Question

A research company desires to know the mean consumption of milk per week among males over age 33. They believe that the milk consumption has a mean of 4.4 liters, and want to construct a 90% confidence interval with a maximum error of 0.08 liters. Assuming a standard deviation of 0.8 liters, what is the minimum number of males over age 33 they must include in their sample?

Answer 1

Solution :

Given that,

standard deviation = =0.8

Margin of error = E = 0.08

At 90% confidence level of z is

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

sample size = n = [Z/2* / E] 2

n = ( 1.645 * 0.8 /0.08 )2

n =270.6025

Sample size = n =271