A doctor want to estimate the HDL cholesterol of all 20- to 29- year old females. how many subjects are needed to estimate the HDL cholesterol within 3 points with 99% confindence assuming s=10.5. Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size required?

Question

A doctor want to estimate the HDL cholesterol of all 20- to 29- year old females. how many subjects are needed to estimate the HDL cholesterol within 3 points with 99% confindence assuming s=10.5. Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size​ required?

John1 · Answer

margin of error = z( s/sqrt n) me^2 = (z^2)(s^2)/n n = (z^2)(s^2)/me^2 n =(2.33)^2 ( 10.5)^2 /3^2 n = 5.43(110.25)/9 n = 66.5 so you need 67 people repeat the problem using 1.28 for z.