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Section 10.2
Hypothesis Testing for Population
Means (s Known)
With valuable content added
by D.R.S., University of Cordele
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Objectives
o Use the rejection region to draw a conclusion.
o Use the p-value to draw a conclusion.
There are two ways to get the answer and we learn BOTH.
• The “Critical Value” method, comparing your z Test Statistic
to the z Critical Value, which was determined by the chosen
α Level of Significance.
• The “p-Value method”, in which your z Test Statistic leads
to a “p-Value”, an area under the normal curve, which is
compared to the chosen α Level of Significance.
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Hypothesis Testing for Population Means
(s Known)
Test Statistic for a Hypothesis Test for a Population Mean
(s Known)
When (1) the population standard deviation is known, (2)
the sample taken is a simple random sample, and (3) either
the sample size is at least 30 or the population distribution
is approximately normal, THEN COMPUTE: the test statistic
for a hypothesis test for a population mean is given by
x 
z
 s 


n
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Hypothesis Testing for Population Means
(s Known)
THEN COMPUTE: the test statistic for a hypothesis test
for a population mean is given by
x 
z
 s 


n
My sample mean is how far away
from the H0 Null Hypothesis mean?
Gauged by the standard deviation for
a sample as defined in the Central
Limit Theorem.
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Hypothesis Testing for Population Means
(s Known)
Test Statistic for a Hypothesis Test for a Population
Mean (s Known) (cont.)
where x ̄ is the sample mean,
 is the presumed value of the population mean from
the null hypothesis,
s is the population standard deviation, and
n is the sample size.
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Rejection Regions
Alternative Hypothesis, Ha
 Value
 Value
 Value
The “rejection region” is
the one-tail area of size α.
Type of Hypothesis Test
Left-tailed test
Right-tailed test
Two-tailed test
The “rejection region” consists
of two tails, each of them
of size α/2, left and right.
And the “Fail to Reject H0” region is the big area of size 1 – α.
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Rejection Regions
Decision Rule for Rejection Regions
Reject the null hypothesis, H0, if the test statistic
calculated from the sample data falls within the
rejection region.
Refer to the next three pictures. They illustrate the
Left-Tailed, Right-Tailed, and Two-Tailed cases.
Be able to draw the right kind of picture for each
Hypothesis Test you perform.
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A Left-Tailed Hypothesis Test: Ha < μ
The critical z value is the negative z value which separates the left tail of area α.
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A Right-Tailed Hypothesis Test: Ha > μ
The critical z value is the positive z value which separates the right tail of area α.
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A Two-Tailed Hypothesis Test: Ha ≠ μ
The critical z values are the +/- z values which separate the two tails, area α/2 each.
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Commonly-occurring critical values
• These values for the α
Level Of Significance (or
c Level Of Confidence)
are typical. You should
already know how to
find them using tables
and using invNorm(). It
is convenient to have
this special table for
reference.
Area in
tail(s),
α, alpha
(and c too)
OneTailed
Test
TwoTailed
Test
0.10 (0.90)
1.28
±1.645
0.05 (0.95)
1.645
±1.96
0.02 (0.98)
2.05
±2.33
0.01 (0.99)
2.33
±2.575
Example 10.10: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Right-Tailed, s Known)
The state education department is considering
introducing new initiatives to boost the reading levels
of fourth graders. The mean reading level of fourth
graders in the state over the last 5 years was a Lexile
reader measure of 800 L. (A Lexile reader measure is a
measure of the complexity of the language that a
reader is able to comprehend.) The developers of a
new program claim that their techniques will raise the
mean reading level of fourth graders by more than 50 L.
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Example 10.10: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Right-Tailed, s Known) (cont.)
To assess the impact of their initiative, the developers
were given permission to implement their ideas in the
classrooms. At the end of the pilot study, a simple
random sample of 1000 fourth graders had a mean
reading level of 856 L. It is assumed that the population
standard deviation is 98 L. Using a 0.05 level of
significance, should the findings of the study convince
the education department of the validity of the
developers’ claim?
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Example 10.10: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Right-Tailed, s Known) (cont.)
Solution
Step 1: State the null and alternative hypotheses.
The developers want to show that their
classroom techniques will raise the fourth
graders’ mean reading level to more than 850
L. This is written mathematically as  > 850,
and since it is the research hypothesis, it will
be Ha. The mathematical opposite is  ≤ 850.
Thus, we have the following hypotheses.
H0 :   850
Ha :   850
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Example 10.10: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Right-Tailed, s Known) (cont.)
Step 2: Determine which distribution to use for the
test statistic, and state the level of significance.
Note that the hypotheses are statements
about the population mean, s is known, the
sample is a simple random sample, and the
sample size is at least 30. Thus, we will use a
normal distribution, which means we need to
use the z-test statistic.
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Example 10.10: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Right-Tailed, s Known) (cont.)
In addition to determining which distribution
to use for the test statistic, we need to state
the level of significance. The problem states
that a = 0.05.
Step 3: Gather data and calculate the necessary
sample statistics.
At the end of the pilot study, a simple random
sample of 1000 fourth graders had a mean
reading level of 856 L. The population standard
deviation is 98 L.
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Example 10.10: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Right-Tailed, s Known) (cont.)
Thus, the test statistic is calculated as follows.
x 
z
 s 


