Question

A marketing research company desires to know the mean consumption of milk per week among people over age 30. They believe that the milk consumption has a mean of 2 liters, and want to construct a 90% confidence interval with a maximum error of 0.09 liters. Assuming a variance of 1.96 liters, what is the minimum number of people over age 30 they must include in their sample?

Answer 1

651 is the minimum number of people over age 30 they must include in their sample.

Explanation:

We are given the following information:

Confidence level = 90%

Significance level = 10%

Maximum error = 0.09

Variance = 1.96

`\text{Standard Deviation} = \sqrt{\text{Variance}} = √(1.96) = 1.4`

Formula:

`\text{Error} = z_(critical)(\sigma)/(√(n))`

Putting all the values,

`z_(critical)\text{ at}~\alpha_(0.05) = \pm 1.64`

`0.09 = 1.64* (1.4)/(√(n))\\\\n = ((1.64* 1.4)/(0.09))^2 = 650.81 \approx 651`

651 is the minimum number of people over age 30 they must include in their sample.