Question

To determine student performance on a national test, a sample of 16 students is given the test. The average for the sample was 493 with a sample standard deviation of 56. Assume performance on the test is usually normally distributed, find a 90% confidence interval for the mean of the test.

Answer 1

Confidence interval = Z*

Here, = 493

Z* for 90% confidence level = 1.645

= 56

Sample size, n = 16

90% confidence interval for the mean of the test = 493 1.645 x

= 493 23.03

= (469.97, 516.03)