A report in a research journal states that the average weight loss of people on a certain drug is 27 lbs with a standard deviation of 5 lbs. If this study is repeated, how large should the sample size be so that the margin of error would be no more than 2.5 lbs with 95% confidence?
Accepted Solution
A:
Answer: 16Step-by-step explanation:We know that the formula to find the sample size is given by :-[tex]n=(\dfrac{z_{\alpha/2}\cdot \sigma}{E})^2[/tex], where [tex]\sigma[/tex] = population standard deviation.E = Margin of error [tex]z_{\alpha/2}[/tex]= Two-tailed z-value for significance level of [tex]\alpha.[/tex]As per given , we have[tex]\alpha=1-0.95=0.05[/tex]Using z-value , table [tex]z_{\alpha/2}=1.96[/tex][tex]\sigma= 5\ lbs[/tex]Margin of error = 2.5 lbsThen, the required sample size would be :_[tex]n=(\dfrac{1.96\cdot 5}{2.5})^2[/tex][tex]n=(3.92)^2=15.3664\approx16[/tex]Hence, the required sample size = 16