(a) From Table $5.2$ and Eq. (5.1),
$$
\begin{aligned}
\bar{y}{\mathrm{st}} &=\frac{1}{N}\left[N{1} \bar{y}{1}+N{2} \bar{y}{2}+\cdots+N{L} \bar{y}{L}\right] \ &=\frac{1}{310}[(155)(33.900)+(62)(25.125)+(93)(19.00)] \ &=27.7 \end{aligned} $$ is the best estimate of the average number of hours per week that all households in the county spend watching television. Also, $$ \begin{aligned} \hat{V}\left(\bar{y}{\mathrm{st}}\right)=& \frac{1}{N^{2}} \sum N_{i}^{2}\left(1-\frac{n_{i}}{N_{i}}\right)\left(\frac{s_{i}^{2}}{n_{i}}\right) \
=& \frac{1}{(310)^{2}}\left[\frac{(155)^{2}(0.871)(5.95)^{2}}{20}+\frac{(62)^{2}(0.871)(15.25)^{2}}{8}\right.\
&\left.+\frac{(93)^{2}(0.871)(9.36)^{2}}{12}\right] \
=& 1.97
\end{aligned}
$$
The estimate of the population mean with an approximate 2-SD bound on the error of estimation is given by
$$
\bar{y}{\mathrm{st}} \pm 2 \sqrt{\hat{V}\left(\bar{y}{\mathrm{st}}\right)} \text { or } 27.675 \pm 2 \sqrt{1.97} \text { or } 27.7 \pm 2.8
$$
STAT0024 COURSE NOTES ：
$$
\sigma_{1} \approx \frac{10}{4}=2.5 \quad \text { and } \quad \sigma_{2} \approx \frac{8}{4}=2.0
$$
From Eq. (5.9),
$$
n_{i}=n\left(\frac{N_{i} \sigma_{i}}{\sum_{k=1}^{2} N_{k} \sigma_{k}}\right)
$$
where
$$
\sum_{i=1}^{2} N_{i} \sigma_{i}=(50)(2.5)+(40)(2.0)=125+80=205
$$
Then
$$
n_{1}=n\left(\frac{N_{1} \sigma_{1}}{\sum_{k=1}^{2} N_{k} \sigma_{k}}\right)=n\left(\frac{125}{205}\right)=0.61 n
$$
and
$$
n_{2}=n\left(\frac{80}{205}\right)=0.39 n
$$