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Section
10.3
Hypothesis Tests
for a Population
Mean- Population
Standard Deviation
Unknown
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Objectives
1. Test the hypotheses about a population mean
with  unknown
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To test hypotheses regarding the population
mean assuming the population standard
deviation is unknown, we use the tdistribution rather than the Z-distribution.
When we replace  with s,
x 
s
n
follows Student’s t-distribution with n-1
degrees of freedom.

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Properties of the t-Distribution
1. The t-distribution is different for different
degrees of freedom.
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Properties of the t-Distribution
1. The t-distribution is different for different
degrees of freedom.
2. The t-distribution is centered at 0 and is
symmetric about 0.
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Properties of the t-Distribution
1. The t-distribution is different for different
degrees of freedom.
2. The t-distribution is centered at 0 and is
symmetric about 0.
3. The area under the curve is 1. Because of the
symmetry, the area under the curve to the right
of 0 equals the area under the curve to the left
of 0 equals 1/2.
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Properties of the t-Distribution
4. As t increases (or decreases) without bound, the
graph approaches, but never equals, 0.
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Properties of the t-Distribution
4. As t increases (or decreases) without bound, the
graph approaches, but never equals, 0.
5. The area in the tails of the t-distribution is a
little greater than the area in the tails of the
standard normal distribution because using s as
an estimate of  introduces more variability to
the t-statistic.
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Properties of the t-Distribution
6. As the sample size n increases, the density
curve of t gets closer to the standard normal
density curve. This result occurs because as the
sample size increases, the values of s get closer
to the values of  by the Law of Large
Numbers.
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Objective 1
• Test hypotheses about a population mean with
 unknown
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Testing Hypotheses Regarding a
Population Mean with σ Unknown
To test hypotheses regarding the population
mean with  unknown, we use the following
steps, provided that:
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. 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: Compute the test statistic
x  0
t0 
s
n
which follows Student’s t-distribution
with n-1 degrees of freedom.

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Classical Approach
Step 4: Use Table VI to determine the critical
value using n-1 degrees of freedom.
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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 5: Compare the critical value with the test
statistic:
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P-Value Approach
Step 4: Use Table VI to estimate the P-value
using n-1 degrees of freedom.
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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
Step 5: If the P-value < , reject the null
hypothesis.
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Step 6: State the conclusion.
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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 =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, we follow the steps for
testing hypotheses about a population mean for large
samples.
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Solution
Step 1: H0: =5710
versus
H1: ≠5710
Step 2: The level of significance is =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
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Solution: Classical Approach
Step 4: Since this is a two-tailed test, we determine the
critical values at the =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.
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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, P-value
= 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.
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Solution
Step 6: There is insufficient evidence at the =0.05
level of significance to conclude that the mean
RMR of males listening to calm classical music
differs from 5710 kJ/day.
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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 =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.
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Assumption of normality appears reasonable.
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No outliers.
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Solution
Step 1: H0: =5.67
versus
H1: ≠5.67
Step 2: The level of significance is =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
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Solution: Classical Approach
Step 4: Since this is a two-tailed test, we determine the
critical values at the =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.
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Solution
Step 6: There is sufficient evidence at the =0.05 level
of significance to conclude that the mean
weight of the state quarters differs from
5.67 grams.
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Summary
Which test to use?
Provided that the population from which the sample is
drawn is normal or that the sample size is large,
• if  is known, use the Z-test procedures from
Section 10.2
• if  is unknown, use the t-test procedures from this
Section.
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