Question

Kim wants to determine a 80 percent confidence interval for the true mean serumHDL cholesterol 20-29 year old females. How large of a sample must she have to get a margin of error less than 2 points? Assume the population standard deviation is 13.4 points.

Answer 1

Solution :

Given that,

standard deviation = = 13.4

margin of error = E = 2

At 99% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z_/2 = Z_0.10 = 1.282

Sample size = n = ((Z_/2 * ) / E)²

= ((1.282 * 13.4) / 2)²

= 73.777

= 74

Sample size = 74