Question

A research company desires to know the mean consumption of milk per week among males over age 45. They believe that the milk consumption has a mean of 4.5 liters, and want to construct a 95% confidence interval with a maximum error of 0.08 liters. Assuming a variance of 1 liters, what is the minimum number of males over age 45 they must include in their sample? Round your answer up to the next integer.

Answer 1

Answer:

The minimum number of males over age 45 they must include in their sample is 601

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population(square root of the variance) and n is the size of the sample.

Assuming a variance of 1 liters, what is the minimum number of males over age 45 they must include in their sample?

This is n when $\sigma = \sqrt{1} = 1, M = 0.08$

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.08 = 1.96*\frac{1}{\sqrt{n}}$

$0.08\sqrt{n} = 1.96$

$\sqrt{n} = \frac{1.96}{0.08}$

$\sqrt{n} = 24.5$

$(\sqrt{n})^{2} = (24.5)^{2}$

$n = 600.25$

Rounding up

The minimum number of males over age 45 they must include in their sample is 601