the probability of a Type II
error decreases, and vice-versa.
10-31
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Objective 3
• State Conclusions to Hypothesis Tests
10-32
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CAUTION!
We never “accept” the null hypothesis,
because, without having access to the entire
population, we don’t know the exact value of
the parameter stated in the null. Rather, we
say that we do not reject the null hypothesis.
This is just like the court system. We never
declare a defendant “innocent”, but rather say
the defendant is “not guilty”.
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Parallel Example 4: Stating the Conclusion
According to a study published in March, 2006 the mean
length of a phone call on a cellular telephone was 3.25
minutes. A researcher believes that the mean length of a
call has increased since then.
a) Suppose the sample evidence indicates that the null
hypothesis should be rejected. State the wording of
the conclusion.
b) Suppose the sample evidence indicates that the null
hypothesis should not be rejected. State the wording
of the conclusion.
10-34
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Solution
a) Suppose the sample evidence indicates that the null
hypothesis should be rejected. State the wording of
the conclusion.
The statement in the alternative hypothesis is that the
mean call length is greater than 3.25 minutes. Since the
null hypothesis (μ = 3.25) is rejected, there is sufficient
evidence to conclude that the mean length of a phone call
on a cell phone is greater than 3.25 minutes.
10-35
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Solution
b) Suppose the sample evidence indicates that the null
hypothesis should not be rejected. State the wording
of the conclusion.
Since the null hypothesis (μ = 3.25) is not rejected, there is
insufficient evidence to conclude that the mean length of a
phone call on a cell phone is greater than 3.25 minutes. In
other words, the sample evidence is consistent with the
mean call length equaling 3.25 minutes.
10-36
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Section 10.3
Hypothesis Tests
for a Population
Mean
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Objectives
1. Test hypotheses about a mean
2. Understand the difference between statistical
significance and practical significance.
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Objective 1
• Test Hypotheses about a Mean
10-39
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To test hypotheses regarding the population
mean assuming the population standard
deviation is unknown, we use the t-distribution
rather than the Z-distribution. When we replace
σ with s,
x 
s
n
follows Student’s t-distribution with n –1
degrees of freedom.

10-40
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Testing Hypotheses Regarding a
Population Mean
To test hypotheses regarding the population
mean, we use the following steps, provided that:
•The sample is obtained using simple random
sampling.
•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).
•The sampled values are independent of each
other.
10-41
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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, α, based
on the seriousness of making a
Type I error.
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Classical Approach
Step 3: Compute the test statistic
x  0
t0 
s
n
which follows the Student’s t-distribution with n – 1
degrees of freedom.
10-44
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Classical Approach
Use Student’s t - table to determine the critical
value.
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Classical Approach
Two-Tailed
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Classical Approach
Left-Tailed
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Classical Approach
Right-Tailed
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Classical Approach
Step 4: Compare the critical value with the test
statistic:
Rejection Region: For a probability a of a Type-I error, we
can reject H0 if
1.
2.
3.
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t > ta
t < - ta
t > ta/2 or t < - ta/2
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P-Value Approach
The p-value associated with the observed value of the
test statistic is the probability of getting the observed
value or a value more extreme (in the direction of Ha),
assuming H0 is true.
To draw conclusions using p-value,
you compare the p-value with aand draw your
conclusions using the following rule.
If the p-value < a,reject H0.
If the p-value  α, fail to reject H0.
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P-Value Approach
By Hand Step 3: Compute the test statistic
x  0
t0 
s
n
which follows the Student’s t-distribution with n – 1
degrees of freedom.
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P-Value Approach
Use Student’s t- table to approximate the Pvalue.
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P-Value Approach
Two-Tailed
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P-Value Approach
Left-Tailed
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P-Value Approach
Right-Tailed
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P-Value Approach
Technology Step 3: Use a statistical
spreadsheet or calculator with statistical
capabilities to obtain the P-value. The
directions for obtaining the P-value using the
TI-83/84 Plus graphing calculator, MINITAB,
Excel, and StatCrunch, are in the Technology
Step-by-Step in the text.
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P-Value Approach
Step 4: If the P-value < α, reject the null
hypothesis.
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Step 5: 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.
10-59
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Parallel Example 1: Testing a Hypothesis about a
Population Mean, Large Sample
Assume the resting metabolic rate (RMR) of healthy males
in complete silence is 5710 kJ/day. Researchers measured
the RMR of 45 healthy males who were listening to calm
classical music and found their mean RMR to be 5708.07
with a standard deviation of 992.05.
At the α = 0.05 level of significance, is there evidence to
conclude that the mean RMR of males listening to calm
classical music is different than 5710 kJ/day?
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Solution
We assume that the RMR of healthy males is 5710
kJ/day. This is a two-tailed test since we are interested
in determining whether the RMR differs from 5710
kJ/day.
