Question

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.

Answer 1

Answer: a) Required minimum sample size= 219

b) Required minimum sample size= 271

Explanation:

As per given , we have

Margin of error : E= 5% =0.05

Critical z-value for 90% confidence interval :
`z_(\alpha/2)=1.645`

a) Prior estimate of true proportion: p=28%=0.28

Formula to find the sample size :-

`n=p(1-p)((z_(\alpha/2))/(E))^2\\\\=0.28(1-0.28)((1.645)/(0.05))^2\\\\=218.213856\approx219`

Required minimum sample size= 219

b) If no estimate of true proportion is given , then we assume p= 0.5

Formula to find the sample size :-

`n=0.5(1-0.5)((z_(\alpha/2))/(E))^2\\\\=0.25((1.645)/(0.05))^2\\\\=270.6025\approx271`

Required minimum sample size= 271