Question

A doctor wants to estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 4 points with 99 % confidence assuming s equals 14.5 based on earlier​ studies? Suppose the doctor would be content with 90 % confidence. How does the decrease in confidence affect the sample size​ required?

A​ 90% confidence level requires ____ subjects. ​(Round up to the nearest​ subject.)

Answer 1

USING 99% confidence interval

z score for 99% confidence level is 2.576 using z distribution table

standard deviation is s = 14.5 and margin of error = 4

sample size n = ((z*s)/ME)^2

this implies

= ((2.576*14.5)/4)^2

= 9.338^2

= 87.198

= 87 (rounded to nearest integer)

USING 90% confidence interval

standard deviation is s = 14.5 and margin of error = 4

z score for 90% confidence level is 1.645 using z distribution table

sample size n = ((z*s)/ME)^2

this implies

= ((1.645*14.5)/4)^2

= 5.9631^2

= 35.56

= 36 (rounded to nearest integer)