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

Answer:

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

Step-by-step explanation:

We are given the following information:

Confidence level = 90%

Significance level = 10%

Maximum error = 0.09

Variance = 1.96

$\text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{1.96} = 1.4$

Formula:

$\text{Error} = z_{critical}\dfrac{\sigma}{\sqrt{n}}$

Putting all the values,

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.64$

$0.09 = 1.64\times \dfrac{1.4}{\sqrt{n}}\\\\n = (\dfrac{1.64\times 1.4}{0.09})^2 = 650.81 \approx 651$

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