Question

A marketing research company desires to know the mean consumption of milk per week among people over age 45. They believe that the milk consumption has a mean of 4.2 liters, and want to construct a 98% confidence interval with a maximum error of 0.08 liters. Assuming a standard deviation of 0.6 liters, what is the minimum number of people over age 45 they must include in their sample

Answer 1

The minimum number of people over age 45 they must include in their sample is 305.

Explanation:

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

`\alpha = (1 - 0.98)/(2) = 0.01`

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

That is z with a pvalue of
`1 - 0.01 = 0.99`, so Z = 2.327.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

Standard deviation of 0.6 liters

This means that
`\sigma = 0.6`

What is the minimum number of people over age 45 they must include in their sample?

This is n for which M = 0.08. So

`M = z(\sigma)/(√(n))`

`0.08 = 2.327(0.6)/(√(n))`

`0.08√(n) = 2.327*0.6`

`√(n) = (2.327*0.6)/(0.08)`

`(√(n))^2 = ((2.327*0.6)/(0.08))^2`

`n = 304.6`

Rounding up:

The minimum number of people over age 45 they must include in their sample is 305.