Question

a random sample of 510 high school students has a normal distribution. The sample mean average ACT exam score was 21 with a 3.2 standard deviation. Construct a 99% confidence interval estimate of the population mean average ACT exam. Find the critical value​

Answer 1

CI = 21 ± 0.365

Explanation:

The confidence interval is:

CI = x ± SE * CV

where x is the sample mean, SE is the standard error, and CV is the critical value (either t score or z score).

Here, x = 21.

The standard error for a sample mean is:

SE = σ / √n

SE = 3.2 / √510

SE = 0.142

The critical value is looked up in a table or found with a calculator. But first, we must find the alpha level and the critical probability.

α = 1 - 0.99 = 0.01

p* = 1 - (α/2) = 1 - (0.01/2) = 0.995

Using a calculator or a z-score table:

P(x<z) = 0.995

z = 2.576

Therefore:

CI = 21 ± 0.142 × 2.576

CI = 21 ± 0.365

Round as needed.