7 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 99% confidenceif(a) she uses a previous estimate of 0.28?(b) she does not use any prior estimates?Table of critical valuesClick the icon to view the table of critical values.(a) n =(Round up to the nearest integer.)Critical Value, zęLevel of Confidence,Area in Each Tail,(1 – a) · 100%(b) n = (Round up to the nearest integer.)90%0.051.64595%0.0251.9699%0.0052.575PrintDone

The sample size(n) is a small part of the population. The appropriate sample size provides the précised point estimate for the population parameter and it reducing the man power and cost of research. sample size (n)=zα22*p(1-p)e2Here, zα2is two sided z -critical value p is the estimated proportion e is the margin of error z - table value : Given Confidence level is 0.99 then, level of significance α=0.01 From the table of critical values zα/2 = z 0.01/2 = z 0.005 =2.575 Part a) Find sample size n, when previous estimate is 0.28 Given values are : Margin of error (e) = 0.03 Estimated proportion (p) = 0.28 then, we have to substitute the values p = 0.28, zα/2 = 2.575 & e = 0.03 in the above formula sample size (n) =2.5752*0.28(1-0.28)0.032 =6.6306*0.20160.0009 =1.33670.0009 =1485.2 ~ 1485 Therefore, sample size (n) is 1485 b) Find sample size n, when there is no previous estimate We can consider p = 0.50, if there is no previous estimated proportion. {Since, if we not sure about proportion we can consider p( success) and p(failure) equally } Thus, substitute the values p = 0.50, zα/2 = 2.575 & e = 0.03 in the above sample size formula sample size (n) =2.5752*0.5(1-0.5)0.032 =6.6306*0.250.0009 =1.65770.0009 =1841.8 ~ 1842 Therefore, sample size (n) is 1842