Question

A fast food restaurant executive wishes to know how many fast food meals teenagers eat each week. They want to construct a 80% confidence interval with an error of no more than 0.07. A consultant has informed them that a previous study found the mean to be 5.8 fast food meals per week and found the standard deviation to be 1.3. What is the minimum sample size required to create the specified confidence interval

Answer 1

The minimum sample size required to create the specified confidence interval is of 565.

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 - 0.8)/(2) = 0.1`

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 - 0.1 = 0.9`, so Z = 1.28.

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.

Found the standard deviation to be 1.3.

This means that
`\sigma = 1.3`

Error of no more than 0.07. What is the minimum sample size required to create the specified confidence interval?

This is n for which M = 0.07. So

`M = z(\sigma)/(√(n))`

`0.07 = 1.28(1.3)/(√(n))`

`0.07√(n) = 1.28*1.3`

`√(n) = (1.28*1.3)/(0.07)`

`(√(n))^2 = ((1.28*1.3)/(0.07))^2`

`n = 565.1`

Rounding up:

The minimum sample size required to create the specified confidence interval is of 565.