Question

A sample of 400 MTH409 students showed that the average time they spent studying Mathematics at home is 30 minutes with a standard deviation of 8 minutes. Find 89.9% confidence interval for estimating the average time a student will spend studying Mathematics. Assume that the sample is taken from a normal population.

Answer 1

$z=\frac{\overline{x}-\mu}{\sigma/\sqrt n}=\frac{30-\mu}{8/20}$

critical values for 89.9% confidence interval:

$z=\pm 1.275$

then:

$-1.275<\frac{30-\mu}{8/20}<1.275$

$-0.51<30-\mu<0.51$

$29.49<\mu<30.51$