n
856  850

 98 


1000 
 1.94
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Example 10.10: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Right-Tailed, s Known) (cont.)
Step 4: Draw a conclusion and interpret the decision.
Remember that we determine the type of test
based on the alternative hypothesis. In this
case, the alternative hypothesis contains “>,”
which indicates that this is a right-tailed test.
To determine the rejection region, we need a
z-value so that 0.05 of the area under the
standard normal curve is to its right. If 0.05 of
the area is to the right then 1  0.05 = 0.95 is
the area to the left.
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Example 10.10: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Right-Tailed, s Known) (cont.)
If we look up 0.9500 in the body of the cumulative
z-table, the corresponding critical z-value is 1.645.
Alternately, we can look up c = 0.95 in the table of
critical z-values for rejection regions. Either way, the
rejection region is z ≥ 1.645.
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Example 10.10: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Right-Tailed, s Known) (cont.)
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Example 10.10: Using a Rejection Region in a Hypothesis Test for
a Population Mean (Right-Tailed, s Known) (cont.)
The z-value of 1.94 falls in the rejection region. So the
conclusion is to reject the null hypothesis. Thus the
evidence collected suggests that the education
department can be 95% sure of the validity of the
developers’ claim that the mean Lexile reader measure
of fourth graders will increase by more than 50 points.
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p-Values
p-value
A p-value is the probability of obtaining a sample
statistic as extreme or more extreme than the one
observed in the data, when the null hypothesis, H0, is
assumed to be true.
We’ll give examples of how to find a p-value by hand.
But usually you’ll get the p-value for free from the TI-84’s ZTest
or TTest feature.
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How to calculate the p value
for your z Test Statistic
You’ve calculated your z = Test Statistic using the
formula.
Here’s how to define your p-value:
• If it’s a left-tailed test, what’s the area to the left of
your z ?
• If it’s a right tailed test, what’s the area to the right
of your z?
• If it’s a two-tailed test, what’s 2x the area to the
left/right of your neg/pos z?
Example 10.11: Calculating the p-Value for a
z-Test Statistic for a Left-Tailed Test
Calculate the p-value for a hypothesis test with the
following hypotheses. Assume that data have been
collected and the test statistic was calculated to be
z = 1.34.
H0 :   0.15
Ha :   0.15
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Example 10.11: Calculating the p-Value for a
z-Test Statistic for a Left-Tailed Test (cont.)
Solution
The alternative hypothesis tells us that this is a lefttailed test. Therefore, the p-value for this situation is
the probability that z is less than or equal to 1.34,
written p-value = P(z  −1.34). Use a table or
appropriate technology to find the area under the
standard normal curve to the left of z = 1.34. Thus,
the p-value ≈ 0.0901.
normalcdf( ________, ________ )
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Example 10.11: Calculating the p-Value for a
z-Test Statistic for a Left-Tailed Test (cont.)
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Example 10.12: Calculating the p-Value for a
z-Test Statistic for a Right-Tailed Test
Calculate the p-value for a hypothesis test with the
following hypotheses. Assume that data have been
collected and the test statistic was calculated to be
z = 2.78.
H0 :   0.43
Ha :   0.43
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Example 10.12: Calculating the p-Value for a
z-Test Statistic for a Right-Tailed Test (cont.)
Solution
The alternative hypothesis tells us that this is a righttailed test. Therefore, the p-value for this situation is
the probability that z is greater than or equal to 2.78,
written p-value = P(z  2.78). Use a table or appropriate
technology to find the area under the standard normal
curve to the right of z = 2.78. Thus, the p-value ≈
0.0027.
normalcdf( ________, ________ )
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Example 10.12: Calculating the p-Value for a
z-Test Statistic for a Right-Tailed Test (cont.)
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Example 10.13: Calculating the p-Value for a
z-Test Statistic for a Two-Tailed Test
Calculate the p-value for a hypothesis test with the
following hypotheses. Assume that data have been
collected and the test statistic was calculated to be
z = 2.15.
H0 :   0.78
Ha :   0.78
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Example 10.13: Calculating the p-Value for a
z-Test Statistic for a Two-Tailed Test (cont.)
Solution
The alternative hypothesis tells us that this is a twotailed test. Thus, the p-value for this situation is the
probability that z is either less than or equal to 2.15 or
greater than or equal to 2.15, which is written
mathematically as p-value  P  z  2.15 . Use a table or
appropriate technology to find the area under the
standard normal curve to the left of z1 = −2.15. The
area to the left of z1 = −2.15 is 0.0158.
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Example 10.13: Calculating the p-Value for a
z-Test Statistic for a Two-Tailed Test (cont.)
Since the standard normal curve is symmetric about its
mean, 0, the area to the right of z2 = 2.15 is also
0.0158. Thus, the p-value is calculated as follows.
p-value   0.0158 2
 0.0316
normalcdf( ________, ________ )*2
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Example 10.13: Calculating the p-Value for a
