Question

A research company desires to know the mean consumption of meat per week among people over age 32. They believe that the meat consumption has a mean of 3 pounds, and want to construct a 80% confidence interval with a maximum error of 0.09 pounds. Assuming a standard deviation of 1 pounds, what is the minimum number of people over age 32 they must include in their sample? Round your answer up to the next integer.

Answer 1

Solution :

Given that,

standard deviation = = 1

margin of error = E = 0.09

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.2

/ 2 = 0.2 / 2 = 0.1

Z_/2 = Z_0.1 = 1.282

Sample size = n = ((Z_/2 * ) / E)²

= ((1.282 * 1) / 0.09)²

= 202.7776

Sample size = 203