Question

15. (8 points) An educator desires to estimate, within 0.03, the true proportion of high school students who study at least 1 hour each school night. He wants to be 98% confident. (a) How large a sample is necessary? Previously, he conducted a study and found that 60% of the 250 students surveyed spent at least 1 hour each school night studying. (b) If no estimate of the sample proportion is available, how large should the sample be?

Answer 1

15. Margin of Error of E=0.03

z value for 98% CI is 2.326 as P(-2.326<z<2.326)=0.98

So we will find n using formula of E

P9 E = 2* n

So $n=(\frac{z}{E})^2*p*q$

a. Here p=0.60

$n=(\frac{z}{E})^2*p*q=(\frac{2.326}{0.03})^2*0.6*0.4=1442.74=1443$

b. Here p=0.5

So $n=(\frac{z}{E})^2*p*q=(\frac{2.326}{0.03})^2*0.5*0.5=1502.85=1503$