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 0.54? (b) she does not use any prior estimates? Click the icon to view the table of critical values. (a) n=(Round up to the nearest integer.)

Answer 1

a)

We know that 90% confidence interval for true proportion is given by:

$\\(p\pm Z_{0.10/2}\sqrt{\frac{p(1-p)}{n}})\\ \\$

We want to find 'n' such that $Z_{0.10/2}\sqrt{\frac{p(1-p)}{n}}\leq 0.03$

$\\\Rightarrow Z_{0.05}\sqrt{\frac{0.54(1-0.54)}{n}}\leq 0.03 \Rightarrow (1.645)\sqrt{\frac{0.54(1-0.54)}{n}}\leq 0.03 \\ \\\Rightarrow n\geq 746.8629$

Required sample size = 747

b)

We know that 90% confidence interval for true proportion is given by:

$\\(p\pm Z_{0.10/2}\sqrt{\frac{p(1-p)}{n}})\\ \\$

We want to find 'n' such that $Z_{0.10/2}\sqrt{\frac{p(1-p)}{n}}\leq 0.03$

Since there is no prior estimate available, so we use the conservative estimate p=0.5.

$\\\Rightarrow Z_{0.05}\sqrt{\frac{0.5(1-0.5)}{n}}\leq 0.03 \Rightarrow (1.645)\sqrt{\frac{0.5(1-0.5)}{n}}\leq 0.03 \\ \\\Rightarrow n\geq 751.6736$

Required sample size = 752