Question

A research company desires to know the mean consumption of meat per week among people over age 27. They believe that the meat consumption has a mean of 2.1 pounds, and want to construct a 99% confidence interval with a maximum error of 0.09 pounds. Assuming a variance of 1.44 pounds, what is the minimum number of people over age 27 they must include in their sample? Round your answer up to the next integer.

Answer 1

Solution :

Given that,

variance = 1.44

standard deviation = = 1.2

margin of error = E = 0.09

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.575

Sample size = n = ((Z_/2 * ) / E)²

= ((2.575 * 1.2) / 0.09)²

= 1178.77 = 1179  

minimum number of people = 1179