When both populations have variance $\sigma^{2}$, the addition rule for variances says that $\bar{x}{1}-\bar{x}{2}$ has variance equal to the sum of the individual variances, which is
$$\frac{\sigma^{2}}{n_{1}}+\frac{\sigma^{2}}{n_{2}}=\sigma^{2}\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)$$
The standardized difference of means in this equal-variance case is therefore
$$z=\frac{\left(\bar{x}{1}-\bar{x}{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{\sigma \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}$$

$$\left(\bar{x}{1}-\bar{x}{2}\right) \pm t^{} s_{p} \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$$ Here $t^{}$ is the value for the $t\left(n_{1}+n_{2}-2\right)$ density curve with area $C$ between $-t^{}$ and $t^{}$.

To test the hypothesis $H_{0}: \mu_{1}=\mu_{2}$, compute the pooled two-sample $t$ statistic
$$t=\frac{\bar{x}{1}-\bar{x}{2}}{s_{p} \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}$$

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