Question

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.03 with 90​% confidence if

​(a) she uses a previous estimate of 0.58​?

​(b) she does not use any prior​ estimates?

Answer 1

Solution:

(a) She uses a previous estimate of 0.58?

Answer:

The formula for finding the sample size is:

$\dpi{100} \large n=\hat{p}(1-\hat{p})\left ( \frac{z_{\frac{0.1}{2}}}{E} \right )^2$

Where:

$\dpi{100} \large \hat{p}=0.58$ is the previous estimate

$\dpi{100} \large z_{\frac{0.1}{2}}=1.645$ is the critical value at 0.1 significance level

$\dpi{100} \large E=0.03$ is the margin of error

$\dpi{100} \large \therefore n=0.58(1-0.58)\left ( \frac{1.645}{0.03} \right )^2$

  $\dpi{100} \large =0.2436 \times 3006.694444$

$\dpi{100} \large =732.43\approx 733$

Therefore, the required sample size is $\dpi{100} \large n=733$

(b) she does not use any prior estimates?

Answer: The formula for finding the sample size is:

$\dpi{100} \large n=0.5(1-0.5)\left ( \frac{z_{\frac{0.1}{2}}}{E} \right )^2$

Where:

$\dpi{100} \large z_{\frac{0.1}{2}}=1.645$ is the critical value at 0.1 significance level

$\dpi{100} \large E=0.03$ is the margin of error

$\dpi{100} \large \therefore n=0.50(1-0.50)\left ( \frac{1.645}{0.03} \right )^2$

  $\dpi{100} \large =0.25\times 3006.694444$

$\dpi{100} \large =751.7\approx 752$

Therefore, the required sample size is $\dpi{100} \large n=752$