Question

A survey of 454 women revealed that 24​% wear flat shoes to work.
a. Use this sample information to develop a 99​% confidence interval for the population proportion of women who wear flat shoes to work.
b. Suppose that the people who administered this survey now wish to estimate the proportion of women who wear athletic shoes to work with a margin of error of 0.01 with 95​% confidence. Determine the sample size required.

Answer 1

Answer with explanation:

As per given , we have

$n= 454$

$p=0.24$

a) s per given , we have

$\ovreline{x}=8.17$

$\sigma=2.2$

n= 64

a)Critical z-value for 99% confidence : $z_{\alpha/2}=2.576$

Confidence interval :

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

$0.24\pm (2.576)\sqrt{\dfrac{0.24(1-0.24)}{454}}$

$\approx0.24\pm0.0516$

$(0.24-0.0516,\ 0.24+0.0516)=(0.1884,\ 0.2916)$

Hence, the 99​% confidence interval for the population proportion of women who wear flat shoes to work = $(0.1884,\ 0.2916)$

b) Margin of error : E= 0.01

Critical z-value for 95% confidence : $z_{\alpha/2}=1.96$

since , the prior estimate of proportion of women who wear athletic shoes to work is unknown.

Thus , sample size = $0.25(\dfrac{z_{\alpha/2}}{E})^2$

$0.25(\dfrac{1.96}{0.01})^2=9604$

Required sample size : 9604