z-Test Statistic for a Two-Tailed Test (cont.)
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Example 10.14: Calculating the p-Value for a z-Test
Statistic Using a TI-83/84 Plus Calculator
Enter normalcdf(856,1û99, 850, 98/ð
(1000)) and press
. But in practice, we’re probably
They demonstrate here how to find the
p-value using normalcdf and “x” values.
Recall that the sample mean was bar_x
= 856, it was a right tailed-test, the null
hypothesis said the mean score was
850, the population standard deviation
was 98, and the sample size was 1000.
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going to use TI-84 Z-Test instead to
give us the p-values, so don’t put a
lot of emphasis on this slide.
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Example 10.14: Calculating the p-Value for a z-Test
Statistic Using a TI-83/84 Plus Calculator (cont.)
The p-value returned is approximately 0.0264, as
shown in the screenshot.
But again,
Z-Test is a
much
better
way !!!!
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Z-Test for this problem.
STAT, TESTS, 1:Z-Test. Since they gave us summary
statistics, we pick “Stats” (not “Data”, which is for a
problem where we have the individuals’ scores).
μ0 comes from H0
σ is the “known” stdev.
bar-x from our sample
n is our sample size
Give the Ha inequality
Highlight Calculate and press ENTER.
Z-Test for this problem.
STAT, TESTS, 1:Z-Test. Here are the results.
It reminds you of what you told it was the alternative
hypothesis. Have a look and make sure it’s what you
wanted.
It tells you the Test Statistic.
It tells you the p-Value.
It reminds you of the sample
data, the sample mean and
sample size you told it.
p-Values
Conclusions Using p-Values
• If p-value ≤ a, then reject the null hypothesis.
• If p-value > a, then fail to reject the null
hypothesis.
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Example 10.15: Determining the Conclusion to a
Hypothesis Test Using the p-Value
For a certain hypothesis test, the p-value is calculated
to be p-value = 0.0146.
a. If the stated level of significance is 0.05, what is the
conclusion to the hypothesis test?
b. If the level of confidence is 99%, what is the
conclusion to the hypothesis test?
Solution
a. The level of significance is a = 0.05. Next, note that
0.0146 < 0.05. Thus, p-value ≤ a, so we reject the
null hypothesis.
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Example 10.15: Determining the Conclusion to a
Hypothesis Test Using the p-Value (cont.)
b. The level of confidence is 99%, so the level of
significance is calculated as follows.
a  1c
 1  0.99
 0.01
Next, note that 0.0146 > 0.01. Thus, p-value > a, so
we fail to reject the null hypothesis.
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Example 10.16: Performing a Hypothesis Test for
a Population Mean (Right-Tailed, s Known)
A researcher claims that the mean age of women in
California at the time of a first marriage is higher than
26.5 years. Surveying a simple random sample of 213
newlywed women in California, the researcher found a
mean age of 27.0 years. Assuming that the population
standard deviation is 2.3 years and using a 95% level of
confidence, determine if there is sufficient evidence to
support the researcher’s claim.
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Example 10.16: TI-84 Z-Test solution
You still have to write the hypotheses:
• Null Hypothesis: ________ Alternative: ________
What is the α Level Of Significance? α = _______
What inputs do you give to the TI-84 Z-Test?
• 𝜇0 = _______, 𝜎 = ________
• 𝑥 = _______, 𝑛 = ________
• 𝜇: ≠ or < or >
The TI-84 Z-Test output gives us a p-value of ________.
Therefore we { reject or fail to reject } H0.
Example 10.16: Performing a Hypothesis Test for a
Population Mean (Right-Tailed, s Known) (cont.)
Solution
Step 1: State the null and alternative hypotheses.
The researcher’s claim is investigating the
mean age at first marriage for women in
California. Therefore, the research hypothesis,
Ha, is that the mean age is greater than 26.5,
 > 26.5. The logical opposite is  ≤ 26.5. Thus,
the null and alternative hypotheses are stated
as follows.
H0 :   26.5
Ha :   26.5
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Example 10.16: Performing a Hypothesis Test for a
Population Mean (Right-Tailed, s Known) (cont.)
Step 2: Determine which distribution to use for the
test statistic, and state the level of significance.
Note that the hypotheses are statements
about the population mean, the population
standard deviation is known, the sample is a
simple random sample, and the sample size is
at least 30. Thus, we will use a normal
distribution and calculate the z-test statistic.
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Example 10.16: Performing a Hypothesis Test for a
Population Mean (Right-Tailed, s Known) (cont.)
We will draw a conclusion by computing the p-value for
the calculated test statistic and comparing the value to
a. For this hypothesis test, the level of confidence is
95%, so the level of significance is calculated as follows.
a  1c
 1  0.95
 0.05
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Example 10.16: Performing a Hypothesis Test for a
Population Mean (Right-Tailed, s Known) (cont.)
Step 3: Gather data and calculate the necessary
sample statistics.
From the information given, we know that the
presumed value of the population mean is
 = 26.5, the sample mean is x ̄ = 27.0, the
population standard deviation is s = 2.3, and
the sample size is n = 213.
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Example 10.16: Performing a Hypothesis Test for a
Population Mean (Right-Tailed, s Known) (cont.)
Thus, the test statistic is calculated as follows.
x 
z
 s 


