Difference between two means: pooled t-testEdexcel A-Level Further Maths: Revision notes

Use the pooled $t$-test to compare the means $\mu_X$ and $\mu_Y$ of two populations when you have two independent random samples, of sizes $n_X$ and $n_Y$, and the population variances are unknown but assumed equal. The populations must be Normal. The common variance $\sigma^2$ is unknown, so it has to be estimated from both samples together. Because $\sigma^2$ is estimated, the standardised difference follows a $t$ distribution rather than a Normal distribution.

Combine the two unbiased sample variances, weighting each by its degrees of freedom: $s^2=\frac{(n_X-1)s_x^2+(n_Y-1)s_y^2}{n_X+n_Y-2}.$ If you are given sums instead of variances, use $S_{xx}=\sum x^2-\frac{(\sum x)^2}{n}$, so that $s^2=\frac{S_{xx}+S_{yy}}{n_X+n_Y-2}$. The pooled value always lies between the two sample variances, nearer to the one from the larger sample.

Under $H_0:\mu_X=\mu_Y$, $T=\frac{\bar{X}-\bar{Y}}{S\sqrt{\frac{1}{n_X}+\frac{1}{n_Y}}}\sim t_{n_X+n_Y-2}.$ More generally $\frac{(\bar{X}-\bar{Y})-(\mu_X-\mu_Y)}{S\sqrt{1/n_X+1/n_Y}}\sim t_{n_X+n_Y-2}$. The degrees of freedom are $n_X+n_Y-2$: one is lost for each sample mean estimated. The denominator $s\sqrt{\frac1{n_X}+\frac1{n_Y}}$ is the estimated standard error of $\bar{X}-\bar{Y}$.

1. State $H_0:\mu_X=\mu_Y$ and $H_1$ (one-tailed $\mu_X>\mu_Y$ or $\mu_X<\mu_Y$, or two-tailed $\mu_X\neq\mu_Y$).
2. Find $s^2$, then $t$.
3. Find the critical value of $t_{n_X+n_Y-2}$ at the stated significance level (halve the level for two tails).
4. Compare, and state the conclusion in the context of the question.

Worked example. $n_X=6$, $\bar{x}=20.5$, $s_x^2=2.4$; $n_Y=8$, $\bar{y}=18.9$, $s_y^2=3.1$. Test $H_1:\mu_X>\mu_Y$ at $5\%$. $s^2=\frac{5(2.4)+7(3.1)}{12}=2.808$, so $t=\frac{1.6}{\sqrt{2.808(\frac16+\frac18)}}=1.77$. The critical value of $t_{12}$ is $1.782$. As $1.77<1.782$, do not reject $H_0$: insufficient evidence that $\mu_X>\mu_Y$.

A $(1-\alpha)$ confidence interval for $\mu_X-\mu_Y$ is $(\bar{x}-\bar{y})\pm t_{n_X+n_Y-2}\;s\sqrt{\frac{1}{n_X}+\frac{1}{n_Y}},$ where $t$ is the two-tailed critical value (for $95\%$, the upper $2.5\%$ point). Using the worked example's data with $t_{12}=2.179$: $1.6\pm2.179\times0.905=(-0.37,\ 3.57)$. If the interval contains $0$, there is no evidence of a difference in means at the corresponding significance level; if it does not, there is.