Question: A recent study of 25 students showed that they spent an average of $18.53 for gasoline per week. The standard deviation of the sample was $3.00. Find the 95% confidence interval of the true mean.

See Answer: Observe that the population standard deviation is not known. This means that the t-statistic will be used. The 95% confidence interval corresponds to the following expression:

\[CI=\left( \bar{X}-{{t}_{\alpha /2}}\times \frac{s}{\sqrt{n}},\,\,\bar{X}+{{t}_{\alpha /2}}\times \frac{s}{\sqrt{n}} \right)\]

In this case we get \({{t}_{\alpha /2}}\) is t-critical value (two-tailed), for our choice of \(\alpha ={0.05}\) and for the number of degrees of freedom as given by the sample size, given by \(25-1=24\). Using the above information, the confidence interval is

\[CI=\left( {18.53}-{2.064}\times \frac{3}{\sqrt{25}},\,\,{18.53}+{2.064}\times \frac{3}{\sqrt{25}} \right)=\left( {17.2917},\,\,{19.7683} \right)\]

This means that we are 95% confident that the actual population mean \(\mu\) is contained by the interval \(\left( {17.2917},\,\text{ }{19.7683} \right)\).

15. For the stress test described in Exercise 14, six women had an average heart rate of 115 beats per minute. The standard deviation of the sample was 6 beats. Find the 95% confidence interval of the true mean for the women.