91Ó°ÊÓ

Given in the question that, Refer to the study on ordering vegetarian considered in Examples 11.8-11.10.

we need to find the common sample size that will ensure a margin of error of at most $0.01$ for a $90\%$ confidence interval without making a guess for the observed values of the sample proportions,

The given values are, $E=0.01$and $90\%$confidence interval.

$n=0.5\left(z\mathrm{Î\pm }2E\right)2$

Here,$n1=n2=n$

Calculated the sample size

The confidence level is $90\%$, with a margin of error of $0.01$.

Determine the value of$\mathrm{Î\pm }$.

Calculate the value of $\mathrm{Î\pm }$,

$$\begin{gathered} 90=100(1-\mathrm{Î\pm })\mathrm{Î\pm } \\ =0.1 \end{gathered}$$

$$\begin{gathered} \mathrm{n}=0.5\left(\mathrm{za}2\mathrm{E}\right)2 \\ =0.5(1.6450.01)2 \\ =0.5(27060.25) \\ =13530.1 \end{gathered}$$

Thus, the required sample size is $13531$.

Given in the question that, Refer to the study on ordering vegetarian considered in Examples 11.8-11.10.

we need to 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.

The given values are, $\mathrm{E}=0.01$and $90\%$confidence interval.

$Cl=\left(p^{âˆ\S }1-p^{âˆ\S }2\right)Â\pm z\mathrm{Î\pm }2p^{âˆ\S }1\left(1-p^{âˆ\S }1\right)n1+p^{âˆ\S }2\left(1-p^{âˆ\S }2\right)n2$

Calculated the $90\%$confidence interval for p1-p2

localid="1651482144629" $$\begin{gathered} \mathrm{Cl}=\left(p^{âˆ\S }1-p^{âˆ\S }2\right)Â\pm z\mathrm{Î\pm }2p^{âˆ\S }1\left(1-p^{âˆ\S }1\right)n1+p^{âˆ\S }2\left(1-p^{âˆ\S }2\right)n2 \\ =(0.383-0.437)Â\pm 1.645(0.383(1-0.383)13531+0.437(1-0.437\left)13531\right) \\ =(-0,054)Â\pm 1.645(0.0059) \\ =-0.054Â\pm 0.0097 \end{gathered}$$

localid="1651482162341" $$\begin{gathered} Cl=(-0.054-0.0097,-0.054+0.0097) \\ =(-0.0637,-0.0443) \end{gathered}$$

Therefore, the$90\%$confidence interval for p1-p2 is$(-0.0637,-0.0443)$.

Given in the question that, Refer to the study on ordering vegetarian considered in Examples 11.8-11.10.

we need to 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 given values are, $E=0.01$and $90\%$confidence interval.

$Cl=\left(p^{âˆ\S }1-p^{âˆ\S }2\right)Â\pm z\mathrm{Î\pm }2p^{âˆ\S }1\left(1-p^{âˆ\S }1\right)n1+p^{âˆ\S }2\left(1-p^{âˆ\S }2\right)n2\text{.}$

From the part (b), the margin of error is$0.0097$

Comparison:

The margin of error in part (c) is smaller than the margin of error in part (a).