Question

A marketing research company desires to know the mean consumption of meat per week among people over age 31. They believe that the meat consumption has a mean of 4.7 pounds, and want to construct a 80% confidence interval with a maximum error of 0.090.09 pounds. Assuming a variance of 0.64 pounds, what is the minimum number of people over age 31 they must include in their sample? Round your answer up to the next integer.

Answer 1

Solution :

Given that,

Variance = = 0.64

Population standard deviation = = 0.8

Margin of error = E = 0.09

At 80% 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

n = (1.282 * 0.8/ 0.09)²

n = 129.86

n = 130

Sample size = 130