Pooled variance procedure

Let $ S_x$ and $ S_y$ be the sample standard deviations constructed from $ X_1,\ldots,X_n$ and $ Y_1,\ldots,Y_m$, respectively. When it is reasonable to assume `` $ \sigma_1^2 = \sigma_2^2$,'' we can construct the pooled sample variance

$\displaystyle S_p^2 = \frac{(n-1)S_x^2 + (m-1)S_y^2}{n+m-2}
$

The test statistic

$\displaystyle T = \frac{\bar{X} - \bar{Y}}{S_p \sqrt{\frac{1}{n} + \frac{1}{m}}}
$

has the $ t$-distribution with $ (n+m-2)$ degrees of freedom under the null hypothesis $ H_0$. Thus, we reject the null hypothesis $ H_0$ with significant level $ \alpha$ when the observed value $ t$ of $ T$ satisfies $ \vert t\vert > t_{\alpha/2,n+m-2}$. Or, equivalently we can compute the $ p$-value

$\displaystyle p^* = 2 \times P(Y \ge \vert t\vert)
$

with $ Y$ having a $ t$-distribution with $ (n+m-2)$ degrees of freedom, and reject $ H_0$ when $ p^* < \alpha$.

Confidence interval. The following table shows the corresponding confidence interval of the population mean difference $ \mu_1 - \mu_2$, when your null hypothesis $ H_0$ is rejected.

Hypothesis testing $ (1-\alpha)$-level confidence interval
   
$ H_0: \mu_1 = \mu_2$ vs. $ H_A: \mu_1 \neq \mu_2$. $ \left(\bar{X} - \bar{Y}
- t_{\alpha/2,n+m-2}S_p
{\textstyle\sqrt{\frac{1}{n}+\...
...r{Y}
+ t_{\alpha/2,n+m-2}S_p
{\textstyle\sqrt{\frac{1}{n}+\frac{1}{m}}}
\right)$
$ H_0: \mu_1 \le \mu_2$ vs. $ H_A: \mu_1 > \mu_2$. $ \left(\bar{X} - \bar{Y}
- t_{\alpha,n+m-2}S_p
{\textstyle\sqrt{\frac{1}{n}+\frac{1}{m}}},\:
\infty\right)$
$ H_0: \mu_1 \ge \mu_2$ vs. $ H_A: \mu_1 < \mu_2$. $ \left(-\infty,\:
\bar{X} - \bar{Y}
+ t_{\alpha,n+m-2}S_p
{\textstyle\sqrt{\frac{1}{n}+\frac{1}{m}}}
\right)$

Example. Suppose that we consider the significant level $ \alpha = 0.01$, and that we have obtained $ \bar{X} = 80.02$ and $ S_x = 0.024$ from the control group of size $ n = 13$, and $ \bar{Y} = 79.98$ and $ S_y = 0.031$ from the experimental group of size $ m = 8$. Here we have assumed that $ \sigma_1^2 = \sigma_2^2$. Then we can compute the square root $ S_p = 0.027$ of the pooled sample variance $ S_p^2$, and the test statistic

$\displaystyle T = \frac{80.02 - 79.98}{0.027 \sqrt{\frac{1}{13} + \frac{1}{8}}}
\approx 3.33.
$

Thus, we can obtain $ p^* = 2 \times P(Y \ge 3.33) \approx 0.0035 < 0.01$, and reject $ H_0$. We conclude that the two population means are significantly different. And the 99% confidence interval for the mean difference is $ (0.006, 0.074)$.



Generated by MATH GO: 2006-03-21