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 0.01 with 90% confidence if :(a) she uses a previous estimate of 0.42? (b) she does not use any prior estimates?
Accepted Solution
A:
Answer:Part A:n=6591.87≅6592Part B:n=6765.06≅6765Step-by-step explanation:In order to calculate the sample size we use the following estimated Sample proportion formula:[tex]n=p*q*\frac{Z^2}{E^2}[/tex]Where:p is the previous estimateq is the 1-pZ is the distributionE is the marginAt 90% Confidence significance level is 0.1/2=0.05Z at 0.05 or 5% =1.645Part A:p=0.42q=1-p=1-0.42=0.58Z=1.645E=0.01[tex]n=0.42*0.58*\frac{1.645^2}{0.01^2}[/tex]n=6591.87≅6592Part B:Since no prior estimate is given we assume p =0.5 and q=0.5[tex]n=0.5*0.5*\frac{1.645^2}{0.01^2}[/tex]n=6765.06≅6765