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244 CHAPTER 7 HYPOTHESIS TESTING
We wish to know if we may conclude, on the basis of these results, that, in general,
persons with thrombosis have, on the average, higher IgG levels than persons without
thrombosis.
Solution:
1. Data. See statement of example.
2. Assumptions. The statistics were computed from two independent samples
that behave as simple random samples from a population of persons
with thrombosis and a population of persons who do not have
thrombosis. Since the population variances are unknown, we will use the
sample variances in the calculation of the test statistic.
3. Hypotheses.
or, alternatively,
H 0 : m T - m NT … 0
H A : m T - m NT 7 0
H 0 : m T … m NT
H A : m T 7 m NT
4. Test statistic. Since we have large samples, the central limit theorem
allows us to use Equation 7.3.5 as the test statistic.
5. Distribution of test statistic. When the null hypothesis is true, the test
statistic is distributed approximately as the standard normal.
6. Decision rule. Let a = .01. This is a one-sided test with a critical
value of z equal to 2.33. Reject if z computed Ú 2.33.
7. Calculation of test statistic.
H 0
z =
59.01 - 46.61
44.89 2
A 53
+ 34.852
54
= 1.59
8. Statistical decision. Fail to reject H 0 , since z = 1.59 is in the nonrejection
region.
9. Conclusion. These data indicate that on the average, persons with
thrombosis and persons without thrombosis may not have differing IgG
levels.
10. p value. For this test, p = .0559. When testing a hypothesis about the
difference between two populations means, we may use Figure 6.4.1 to
decide quickly whether the test statistic should be z or t.
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We may use MINITAB to perform two-sample t tests. To illustrate, let us
refer to the data in Table 7.3.1. We put the data for control subjects and spinal cord