Transcript Chapter 7: Hypothesis Testing with One Sample

Chapter 7
Hypothesis Testing
with One Sample
§ 7.3
Hypothesis Testing for
the Mean
(Small
Samples)
Critical Values in a t-Distribution
Finding Critical Values in a t-Distribution
1. Identify the level of significance .
2. Identify the degrees of freedom d.f. = n – 1.
3. Find the critical value(s) using Table 5 in Appendix B in
the row with n – 1 degrees of freedom. If the hypothesis
test is
a. left-tailed, use “One Tail,  ” column with a negative
sign,
b. right-tailed, use “One Tail,  ” column with a positive
sign,
c. two-tailed, use “Two Tails,  ” column with a
negative and a positive sign.
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Finding Critical Values for t
Example:
Find the critical value t0 for a right-tailed test given  = 0.01
and n = 24.
The degrees of freedom are d.f. = n – 1 = 24 – 1 = 23.
To find the critical value, use Table 5 with d.f. = 23 and 0.01
in the “One Tail,  “ column. Because the test is a right-tail
test, the critical value is positive.
t0 = 2.500
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Finding Critical Values for t
Example:
Find the critical values t0 and t0 for a two-tailed test given
 = 0.10 and n = 12.
The degrees of freedom are d.f. = n – 1 = 12 – 1 = 11.
To find the critical value, use Table 5 with d.f. = 11 and 0.10
in the “Two Tail,  “ column. Because the test is a two-tail
test, one critical value is negative and one is positive.
t0 =  1.796 and t0 = 1.796
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t-Test for a Mean μ (n < 30,  Unknown)
The t-test for the mean is a statistical test for a population
mean. The t-test can be used when the population is
normal or nearly normal,  is unknown, and n < 30.
The test statistic is the sample mean
test statistic is t.
and the standardized
t x μ
s n
The degrees of freedom are d.f. = n – 1 .
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t-Test for a Mean μ (n < 30,  Unknown)
Using the t-Test for a Mean μ (Small Sample)
In Words
In Symbols
1. State the claim mathematically
and verbally. Identify the null
and alternative hypotheses.
State H0 and Ha.
2. Specify the level of significance.
Identify .
3. Identify the degrees of freedom
and sketch the sampling
distribution.
d.f. = n – 1.
4. Determine any critical values.
Use Table 5 in
Appendix B.
5. Determine any rejection region(s).
Continued.
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t-Test for a Mean μ (n < 30,  Unknown)
Using the t-Test for a Mean μ (Small Sample)
In Words
In Symbols
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.
t x μ
s
n
If t is in the rejection
region, reject H0.
Otherwise, fail to
reject H0.
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Testing μ Using Critical Values
Example:
A local telephone company claims that the average
length of a phone call is 8 minutes. In a random sample
of 18 phone calls, the sample mean was 7.8 minutes and
the standard deviation was 0.5 minutes. Is there
enough evidence to support this claim at  = 0.05?
H0:  = 8 (Claim)
H a:   8
The level of significance is  = 0.05.
The test is a two-tailed test.
Degrees of freedom are d.f. = 18 – 1 = 17.
The critical values are t0 = 2.110 and t0 = 2.110
Larson & Farber, Elementary Statistics: Picturing the World, 3e
Continued.
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Testing μ Using Critical Values
Example continued:
A local telephone company claims that the average
length of a phone call is 8 minutes. In a random sample
of 18 phone calls, the sample mean was 7.8 minutes and
the standard deviation was 0.5 minutes. Is there
enough evidence to support this claim at  = 0.05?
Ha:   8
H0:  = 8 (Claim)
The standardized test statistic is
The test statistic falls in
the nonrejection region,
so H0 is not rejected.
t  x  μ  7.8  8
s n
0.5 18
 1.70.
z0 = 2.110
0
z0 = 2.110
z
At the 5% level of significance, there is not enough evidence to reject the
claim that the average length of a phone call is 8 minutes.
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Testing μ Using P-values
Example:
A manufacturer claims that its rechargeable batteries
have an average life greater than 1,000 charges. A
random sample of 10 batteries has a mean life of 1002
charges and a standard deviation of 14. Is there enough
evidence to support this claim at  = 0.01?
H0:   1000
Ha:  > 1000 (Claim)
The level of significance is  = 0.01.
The degrees of freedom are d.f. = n – 1 = 10 – 1 = 9.
The standardized test statistic is
t  x  μ  1002  1000
s n
14 10
 0.45
Larson & Farber, Elementary Statistics: Picturing the World, 3e
Continued.
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Testing μ Using P-values
Example continued:
A manufacturer claims that its rechargeable batteries
have an average life greater than 1,000 charges. A
random sample of 10 batteries has a mean life of 1002
charges and a standard deviation of 14. Is there enough
evidence to support this claim at  = 0.01?
Ha:  > 1000 (Claim)
H0:   1000
t  0.45
0
0.45
z
Using the d.f. = 9 row from Table 5, you can
determine that P is greater than  = 0.25 and is
therefore also greater than the 0.01 significance
level. H0 would fail to be rejected.
At the 1% level of significance, there is not enough evidence to
support the claim that the rechargeable battery has an average
life of at least 1000 charges.
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