Eating Out Vegetarian. Repeat Exercise $11.136$by applying the formula in Exercise 11.134(a) if you can reasonably presume that at most $41\%$of the men sampled and at most $49\%$of the women sampled will be people who sometimes order veg. Compare the results obtained in Exercise $11.136$to those obtained here.

1. Without making a guess for the observed values of the sample proportions, find the common sample size that will ensure a margin of error of at most $0.01$for a $90\%$confidence interval. Hint: Use Exercise $11.133.$
2. Find a $90\%$confidence interval for $p_1-p_2$if, for samples of the size determined in part (a),$38.3\%$of the men and $43.7\%$of the women sometimes order veg.
3. Determine the margin of error for the estimate in part (b), and compare it to the required margin of error specified in part (a).

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 ${\hat{p}}_{2g}$

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

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

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:

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

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

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

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:

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