A researcher wishes to estimate, with 9090% confidence, the proportion of adults who have high-speed Internet access. Her estimate must be accurate within 55% of the true proportion. a) Find the minimum sample size needed, using a prior study that found that 2828% of the respondents said they have high-speed Internet access. b) No preliminary estimate is available. Find the minimum sample size needed.
Accepted Solution
A:
Answer: a) Required minimum sample size= 219b) Required minimum sample size= 271Step-by-step explanation:As per given , we haveMargin of error : E= 5% =0.05Critical z-value for 90% confidence interval : [tex]z_{\alpha/2}=1.645[/tex]a) Prior estimate of true proportion: p=28%=0.28Formula to find the sample size :-[tex]n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.28(1-0.28)(\dfrac{1.645}{0.05})^2\\\\=218.213856\approx219[/tex]Required minimum sample size= 219b) If no estimate of true proportion is given , then we assume p= 0.5Formula to find the sample size :-[tex]n=0.5(1-0.5)(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.25(\dfrac{1.645}{0.05})^2\\\\=270.6025\approx271[/tex]Required minimum sample size= 271