Asked by Makaila
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?
Answers
Answered by
John1
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.
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.
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