$100(1-\alpha) \%$ confidence interval for $\mu_{1}-\mu_{2}$, independent samples; $y_{1}$ and $y_{2}$ approximately normal; $\sigma_{1}^{2}=\sigma_{2}^{2}$
$$
\bar{y}{1}-\bar{y}{2} \pm t_{\alpha / 2} s_{p} \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}
$$
where
$$
s_{p}=\sqrt{\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}} \quad \text { and } \quad \mathrm{df}=n_{1}+n_{2}-2
$$
$t$ test for $\mu_{1}-\mu_{2}$, independent samples; $y_{1}$ and $y_{2}$ approximately normal; $\sigma_{1}^{2}=\sigma_{2}^{2}$
T.S.: $t=\frac{\bar{y}{1}-\bar{y}{2}-D_{0}}{s_{p} \sqrt{1 / n_{1}+1 / n_{2}}} \quad \mathrm{df}=n_{1}+n_{2}-2$
MATH0099 COURSE NOTES ：
The $99 \%$ confidence interval for $\sigma$ is then
$$
\sqrt{\frac{29(3.433)^{2}}{52.34}}<\sigma<\sqrt{\frac{29(3.433)^{2}}{13.12}}
$$
or
$$
2.56<\sigma<5.10
$$
Thus, we are $99 \%$ confident that the standard deviation in the weights of coffee cans lies between $2.56$ and $5.10$ grams. The designed value for $\sigma, 4$ grams, falls within our confidence interval. Using our results from , a $99 \%$ confidence interval for $\mu$ is
$$
500.453 \pm 2.756 \frac{3.433}{\sqrt{30}} \quad 500.453 \pm 1.73
$$
or
$$
498.7<\mu<502.2
$$