91Ó°ÊÓ

The observed values are:

$$\begin{gathered} {\hat{p}}_{1g}=0.41, \\ {\hat{p}}_{2g}=0.49, \\ {z}_{\frac{a}{2}}=1.645, \\ E=0.01\text{and}90\%\text{confidence interval.} \end{gathered}$$

The sample size formula is as follows:

$$\begin{gathered} n=n_1 \\ =n_2 \end{gathered}$$

$=\left({\hat{p}}_{1g}\left(1-{\hat{p}}_{1_g}\right)+{\hat{p}}_{2g}\left(1-{\hat{p}}_{2g}\right)\right){\left(\frac{z\frac{a}{2}}{E}\right)}^2$

The sample size was calculated as follows:

$$\begin{gathered} n=n_1 \\ =n_2 \end{gathered}$$

$=\left({\hat{p}}_{1g}\left(1-{\hat{p}}_{1g}\right)+{\hat{p}}_{2g}\left(1-{\hat{p}}_{2g}\right)\right){\left(\frac{za}{E}\right)}^2$

Replace $0.41$for ${\hat{p}}_{1g}$

Replace $0.49$for ${\hat{p}}_{2g}$

$1.645\text{for}{z}_{\frac{\mathrm{Î\pm }}{2}}\text{and}0.01\text{for}E\text{in above equation}$

$$\begin{gathered} n=(0.41(1-0.41)+0.49(1-0.49\left)\right){\left(\frac{1.645}{0.01}\right)}^2 \\ =(02419+0.2499)(27060.25) \\ =13308.231 \\ =13309 \end{gathered}$$

Given in the question that, the samples of the size determined in part (a), $38.3\%$of the men and $43.7\%$of the women sometimes order veg.

$$\begin{gathered} {\hat{p}}_{1g}=0.41, \\ {\hat{p}}_{2g}=0.49, \\ {z}_{\frac{a}{2}}=1.645, \\ E=0.01\text{and}90\%\text{confidence interval.} \end{gathered}$$

The confidence interval formula is as follows:

$CI=\left({\hat{p}}_1-{\hat{p}}_2\right)Â\pm z_{\frac{a}{2}}\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}.$

For $P_1-P_2$calculated the $90\%$confidence interval as follows:

$$\begin{gathered} CI=\left({\hat{p}}_1-{\hat{p}}_2\right)Â\pm z_{\frac{a}{2}}\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}} \\ =(0.383-0.437)Â\pm 1.645\sqrt{\frac{0.383(1-0.383)}{13309}+\frac{0.437(1-0.437)}{13309}} \\ =(-0.054)Â\pm 0.0099030900 \\ =(-0.0639,-0.0441) \end{gathered}$$

The observed values are:

$$\begin{gathered} {\hat{p}}_{1g}=0.41, \\ {\hat{p}}_{2g}=0.49, \\ {z}_{\frac{a}{2}}=1.645, \\ E=0.01\text{and}90\%\text{confidence interval.} \end{gathered}$$

The margin of error formula is as follow:

$$\begin{gathered} E=z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}} \\ =1.645\sqrt{\frac{0.383(1-0.383)}{13309}+\frac{0.437(1-0.437)}{13309}} \\ =1.645(0.00602) \\ =0.0099 \end{gathered}$$