n
27.0  26.5

 2.3 


213 
 3.17
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Example 10.16: Performing a Hypothesis Test for a
Population Mean (Right-Tailed, s Known) (cont.)
Step 4: Draw a conclusion and interpret the decision.
The alternative hypothesis tells us that we
have a right-tailed test. Therefore, the p-value
for this test statistic is the probability of
obtaining a test statistic greater than or equal
to z = 3.17, written as p-value  P  z  3.17 .
To find the p-value, we need to find the area
under the standard normal curve to the right
of z = 3.17.
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Example 10.16: Performing a Hypothesis Test for a
Population Mean (Right-Tailed, s Known) (cont.)
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Example 10.16: Performing a Hypothesis Test for a
Population Mean (Right-Tailed, s Known) (cont.)
Using a normal distribution table or appropriate
technology, we find that the area is p-value ≈ 0.0008.
Comparing this p-value to the level of significance, we
see that 0.0008 < 0.05, so p-value ≤ a. Thus, the
conclusion is to reject the null hypothesis. Therefore,
we can say with 95% confidence that there is sufficient
evidence to support the researcher’s claim that the
mean age at first marriage for women in California is
higher than 26.5 years.
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Example 10.17: Performing a Hypothesis Test for
a Population Mean (Two-Tailed, s Known)
A recent study showed that the mean number of
children for women in Europe is 1.5. A global watch
group claims that German women have a mean fertility
rate that is different from the mean for all of Europe. To
test its claim, the group surveyed a simple random
sample of 128 German women and found that they had
a mean fertility rate of 1.4 children. The population
standard deviation is assumed to be 0.8. Is there
sufficient evidence to support the claim made by the
global watch group at the 90% level of confidence?
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Example 10.17: TI-84 Z-Test solution
You still have to write the hypotheses:
• Null Hypothesis: ________ Alternative: ________
What is the α Level Of Significance? α = _______
What inputs do you give to the TI-84 Z-Test?
• 𝜇0 = _______, 𝜎 = ________
• 𝑥 = _______, 𝑛 = ________
• 𝜇: ≠ or < or >
The TI-84 Z-Test output gives us a p-value of ________.
Therefore we { reject or fail to reject } H0.
Example 10.17: Performing a Hypothesis Test for a
Population Mean (Two-Tailed, s Known) (cont.)
Solution
Step 1: State the null and alternative hypotheses.
The watch group is investigating whether the
mean fertility rate for German women is
different from the mean for all of Europe.
Thus, they need to find evidence that the
mean fertility rate is not equal to 1.5.
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Example 10.17: Performing a Hypothesis Test for a
Population Mean (Two-Tailed, s Known) (cont.)
So the research hypothesis, Ha, is that the mean does
not equal 1.5,  ≠ 1.5. The logical opposite is  = 1.5.
Thus, the null and alternative hypotheses are stated as
follows.
H0 :   1.5
Ha :   1.5
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Example 10.17: Performing a Hypothesis Test for a
Population Mean (Two-Tailed, s Known) (cont.)
Step 2: Determine which distribution to use for the
test statistic, and state the level of significance.
Note that the hypotheses are statements
about the population mean, s is known, the
sample is a simple random sample, and the
sample size is at least 30. Thus, we will use a
normal distribution and calculate the z-test
statistic.
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Example 10.17: Performing a Hypothesis Test for a
Population Mean (Two-Tailed, s Known) (cont.)
We will draw a conclusion by computing the p-value for
the calculated test statistic and comparing the value to
a. For this hypothesis test, the level of confidence is
90%, so the level of significance is calculated as follows.
a  1c
 1  0.90
 0.10
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Example 10.17: Performing a Hypothesis Test for a
Population Mean (Two-Tailed, s Known) (cont.)
Step 3: Gather data and calculate the necessary
sample statistics.
From the information given, we know that the
presumed value of the population mean is
 = 1.5, the sample mean is x  1.4, the
population standard deviation is s = 0.8, and
the sample size is n = 128.
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Example 10.17: Performing a Hypothesis Test for a
Population Mean (Two-Tailed, s Known) (cont.)
Thus, the test statistic is calculated as follows.
x 
z
 s 


