Testing Equality of Means for Two Independent Normal Populations

We want to test whether the means of two independent normal populations are equal. The null and alternative hypotheses for two‑tailed and one‑tailed tests are presented.

Before performing the test we distinguish two cases: (1) population variances are unknown but assumed equal, (2) variances are unknown and assumed unequal.

For case (1) we use the pooled‑variance t‑test with degrees of freedom ν = n₁ + n₂ – 2. The test statistic is t = (\bar X₁ – \bar X₂) / \sqrt{ s_p^2 (1/n₁ + 1/n₂) }, where s_p^2 is the pooled variance calculated from the sample variances s₁² and s₂² and sample sizes n₁ and n₂.

Example 1: Compare the average monthly returns of the S&P 500 in the 1970s and 1980s. The data suggest we cannot reject the hypothesis that the two decades have the same population variance. The null hypothesis H₀: μ₁ = μ₂ and the alternative H₁: μ₁ ≠ μ₂ are set. The test statistic is computed, and the critical values at the 0.05 and 0.01 significance levels are obtained. In both cases the null hypothesis is not rejected.

When the variances are assumed unequal, we use Welch’s t‑test with adjusted degrees of freedom ν = \frac{ (s₁²/n₁ + s₂²/n₂)^2 }{ (s₁²/n₁)^2/(n₁-1) + (s₂²/n₂)^2/(n₂-1) }.

Example 2: Recovery rates of defaulted bonds for public‑utility companies versus non‑utility firms. Because the utility sample has a smaller standard deviation, the equal‑means test is inappropriate, so Welch’s test is applied. The test statistic is calculated, the adjusted degrees of freedom are ν = 65, and the critical value at the 0.10 significance level is found. The result leads to rejection of the null hypothesis, indicating a significant difference in recovery rates between the two groups.