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

$n=400, \space \bar{x}=30,\space s=8$

Since we do not know the population variance, we shall apply the student's t distribution to construct the confidence interval.

A 89.9% confidence for the mean is given as,

$C.I=\bar{x}\pm t_{(\alpha/2,n-1)}(s/\sqrt n)$

Now,

$t_{\alpha/2,n-1}$ is the table value at $\alpha=0.101$ with $(n-1)=400-1=399$ degrees of freedom.

The table value $t_{0.101/2,399}=t_{0.0505,399}$ can be found using the following command in $R$

> qt(0.9495,399)

[1] 1.643825

Thus, the table value is $t_{0.0505,399}=1.643825$ and the confidence interval is,

$C.I=30\pm1.643825*(8/\sqrt{400})=30\pm(1.643825*0.4)=30\pm 0.65753 =(29.34247,\space 30.65753)$

A 89.9% confidence interval  for estimating the average time a student will spend studying Mathematics is, (29.34247, 30.65753)