1
SECTION 7.2
Larson/Farber 4th
ed.
Hypothesis Testing for the Mean (Large Samples)
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
2
Larson/Farber 4th ed.
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.
2.
If P  , then reject H0.
If P > , then fail to reject H0.
3
Larson/Farber 4th ed.
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.
4
Larson/Farber 4th ed.
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).
5
Larson/Farber 4th ed.
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
z
0
Because 0.0129 > 0.01, you should fail to reject H0
6
Larson/Farber 4th ed.
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)
0.983
8 0
1 – 0.9838
= 0.0162
2.14
P = 2(0.0162)
= 0.0324
z
Because 0.0324 < 0.05, you should reject H0
7
Larson/Farber 4th ed.
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 meanx
The standardized test statistic is z
z

x 
 n
  standard error  
x
n
When n  30, the sample standard deviation s can
be substituted for .
8
Larson/Farber 4th ed.
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.
In Symbols
State H0 and Ha.
2. Specify the level of significance.
Identify .
3. Determine the standardized test
statistic.
z
4. Find the area that corresponds
to z.
x 
 n
Use Table 4 in
Appendix B.
9
Larson/Farber 4th ed.
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).
Reject H0 if P-value
6. Make a decision to reject or
is less than or equal
fail to reject the null hypothesis.
to . Otherwise,
fail to reject H0.
7. Interpret the decision in the
context of the original claim.
10
Larson/Farber 4th ed.
Example: Hypothesis Testing Using Pvalues
In an advertisement, a pizza shop claims that its mean
delivery 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.
11
Larson/Farber 4th ed.
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
• 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.
12
Larson/Farber 4th ed.
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.
13
Larson/Farber 4th ed.
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
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.
14 Larson/Farber 4th ed.
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.
15
Larson/Farber 4th ed.
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
region(s).
16 rejection
Larson/Farber 4th
ed.
Example: Finding Critical Values
Find the critical value and rejection region for a twotailed test with  = 0.05.
Solution:
1 – α = 0.95
½α = 0.025
½α = 0.025
0 z0 =z0
-z0 = z-1.96
0
1.96
z
The rejection regions are to the left of -z0 = -1.96 and
to the right of z0 = 1.96.
17
Larson/Farber 4th ed.
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
0
Fail to reject H0
Left-Tailed Test
Reject H0
z < -z0 z0
0
z0
Right-Tailed Test
Reject H0
z
z0 z > z0
Two-Tailed Test
z > z0
18
Larson/Farber 4th ed.
z
Using Rejection Regions for a z-Test for
a Mean μ
In Words
In Symbols
State H0 and Ha.
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
Identify .
2. Specify the level of significance.
3. Sketch the sampling distribution.
Use Table 4 in
Appendix B.
4. Determine the critical value(s).
5. Determine the rejection region(s).
19
Larson/Farber 4th ed.
Using Rejection Regions for a z-Test for
a Mean μ
In Words
In Symbols
x 
or if n  30
 n
use   s.
z
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.
If z is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
20
Larson/Farber 4th ed.
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.
21
Larson/Farber 4th ed.
Solution: Testing with Rejection Regions
•
•
•
•
• Test Statistic
x   43,500  45, 000
z

 n
5200 30
H0: μ ≥ $45,000
Ha: μ < $45,000
 = 0.05
Rejection Region:
0.05
-1.645 0
-1.58
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.
22
Larson/Farber 4th ed.
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)
23
Larson/Farber 4th ed.
Solution: Testing with Rejection Regions
•
•
•
•
• Test Statistic
x   10,345  10, 460
z

 n
1540 900
H0: μ = $10,460
Ha: μ ≠ $10,460
 = 0.05
Rejection Region:
0.025
-1.96
-2.24
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.
24
Larson/Farber 4th ed.
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
Larson/Farber 4th ed.