It's believed that as many as 20 % of adults over 50 never graduated from high school. We wish to see if this percentage is the same among the 25 to 30 age group. What sample size would allow us to increase our confidence level to 95% while reducing the margin of error to only 3 %?n = ?? (Round up to the nearest integer.)
Accepted Solution
A:
Answer: 683Step-by-step explanation:Formula to calculate the minimum sample size is given by :-[tex]n=p(1-p)(\dfrac{z_c}{E})^2[/tex], where p = prior estimate of population proportion.E= Margin of error.[tex]z_c[/tex] = z-value for confidence interval of c.Given : It's believed that as many as 20 % of adults over 50 never graduated from high school.We wish to see if this percentage is the same among the 25 to 30 age group.i.e. Prior estimate of proportion of adults among the 25 to 30 age group : p=20%=0.20Confidence interval : 95%From the standard normal z-value table , the z-value for 95% confidence interval = [tex]z_c=1.96[/tex]Margin of error : E= 3%=0.03Required minimum sample size would be :-[tex]n=(0.20)(1-0.20)(\dfrac{1.96}{0.03})^2[/tex]Simplify ,[tex]n=(0.20)(0.80)(\dfrac{1.96}{0.03})^2\\\\=0.16\times4268.44\\\\=682.9504\approx683[/tex]Thus , the minimum sample size required = 683