Question

Scores on a certain test are normally distributed with a variance of 22. A researcher wishes to estimate the mean score achieved by all adults on the test. Find the sample size needed to assure with 95 percent confidence that the sample mean will not differ from the population mean by more than 2 units.

Answer 1

Answer:

If the variance is 22, the standard deviation is sqrt(22) = 4.6904

For us to be 95% confident, we need to look up where P(z<?) = .975 (2.5% of the data will be excluded on the right; 2.5% will be excluded on the left). On a z-table, this is z = 1.96.

So from there, we know everything we need.
Margin of error = z-score * sigma / sqrt(n)
2 = 1.96 * 4.6904 / sqrt(n)

sqrt(n)=9.193184/2
sqrt(n) = 4.6
n = 21.16

Since you can't have .16 of a person, though, round up. n = 22 people.