#### Transcript Ch7-Sec7.2

```Section 7.2
Hypothesis Testing for the Mean (Large Samples)
1
Section 7.2 Objectives
 Find P-values and use them to test a mean μ
 Use P-values for a z-test
 Find critical values and rejection regions in a normal
distribution
 Use rejection regions for a z-test
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Using P-values to Make a Decision
Decision Rule Based on P-value
 To use a P-value to make a conclusion in a hypothesis test,
compare the P-value with .
1. If P  , then reject H0.
2. If P > , then fail to reject H0.
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Example: Interpreting a P-value
The P-value for a hypothesis test is P = 0.0237. What is your
decision if the level of significance is
1. 0.05?
Solution:
Because 0.0237 < 0.05, you should reject the null
hypothesis.
2.
0.01?
Solution:
Because 0.0237 > 0.01, you should fail to reject the
null hypothesis.
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Finding the P-value
After determining the hypothesis test’s standardized test statistic and
the test statistic’s corresponding area, do one of the following to find
the P-value.
a. For a left-tailed test, P = (Area in left tail).
b. For a right-tailed test, P = (Area in right tail).
c. For a two-tailed test, P = 2(Area in tail of test statistic).
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Example: Finding the P-value
Find the P-value for a left-tailed hypothesis test with a test statistic
of z = -2.23. Decide whether to reject H0 if the level of significance
is α = 0.01.
Solution:
For a left-tailed test, P = (Area in left tail)
P = 0.0129
-2.23
0
z
Because 0.0129 > 0.01, you should fail to reject H0
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Example: Finding the P-value
Find the P-value for a two-tailed hypothesis test with a test statistic
of z = 2.14. Decide whether to reject H0 if the level of significance
is α = 0.05.
Solution:
For a two-tailed test, P = 2(Area in tail of test statistic)
1 – 0.9838 =
0.0162
0.9838
0
2.14
P = 2(0.0162)
= 0.0324
z
Because 0.0324 < 0.05, you should reject H0
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Z-Test for a Mean μ
 Can be used when the population is normal and  is known,
or for any population when the sample size n is at least 30.
 The test statistic is the sample mean x
 The standardized test statistic is z
x   where   standard error  
z
x
n
 n
 When n  30, the sample standard deviation s can be
substituted for .
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Using P-values for a z-Test for Mean μ
In Words
1. State the claim mathematically and
verbally. Identify the null and
alternative hypotheses.
2. Specify the level of significance.
3. Determine the standardized test
statistic.
4. Find the area that corresponds to z.
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In Symbols
State H0 and Ha.
Identify .
z
x 
 n
Use Table 4 in
Appendix B.
Using P-values for a z-Test for Mean
μ
In Words
In Symbols
5. Find the P-value.
a. For a left-tailed test, P = (Area in left tail).
b. For a right-tailed test, P = (Area in right tail).
c. For a two-tailed test, P = 2(Area in tail of test statistic).
6. Make a decision to reject or
fail to reject the null hypothesis.
7. Interpret the decision in the
context of the original claim.
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Reject H0 if P-value
is less than or equal
to . Otherwise,
fail to reject H0.
Example: Hypothesis Testing Using Pvalues
time is less than 30 minutes. A random selection of 36 delivery
times has a sample mean of 28.5 minutes and a standard
deviation of 3.5 minutes. Is there enough evidence to support
the claim at  = 0.01? Use a P-value.
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Solution: Hypothesis Testing Using Pvalues
•
•
•
•
H0: μ ≥ 30 min
Ha: μ < 30 min
 = 0.01
Test Statistic:
z
x 
 n
28.5  30

3.5 36
 2.57
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• P-value
0.0051
-2.57
0
z
• Decision: 0.0051 < 0.01
Reject H0
At the 1% level of significance, you
have sufficient evidence to conclude
the mean delivery time is less than 30
minutes.
Example: Hypothesis Testing Using Pvalues
You think that the average franchise investment information
shown in the graph is incorrect, so you randomly select 30
franchises and determine the necessary investment for each.
The sample mean investment is \$135,000 with a
standard deviation of \$30,000. Is
there enough evidence to support
your claim at  = 0.05? Use a
P-value.
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Solution: Hypothesis Testing Using Pvalues
•
•
•
•
H0: μ = \$143,260
Ha: μ ≠ \$143,260
 = 0.05
Test Statistic:
z

x 

n
135, 000  143, 260
30, 000
 1.51
14
30
• P-value
P = 2(0.0655)
= 0.1310
0.0655
-1.51
0
z
• Decision: 0.1310 > 0.05
Fail to reject H0
At the 5% level of significance, there
is not sufficient evidence to conclude
the mean franchise investment is
different from \$143,260.