n
1.4  1.5

 0.8 


128 
 1.41
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Example 10.17: Performing a Hypothesis Test for a
Population Mean (Two-Tailed, s Known) (cont.)
Step 4: Draw a conclusion and interpret the decision.
The alternative hypothesis tells us that we
have a two-tailed test. Therefore, the p-value
for this test statistic is the probability of
obtaining a test statistic that is either less than
or equal to z1 = −1.41 or greater than or equal
to z2 = 1.41, which is written mathematically as
p-value  P  z  1.41 . To find the p-value, we
need to find the sum of the areas under the
standard normal curve to the left of z1 = −1.41
and to the right of z2 = 1.41.
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Example 10.17: Performing a Hypothesis Test for a
Population Mean (Two-Tailed, s Known) (cont.)
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Example 10.17: Performing a Hypothesis Test for a
Population Mean (Two-Tailed, s Known) (cont.)
By looking up z1 = −1.41 in the cumulative normal
distribution table, we find that the area to the left is
equal to 0.0793. Since the standard normal curve is
symmetric about its mean, 0, the area to the right of
z2 = 1.41 is also 0.0793. Thus the p-value is calculated
as follows.
p-value   0.0793 2
 0.1586
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Example 10.17: Performing a Hypothesis Test for a
Population Mean (Two-Tailed, s Known) (cont.)
Comparing the p-value to the level of significance, we
see that 0.1586 > 0.10, so p-value > a. Thus, the
conclusion is to fail to reject the null hypothesis. This
means that, at a 90% level of confidence, the evidence
does not support the watch group’s claim that the
fertility rate of German women is different from the
mean for all of Europe.
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Example 10.18: Performing a Hypothesis Test for a Population
Mean Using a TI-83/84 Plus Calculator (Two-Tailed, s Known)
A recent study showed that the mean number of
children for women in Europe is 1.5. A global watch
group claims that German women have a mean fertility
rate that is different from the mean for all of Europe. To
test its claim, the group surveyed a simple random
sample of 128 German women and found that they had
a mean fertility rate of 1.4 children. The population
standard deviation is assumed to be 0.8. Is there
sufficient evidence to support the claim made by the
global watch group at the 90% level of confidence? Use
a TI-83/84 Plus calculator to answer the question.
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Example 10.18: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Two-Tailed, s Known) (cont.)
Solution
Even when working with technology, Steps 1 and 2 are
the same, so we will just reiterate the important
information for these two steps from Example 10.17.
Step 1: State the null and alternative hypotheses.
The hypotheses are stated as follows.
H0 :   1.5
Ha :   1.5
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Example 10.18: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Two-Tailed, s Known) (cont.)
Step 2: Determine which distribution to use for the
test statistic, and state the level of significance.
Since s is known, the sample is a simple
random sample, and the sample size is at least
30, we use the normal distribution and z-test
statistic for this hypothesis test for the
population mean. The level of significance is
a = 0.10.
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Example 10.18: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Two-Tailed, s Known) (cont.)
