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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 98% confidence interval to 5 points, how many students should the administrator sample? Make sure to give a whole number answer.

Answer 1

We are given that the standard deviation of SAT scores is 300 points, and the administrator wants to estimate the average SAT score with a margin of error of no more than 5 points. The confidence level is 98 percent. We are asked to find how many students should be sampled.

Since we are estimating a population mean and we know the population standard deviation, we use the sample size formula for a confidence interval for the mean:

Sample size equals = (Z-score times the population standard deviation divided by the margin of error) squared.

Now let's break that down.

The Z-score corresponds to the confidence level. For a 98 percent confidence level, 1 percent is in the lower tail and 1 percent is in the upper tail, leaving 98 percent in the middle. So we want the Z-score that leaves 99 percent to the left (since 1 percent is still above it). Using a Z-table or calculator, this Z-score is about 2.326.

Now plug in the values:

1. Z-score: 2.326
2. Standard deviation: 300
3. Margin of error: 5

So the sample size equals:

(2.326 times 300 divided by 5), and then square that whole result.

First, multiply 2.326 by 300, which gives 697.8.

Then divide 697.8 by 5, which gives 139.56.

Now square 139.56, which means multiplying 139.56 by itself. That gives approximately 19,476.6.

Because you can't sample a fraction of a person, you always round up, not to the nearest number, but up to make sure the margin of error is still within the limit.

So the final answer is:

The administrator should sample 19,477 students.

Answer 2

Hi Emery — no worries, I’ve got you!

You’re being asked to find the minimum sample size needed for a 98% confidence interval with a margin of error of 5, assuming SAT scores are normally distributed with a standard deviation of 300.

Use the sample size formula for estimating a population mean:

n = (z* × σ / E)²

Where:

1. z* = 2.326 (critical value for 98% confidence)
2. σ = 300
3. E = 5

Now plug in the values:

n = (2.326 × 300 / 5)²

n = (697.8 / 5)²

n = (139.56)²

n ≈ 19,476.6

Since sample size must be a whole number, always round up:

Final Answer: 19,477 students

Let me know if you want help understanding where that z*-value comes from or how confidence intervals work — happy to help!

Best regards,

Sid Bhatia

Answer 3

Hi Emery,

You are interested in the minimum sample size, n. Formula for this is:

n = (z*sigma/E)^2

z* = z critical value = 2.326 for 98%

Note: You can get these from the z-table, but it is probably worth memorizing the following:

z* (90%) = 1.645

z* (95%) = 1.96

z* (98%) = 2.326

z* (99%) = 2.576

sigma = standard deviation = 300

E = margin of error = 5

Thus:

n = ((2.326*300)/5))^2

n= 19477

I hope this helps!