Question

SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is interested in estimating the average SAT score of first-year students. If the administrator would like to limit the margin of error of the 88% confidence interval to 15 points, how many students should the administrator sample? Make sure to give a whole number answer.

Answer 1

The administrator should sample 968 students.

Explanation:

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

`\alpha = (1 - 0.88)/(2) = 0.06`

Now, we have to find z in the Z-table as such z has a p-value of
`1 - \alpha`.

That is z with a p-value of
`1 - 0.06 = 0.94`, so Z = 1.555.

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 300.

This means that
`n = 300`

If the administrator would like to limit the margin of error of the 88% confidence interval to 15 points, how many students should the administrator sample?

This is n for which M = 15. So

`M = z(\sigma)/(√(n))`

`15 = 1.555(300)/(√(n))`

`15√(n) = 300*1.555`

Dividing both sides by 15

`√(n) = 20*1.555`

`(√(n))^2 = (20*1.555)^2`

`n = 967.2`

Rounding up:

The administrator should sample 968 students.