Step 3: Gather data and calculate the necessary
sample statistics.
This is where we begin to use a TI-83/84 Plus
calculator. Let’s start by writing down the
information from the problem, as we will need
to enter these values into the calculator. We
know that the presumed value of the
population mean is  = 1.5, the sample mean
is x ̄ = 1.4, the population standard deviation is
s = 0.8, and the sample size is n = 128.
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Example 10.18: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Two-Tailed, s Known) (cont.)
Press
, scroll to TESTS, and choose option 1:ZTest. Since we know the sample statistics, choose
Stats instead of Data. For À, enter the value from
the null hypothesis, thus enter 1.5 for À. Enter the
rest of the given values, as shown in the following
screenshot on the left. Choose the alternative
hypothesis øÀ. Highlight Calculate and press
.
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Example 10.18: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Two-Tailed, s Known) (cont.)
The output screen, shown on the right, displays the
alternative hypothesis, calculates the z-test statistic and
the p-value, and then reiterates the sample mean and
sample size that were entered.
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Example 10.18: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Two-Tailed, s Known) (cont.)
Step 4: Draw a conclusion and interpret the decision.
The p-value given by the calculator is
approximately 0.1573. It is not identical to the
one we found when using the table because
there were several intermediate steps when
we found the p-value by hand that reduced
the accuracy of the p-value. However, the
conclusion is the same since the two p-values
are extremely close.
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Example 10.18: Performing a Hypothesis Test for a Population Mean
Using a TI-83/84 Plus Calculator (Two-Tailed, s Known) (cont.)
Since p-value > a, the conclusion is to fail to reject the
null hypothesis. This means that at a 90% level of
confidence, the evidence does not support the watch
group’s claim that the fertility rate of German women is
different from the mean for all of Europe.
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Excel’s Z.TEST(data array, μ, σ) function
•
•
•
•
Give it the cells (“range”) of the sample data values.
And the μ mean from the null hypothesis.
And the σ population standard deviation.
It gives you back the one-tailed p-value.
• So it calculates a z = test statistic based on the
sample.
• And then it calculates the area in the tail beyond
that z
• Either left tail beyond negative z
• Or right tail beyond positive z, it knows which one.
Excel’s Z.TEST(data array, μ, σ) function
•
•
•
•
•
Give it the cells (“range”) of the sample data values.
And the μ mean from the null hypothesis.
And the σ population standard deviation.
It gives you back the one-tailed p-value.
You compare that p-value to the α level of
significance that was pre-selected in the design.
• If p-value < α, then reject H0.
• If p-value > α, then fail to reject H0.
Excel’s Z.TEST(data array, μ, σ) if you have
a Two-Tailed Situation
• Recall the two-tailed situation: Ha has a ≠.
• So the rejection regions are two tails of area α/2.
• Excel Help says to do this to get the p-value:
2 * MIN(Z.TEST(data , μ, σ), 1 – Z.TEST(data , μ, σ))
(whereas we tell TI-84 it’s two-tailed and TI-84 takes
care of doubling it transparently, thank you.)
• And then decide as usual:
• If p-value < α, then reject H0.
• If p-value > α, then fail to reject H0.