Module 27 Two Sample ttests With Unequal Variances

The General Situation Earlier we tested hypotheses about two population means, based on data

We then tested the null hypothesis: H 0 : µ 1= µ 2 vs.

What if σ12 ≠ σ22 If the two population variances are not the same,

When we have two independent random samples from two different populations and there is

To deal with 12 ≠ 22, we are adjusting the degrees of freedom in

Example Independent random samples were taken from populations of histidine levels for males and

Test the hypothesis H 0: M = F vs. H 1: M F 1.

5. The critical region: t 0. 975(f), where Reject if t is not between

6. The result: 7. The conclusion: The value calculated for t above is very

Example: Am J Public Health, 1996; Oct: 1436 13

Example: Am J Public Health, 1999; Dec: 1692 Table 1 in the article by

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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 requirement that the two populations have the same variance is not met. REVIEWED 19 July 05 /MODULE 27 1

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: 2

We then tested the null hypothesis: H 0 : µ 1= µ 2 vs. H 1: µ 1 ≠ µ 2 with the test statistic: 3

What if σ12 ≠ σ22 If the two population variances are not the same, then they are estimated separately and: is estimated by: 4

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: 5

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: 6

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: n s s 2 Males 5 300. 8 15, 291. 7 Females 10 153. 2 2, 484. 62 7

Test the hypothesis H 0: M = F vs. H 1: M F 1. The hypothesis: H 0: M = F vs H 1: M F 2. The assumptions: Independent samples, normal distributions 3. The - level: = 0. 05 4. The test statistic: 8

5. The critical region: t 0. 975(f), where Reject if t is not between so that f = 6. 99 -2 = 4. 99. We will use f = 5, and t 0. 975(5) = 2. 57058. 9

6. The result: 7. The conclusion: The value calculated for t above is very close to the cut point so that P 0. 05 10

99% Confidence Interval 11

We have 12

Example: Am J Public Health, 1996; Oct: 1436 13

Example: Am J Public Health, 1999; Dec: 1692 Table 1 in the article by Sargent JD et al provides data on blood lead levels for samples from two communities. n SD Providence (1) 136 6. 7 2. 06 Worcester (2) 153 5. 4 1. 10 Test the hypothesis that 15