Average SAT scores of first-year students

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 96% confidence interval to 5 points, how many students should the administrator sample? Make sure to give a whole number answer.

Answer 1

Hi Bailey,

We have normally distributed data with known standard deviation, so we can use z to seek sample size. Now, let's look at common formulas and evaluate what we are given: Recall:

MoE= z*SE

z*=z-critical value

SE=standard error

SE=sigma/sqrt(n)

sigma=standard deviation

n=sample size

Ultimately, the goal is to find n. We have sigma, but no SE yet. We do, however, have a margin of error.

MoE=5, so:

5=z*SE

Now, let's talk about z*. For a 96% confidence interval, we need to subtract:

(1-0.96)=0.04

Divide by 2:

0.02

Subtract from 1 again:

(1-0.02)=0.98

Look in the interior of the z-table for 0.98; closest I have is 0.9798 at:

z*=2.05

Substituting into Margin of Error equation:

MoE=z*SE

MoE=5

z*=2.05

5=2.05*SE

SE=2.44

Now that we have standard error, we can substitute into the equation for sample size:

SE=sigma/sqrt(n)

SE=2.44

sigma=300

n=n, the sample size we are looking for

2.44=300/sqrt(n)

2.44sqrt(n)=300

sqrt(n)=(300/2.44)

n=(300/2.44)²

n=15117

I hope this helps.