Transcript Testing when unknown

Hypothesis Testing
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Section
9.2
Testing the Mean 
Copyright © Cengage Learning. All rights reserved.
Focus Points
•
Test  when  is unknown using a Student’s t
distribution.
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Part II: Testing  when 
is unknown
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Part II: Testing  when  is unknown
In many real-world situations, you have only a random
sample of data values. In addition, you may have some
limited information about the probability distribution of your
data values.
Can you still test  under these circumstances? In most
cases, the answer is yes!
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Part II: Testing  when  is unknown
Procedure:
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Part II: Testing  when  is unknown
Procedure:
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Part II: Testing  when  is unknown
We used Table 4 of the Appendix, “Critical Values for
Student’s t Distribution”, to find critical values tc for
confidence intervals. The critical values are in the body of
the table.
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Example 4 – Testing  when  unknown
The drug 6-mP (6-mercaptopurine) is used to treat
leukemia. The following data represent the remission times
(in weeks) for a random sample of 21 patients using 6-mP
(Reference: E. A. Gehan, University of Texas Cancer
Center).
The sample mean is  17.1 weeks, with sample standard
deviation s  10.0. Let x be a random variable representing
the remission time (in weeks) for all patients using 6-mP.
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Example 4 – Testing  when  unknown
cont’d
Assume the x distribution is mound-shaped and symmetric.
A previously used drug treatment had a mean remission
time of  = 12.5 weeks.
Do the data indicate that the mean remission time using the
drug 6-mP is different (either way) from 12.5 weeks? Use
 = 0.01.
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Example 4 – Solution
(a) Establish the null and alternate hypotheses.
Since we want to determine if the drug 6-mP provides a
mean remission time that is different from that provided
by a previously used drug having  = 12.5 weeks,
H0:  = 12.5 weeks
and
H1:   12.5 weeks
(b) Check Requirements What distribution do we use for
the sample test statistic t? Compute the sample test
statistic from the sample data.
The x distribution is assumed to be mound-shaped and
symmetric.
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Example 4 – Solution
cont’d
Because we don’t know , we use a Student’s t
distribution with d.f. = 20. Using  17.1 and s  10.0
from the sample data,  = 12.5 from H0, and n = 21
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Example 4 – Solution
cont’d
(c) Find the P-value or the interval containing the P-value.
Figure 9-5 shows the P-value. Using Table 4 of the
Appendix, we find an interval containing the P-value.
P-value
Figure 9-5
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Example 4 – Solution
cont’d
Since this is a two-tailed test, we use entries from the
row headed by two-tail area. Look up the t value in the
row headed by d.f. = n – 1 = 21 – 1 = 20.
The sample statistic t = 2.108 falls between 2.086 and
2.528. The P-value for the sample t falls between the
corresponding two-tail areas 0.050 and 0.020.
(See Table 9-5.) 0.020 < P-value < 0.050
Excerpt from Student’s t Distribution (Table 4, Appendix)
Table 9-5
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Example 4 – Solution
cont’d
(d) Conclude the test.
The following diagram shows the interval that contains
the single P-value corresponding to the test statistic.
Note that there is just one P-value corresponding to the
test statistic. Table 4 of the Appendix does not give that
specific value, but it does give a range that contains that
specific P-value.
As the diagram shows, the entire range is greater than
. This means the specific P-value is greater than , so
we cannot reject H0.
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Example 4 – Solution
cont’d
Note:
Using the raw data, computer software gives
P-value  0.048. This value is in the interval we
estimated. It is larger than the  value of 0.01, so we do
not reject H0.
(e) Interpretation Interpret the results in the context of the
problem. At the 1% level of significance, the evidence is
not sufficient to reject H0. Based on the sample data, we
cannot say that the drug 6-mP provides a different
average remission time than the previous drug.
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