of Two Independent Normal Populations

Consider two independent normally distributed populations W1 and W2 having unknown means µ1 and µ2 respectively. We wish to test, with level of significance a, the following three null hypotheses about the difference in means µ1 - µ2:

As in the previous section, we base our decision for each test on the difference in sample means Xbar - Ybar from independent random samples, of sizes n and m respectively, conducted on each population.

If we assume the populations have a common variance, then we define a "pooled deviation" Sp to approximate the common standard deviation by

where Sx and Sy are the respective sample deviations. Then we define a test statistic by

which follows a T distribution with n + m - 2 degrees of freedom: X ~ T(n+m-2).

If we do not assume that the populations have a common variance, then the
data still forms a T distribution with r degrees of freedom, where r is the greatest integer less than or equal to

The test statistic is then given by

In either case, we use the appropriate T distribution to compute the right and left tail probability values created by the test statistic, P(X >= x) and P(X <= x).

We reject the first hypothesis µx - µy <= M when Xbar - Ybar is too large, which means the right tail value will be too small: P(X >= x) < a. This test is equivalent to the test Ho: µx - µy = M with the one-sided alternative Ha: µx - µy > M.

Likewise, we reject the the second hypothesis µx - µy >= M if Xbar - Ybar is too small, which means the left tail value will be too small: P(X <= x) < a. This test is equivalent to the test Ho: µx - µy = M with the one-sided alternative Ha: µx - µy < M.

If P(X >= x) < a / 2 or P(X <= x) < a / 2, then we reject the third hypothesis µx - µy = M. The p-value for this two-sided test is always given by twice the smallest tail value.

The **T2MNTEST** program can be used to perform these hypothesis tests. To execute the program, we enter the value of the difference M to be tested, the sample sizes, sample averages, sample deviations, and the level of significance. To denote that we are using the pooled sample deviation for populations having a common variance, enter **1** for **POOLED?**. Otherwise enter **0** for **POOLED?**. The program displays the conclusion for each test, the test statistic, and the left and right tail values.

** Example.** A study on the IQ's of a group of 29 adopted children yielded an average IQ of 97 with a standard deviation of 13. Suppose another group of 68 children yielded an average IQ of 109 with a standard deviation of 11. Assume that IQ scores are normally distributed, that these groups were independent, and that there is equality of the true variances of scores within the entire populations of children from which these two samples came.

Test, at the 0.10 level of significance, the following three hypotheses about the true difference in average IQ scores between all children similar to the first group and all children similar to the second group:

*Solution.* After calling up the **T2MNTEST** program, enter **-10** for **TEST DIFFERENCE**, **29** for **X SAMPLE SIZE**, **97** for **XBAR**, **13** for **X SAMPLE DEV.**, **68** for **Y SAMPLE SIZE**, **109** for YBAR, **11** for **Y SAMPLE DEV.**, and **.10** for **LEVEL OF SIG.** Finally enter **1** for **POOLED?**.

We obtain a left tail value of 0.2199 from a test statistic of -0.7757, and we do not reject any of the hypotheses.

Based on this data, we can accept the claim that µ1 - µ2 = -10, or that µ2 - µ1 = 10. That is, the second population of children could possibly average 10 points higher in IQ than the first population of children.

We note that Xbar - Ybar = -12. But if µ1 - µ2 = -10, then there is still a 21.99% chance of Xbar - Ybar being as low as -12 with random samples of these sizes; thus we do not have enough evidence to reject the claim of µ1 - µ2 = - 10. (The p-value for this two-sided test is 2*.2199 = 0.4398.)

1. Consider the following independent random samples of Verbal SAT scores:

Assume that the scores are from normally distributed populations with girls' scores and boys' scores having equal variance. If the true mean for girls is µ1 and the true mean for boys is µ2, test the following three hypotheses at the 0.10 level of significance.

2. Consider now the following random samples of Math SAT scores for which we no longer assume equal variances among boys' scores and girls' scores.

State and conduct a test on whether or not the average scores of boys and girls are the same.

1. Enter the data into appropriate lists to compute the summary statistics: Xbar = 507.1428571, Sx = 110.2788673, Ybar = 490.4761905, Sy= 65.68684727 for these samples of size 21. Now we can either enter these value into the **T2MNTEST** program directly, or we can access them symbolically from the menus.

After executing the **T2MNTEST** program with a test difference of **70** and with a pooled variance, we obtain a left tail value of 0.0321 from a test statistic of -1.904057, and we reject only the second and third hypotheses.

Based on this data, we can accept that µ1 - µ2 < 70. That is, the average girls' score should be less than 70 points higher than the average boys' score.

We note that Xbar - Ybar = 16.667. If µ1 - µ2 >= 70, then there would be at most a 3.21% chance of Xbar - Ybar being as small as it is with samples of these sizes. (The p-value for the two-sided test is 2*.0321 = 0.0642, which is still below the level of significance.) Thus, we reject the second and third hypotheses.

2. Let µ1 be the true mean for boys and let µ2 be the true mean for girls. We shall test the hypothesis Ho: µ1 = µ2, with the alternative Ha: µ1 > µ2, at the 0.05 level of significance. This test is equivalent to Ho: µ1 - µ2 = 0, with a one-sided alternative Ha: µ1 - µ2 > 0.

After entering the data and computing the statistics, we execute the **T2MNTEST** program with **0** for **TEST DIFFERENCE** and **0** for **POOLED?**. We receive a right-tail value of 0.0063, so we must reject the null hypothesis. For if µ1 - µ2 = 0, then there would be at most a 0.63% chance of Xbar - Ybar being as large as 48.125 with samples of these sizes.

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