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

Answer:

The administrator should sample 968 students.

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.88}{2} = 0.06$

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

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

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

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

$15 = 1.555\frac{300}{\sqrt{n}}$

$15\sqrt{n} = 300*1.555$

Dividing both sides by 15

$\sqrt{n} = 20*1.555$

$(\sqrt{n})^2 = (20*1.555)^2$

$n = 967.2$

Rounding up:

The administrator should sample 968 students.