Rejection Regions and Critical
Values
Rejection region (or critical region)
 The range of values for which the null hypothesis is not probable.
 If a test statistic falls in this region, the null hypothesis is rejected.
 A critical value z0 separates the rejection region from the
nonrejection region.
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Rejection Regions and Critical Values
Finding Critical Values in a Normal Distribution
1. Specify the level of significance .
2. Decide whether the test is left-, right-, or two-tailed.
3. Find the critical value(s) z0. If the hypothesis test is
a. left-tailed, find the z-score that corresponds to an area
of ,
b. right-tailed, find the z-score that corresponds to an area
of 1 – ,
c. two-tailed, find the z-score that corresponds to ½ and
1 – ½.
4. Sketch the standard normal distribution. Draw a vertical
line at each critical value and shade the rejection region(s).
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Example: Finding Critical Values
Find the critical value and rejection region for a two-tailed test
with  = 0.05.
Solution:
1 – α = 0.95
½α = 0.025
z0
-z0 = -1.96
½α = 0.025
0 z0 =z01.96
z
The rejection regions are to the left of -z0 = -1.96 and
to the right of z0 = 1.96.
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Decision Rule Based on Rejection
Region
To use a rejection region to conduct a hypothesis test,
calculate the standardized test statistic, z. If the
standardized test statistic
1. is in the rejection region, then reject H0.
2. is not in the rejection region, then fail to reject H0.
Fail to reject Ho.
Fail to reject H0.
Reject H0.
z < z0
Reject Ho.
z0
z
0
Fail to reject H0
Left-Tailed Test
Reject H0
z < -z0 z0
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0
0
z > z0
Right-Tailed Test
Reject H0
z
z0 z > z0
Two-Tailed Test
z0
z
Using Rejection Regions for a z-Test
for a Mean μ
In Words
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
2. Specify the level of significance.
In Symbols
State H0 and Ha.
Identify .
3. Sketch the sampling distribution.
4. Determine the critical value(s).
5. Determine the rejection region(s).
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Use Table 4 in
Appendix B.
Using Rejection Regions for a z-Test
for a Mean μ
In Words
6. Find the standardized test
statistic.
7. Make a decision to reject or fail
to reject the null hypothesis.
8. Interpret the decision in the
context of the original claim.
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In Symbols
x 
or if n  30
 n
use   s.
z
If z is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
Example: Testing with Rejection
Regions
Employees in a large accounting firm claim that the mean salary
of the firm’s accountants is less than that of its competitor’s,
which is \$45,000. A random sample of 30 of the firm’s
accountants has a mean salary of \$43,500 with a standard
deviation of \$5200. At
α = 0.05, test the employees’ claim.
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Solution: Testing with Rejection
Regions
•
•
•
•
H0: μ ≥ \$45,000
Ha: μ < \$45,000
 = 0.05
Rejection Region:
• Test Statistic
x   43,500  45, 000
z

 n
5200 30
0.05
-1.645 0
-1.58
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z
 1.58
• Decision: Fail to reject H0
At the 5% level of significance,
there is not sufficient evidence to
support the employees’ claim that
the mean salary is less than
\$45,000.
Example: Testing with Rejection
Regions
The U.S. Department of Agriculture reports that the mean cost of
raising a child from birth to age 2 in a rural area is \$10,460.You
believe this value is incorrect, so you select a random sample of 900
children (age 2) and find that the mean cost is \$10,345 with a
standard deviation of \$1540. At α = 0.05, is there enough
evidence to conclude that the mean cost is different from
\$10,460? (Adapted from U.S. Department of Agriculture Center
for Nutrition Policy and Promotion)
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Solution: Testing with Rejection
Regions
•
•
•
•
H0: μ = \$10,460
Ha: μ ≠ \$10,460
 = 0.05
Rejection Region:
0.025
-1.96
-2.24
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• Test Statistic
x   10,345  10, 460
z

 n
1540 900
0.025
0
1.96
z
 2.24
• Decision: Reject H0
At the 5% level of significance, you
have enough evidence to conclude
the mean cost of raising a child
from birth to age 2 in a rural area
is significantly different from
\$10,460.
Section 7.2 Summary
 Found P-values and used them to test a mean μ
 Used P-values for a z-test
 Found critical values and rejection regions in a normal
distribution
 Used rejection regions for a z-test
25
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