Since the sample size is large (n>30), we follow the
steps for testing hypotheses about a population mean for
large samples.
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Solution
Step 1: H0: μ = 5710
versus
H1: μ ≠ 5710
Step 2: The level of significance is α = 0.05.
Step 3: The sample mean is x = 5708.07 and the
sample standard deviation is s = 992.05. The test
statistic is

5708.07
 5710
t0 
 0.013
992.05 45
10-62
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Solution: Classical Approach
Step 4: Since this is a two-tailed test, we determine the
critical values at the α = 0.05 level of
significance with n –1 = 45 – 1 = 44 degrees of
freedom to be approximately –t0.025 = –2.021
and
t0.025 = 2.021.
Step 5: Since the test statistic, t0 = –0.013, is between
the critical values, we fail to reject the null
hypothesis.
10-63
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Solution: P-Value Approach
Step 4: Since this is a two-tailed test, the P-value is the
area under the t-distribution with n –1 = 45 – 1
= 44
degrees of freedom to the left of –t0.025= –0.013
and to the right of t0.025 = 0.013. That is, Pvalue = P(t < -0.013) + P(t > 0.013) = 2 P(t > 0.013).
0.50 < P-value.
Step 5: Since the P-value is greater than the level of
significance (0.05<0.5), we fail to reject the
null hypothesis.
10-64
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Solution
Step 6: There is insufficient evidence at the α = 0.05
level of significance to conclude that the mean
RMR of males listening to calm classical music
differs from 5710 kJ/day.
10-65
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Parallel Example 3: Testing a Hypothesis about a
Population Mean, Small Sample
According to the United States Mint, quarters weigh 5.67
grams. A researcher is interested in determining whether the
“state” quarters have a weight that is different from 5.67
grams. He randomly selects 18 “state” quarters, weighs
them and obtains the following data.
5.70
5.67
5.73
5.61
5.70
5.67
5.65
5.62
5.73
5.65
5.79
5.73
5.77
5.71
5.70
5.76
5.73
5.72
At the α = 0.05 level of significance, is there evidence to
conclude that state quarters have a weight different than
5.67 grams?
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Solution
We assume that the weight of the state quarters is 5.67
grams. This is a two-tailed test since we are interested
in determining whether the weight differs from 5.67
grams.
Since the sample size is small, we must verify that the
data come from a population that is normally distributed
with no outliers before proceeding to Steps 1-6.
10-67
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Solution
Step 1: H0: μ = 5.67
versus
H1: μ ≠ 5.67
Step 2: The level of significance is α = 0.05.
Step 3: From the data, the sample mean is calculated to
be 5.7022 and the sample standard deviation is
s = 0.0497. The test statistic is
5.7022  5.67
t0 
 2.75
.0497 18
10-68
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Solution: Classical Approach
Step 4: Since this is a two-tailed test, we determine the
critical values at the α = 0.05 level of
significance with n – 1 = 18 – 1 = 17 degrees of
freedom to be –t0.025 = –2.11 and t0.025 = 2.11.
Step 5: Since the test statistic, t0 = 2.75, is greater than
the critical value 2.11, we reject the null
hypothesis.
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Solution: P-Value Approach
Step 4: Since this is a two-tailed test, the P-value is the
area under the t-distribution with n – 1 = 18 – 1
= 17
degrees of freedom to the left of –t0.025 = –2.75
and to the right of t0.025 = 2.75. That is, P-value
= P(t < -2.75) + P(t > 2.75) = 2 P(t > 2.75).
0.01 < P-value < 0.02.
Step 5: Since the P-value is less than the level of
significance (0.02<0.05), we reject the null
hypothesis.
10-70
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Solution
Step 6: There is sufficient evidence at the α = 0.05 level
of significance to conclude that the mean
weight of the state quarters differs from
5.67 grams.
10-71
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Objective 5
• Distinguish between Statistical Significance
and Practical Significance
10-72
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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 with a
standard deviation of 4.8, determine whether the mean
age has increased using a significance level of
α = 0.05.
10-75
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Solution
Step 1: To determine whether the mean age has
increased, this is a right-tailed test with
H0: μ = 25.2
versus
H1: μ > 25.2.
Step 2: The level of significance is α = 0.05.
Step 3: Recall that the sample mean is 25.5. The
test statistic is then
t0 
10-76
25.5  25.2
4.8
 2.17
1200
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Solution
Step 4: Since this is a right-tailed test, the critical value
with α = 0.05 and 1200 – 1 = 1199 degrees of freedom is
P-value = P(t > 2.17).
From Table VI: .01 < P-value < .02
Step 5: Because the P-value is less than the level of
significance, 0.05, we reject the null
hypothesis.
10-77
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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).
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Large Sample Sizes
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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