Question

Refer to the study on ordering vegetarian considered in Examples 11.8-11.10.

a. 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.

b. 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.

c. Determine the margin of error for the estimate in part (b), and compare it to the required margin of error specified in part (a).

Short answer

(a) The required sample size is $13531$.

(b) The $90\%$confidence interval for $\mathrm{p}1-\mathrm{p}2$is $(-0.0637,-0.0443)$.

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

Step-by-step solution

Part (a) Step 1: Given information

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,

Part(a) Step 2: Explanation

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

$n=0.5\left(z\alpha 2E\right)2$

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$\alpha$.

Calculate the value of $\alpha$,

$$\begin{gathered} 90=100(1-\alpha )\alpha \\ =0.1 \end{gathered}$$

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

Thus, the required sample size is $13531$.

Part (b) Step 1: Given information

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.

Part(b) Step 2: Explanation

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

$Cl=\left(p^{\wedge }1-p^{\wedge }2\right)\pm z\alpha 2p^{\wedge }1\left(1-p^{\wedge }1\right)n1+p^{\wedge }2\left(1-p^{\wedge }2\right)n2$

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

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

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$(-0.0637,-0.0443)$.

Part (c) Step 1: Given information

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).

Part (c) Step 2: Explanation

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

$Cl=\left(p^{\wedge }1-p^{\wedge }2\right)\pm z\alpha 2p^{\wedge }1\left(1-p^{\wedge }1\right)n1+p^{\wedge }2\left(1-p^{\wedge }2\right)n2\text{.}$

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).