Two-Sample t Test (Independent Samples with a Common Variance)

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Chapter: Biostatistics for the Health Sciences: Tests of Hypotheses

Power function for a test that a normal population has mean zero versus a two-sided alternative

TWO-SAMPLE t TEST (INDEPENDENT SAMPLES WITH A COMMON VARIANCE)

Recall from Section 8.5 the use of the appropriate t statistic for a confidence interval under the following circumstances: the parent populations have normal distributions and common variance that is unknown. In this situation, we used the pooled variance estimate, s2p, calculated by the formula Sp2 = {St2(nt – 1) + Sc2(nc – 1)}/[nt + nc – 2]

Figure 9.2. Power function for a test that a normal population has mean zero versus a two-sided alternative when the sample size n = 25, n = 100, and the significance level α = 0.05.

Suppose we want to evaluate whether the means of two independent samples selected from two parent populations are significantly different. We will use a t test with sp2 as the pooled variance estimate. The corresponding t statistic is t. The formula for t is obtained by replacing the common in the formula for the two sample Z test with the pooled estimate Sp. The resulting statistic has Student’s t distribution with nt + nc – 2 degrees of freedom. This sample t statistic is used for hypothesis testing. For a two-sided test the steps are as follows:

1. State the null hypothesis H0: μt = μc versus the alternative hypothesis H1: μt

≠ μc.

2. Choose a significance level α= α0 (often we take α0 = 0.05 or 0.01).

3. Determine the critical region, that is, the region of values of t in the upper and lower α/2 tails of the sampling distribution for Student’s t distribution with nt + nc – 2 degrees of freedom when μt = μc (i.e., the sampling distribution when the null hypothesis is true).

4. Compute the t statistic:  for the given sample and sample sizes nt and nc, where Xt is the sample mean for the treatment group, c is the sample mean for the control group, and Sp is the pooled sample standard deviation.

5. Reject the null hypothesis if the test statistic t (computed in step 4) falls in the rejection region for this test; otherwise, do not reject the null hypothesis.

We will apply these steps to the pig blood loss data from Section 8.7, Table 8.1. Recall that Sp2 = {S2t(nt – 1) + Sc2(nc – 1)}/[nt + nc – 2] = {(717.12)2 9 + (1824.27)2 9}/18, since nt = nc = 10, St = 717.12, and Sc = 1824.27. So Sp2 = 2178241.61 and taking the square root we find Sp = 1475.89. As the degrees of freedom are nt + nc – 2 = 18, we find that the constant C from the table of the Student’s t distribution is 2.101. Applying steps 1–5 to the pig blood loss data for a two-tailed (two-sided) test, we have:

1. State the null hypothesis H0: μt = μc versus the alternative hypothesis H1: μt ≠ μc.

2. Choose a significance level α = α0 = 0.05.

3. Determine the critical region, that is, the region of values of t in the upper and lower 0.025 tails of the sampling distribution for Student’s t distribution with 18 degrees of freedom when μt/μc (i.e., the sampling distribution when the null hypothesis is true).

4. Compute the t statistic:  We are given that the sample sizes are nt = 10 and nc = 10, respectively. Under the null hypothesis, μtμc = 0 and tXc = 1085.9–2187.4 = –1101.5 and sp, the pooled sample standard deviation, is 1475.89. Since {(1/nt) + (1/nc)]} = (2/20) = 0.1 = 0.316, t = –1101.5/(1475.89)0.316 = –2.362.

5. Now, since –2.362 < –C = –2.101, we reject H0.

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