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Module 27: Two Sample t-tests With Unequal Variances

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Module 27: Two Sample t-tests With Unequal Variances This module shows how to test the hypothesis that two population means are equal when there is evidence that the ... – PowerPoint PPT presentation

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Title: Module 27: Two Sample t-tests With Unequal Variances


1
Module 27 Two Sample t-tests With Unequal
Variances
This module shows how to test the hypothesis that
two population means are equal when there is
evidence that the requirement that the two
populations have the same variance is not met.
REVIEWED 19 July 05 /MODULE 27
2
The General Situation
Earlier we tested hypotheses about two population
means, based on data from two independent samples
under the assumption that the two populations had
the same variance. For this situation, we
calculated the pooled estimate of the common
variance with
3
We then tested the null hypothesis H0 µ1 µ2
vs. H1 µ1 ? µ2 with the test statistic
4
What if s12 ? s22
If the two population variances are not the same,
then they are estimated separately and is
estimated by
5
When we have two independent random samples from
two different populations and there is evidence
that the two population variances differ, i.e.
?12 ? ?22, then we can base a test about
the difference between the population means ?1
and ?2 on the following
6
To deal with ?12 ? ?22, we are adjusting the
degrees of freedom in order to obtain a better
approximation than would otherwise be the case.
This adjustment requires that we calculate
7
Example
Independent random samples were taken from
populations of histidine levels for males and
females. The measurements were approximately
normally distributed . The statistics from the
two samples were

Males Females
n 5 10
s 300.8 153.2
s2 15,291.7 2,484.62
8
Test the hypothesis H0 ?M ?F vs. H1 ?M ? ?F
  • The hypothesis H0 ?M ?F vs H1 ?M ? ?F
  • The assumptions Independent samples, normal
    distributions
  • The ? - level ? 0.05
  • 4. The test statistic


9
  • The critical region Reject if t is not between
    ? t0.975(f), where
  • so that f 6.99-2 4.99. We will use f 5,
    and t0.975(5) 2.57058.

10
6. The result 7. The conclusion The
value calculated for t above is very close to the
cut point so that P ? 0.05

11
99 Confidence Interval
12
We have

13
Example Am J Public Health, 1996Oct1436
14
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15
Example Am J Public Health, 1999Dec1692
Table 1 in the article by Sargent JD et al
provides data on blood lead levels for samples
from two communities.
Providence (1) Worcester (2)
n 136 153
6.7 5.4
SD 2.06 1.10
Test the hypothesis that Test the hypothesis
that Test the hypothesis that
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