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)

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