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

Step-by-step 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 :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.28(1-0.28)(\dfrac{1.645}{0.05})^2\\\\=218.213856\approx219$

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 :-

$n=0.5(1-0.5)(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.25(\dfrac{1.645}{0.05})^2\\\\=270.6025\approx271$

Required minimum sample size= 271