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

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 s_p² 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 S_p. The resulting statistic has Student’s t distribution with n_t + n_c – 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: μ_t = μ_c versus the alternative hypothesis H₁: μ_t

≠ μ_c.

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 n_t + n_c – 2 degrees of freedom when μ_t = μ_c (i.e., the sampling distribution when the null hypothesis is true).

4. Compute the t statistic:  for the given sample and sample sizes n_t and n_c, where X_t is the sample mean for the treatment group, _c is the sample mean for the control group, and S_p 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 S_p² = {S²_t(n_t – 1) + S_c²(n_c – 1)}/[n_t + n_c – 2] = {(717.12)² 9 + (1824.27)² 9}/18, since n_t = n_c = 10, S_t = 717.12, and S_c = 1824.27. So S_p² = 2178241.61 and taking the square root we find S_p = 1475.89. As the degrees of freedom are n_t + n_c – 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: μ_t = μ_c versus the alternative hypothesis H₁: μ_t ≠ μ_c.

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 μ_t/μ_c (i.e., the sampling distribution when the null hypothesis is true).

4. Compute the t statistic:  We are given that the sample sizes are n_t = 10 and n_c = 10, respectively. Under the null hypothesis, μ_t – μ_c = 0 and _t – X_c = 1085.9–2187.4 = –1101.5 and s_p, the pooled sample standard deviation, is 1475.89. Since √{(1/n_t) + (1/n_c)]} = √(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.