Chapter MPS, Problem 12.10MPS

To determine

Determine a $95\%$ confidence interval.

Answer to Problem 12.10MPS

  $84.4283\le \mu \le 85.5717$

Explanation of Solution

Given information:

A sample of $144$ students was asked for the average amount of time they spent talking on the phone every day. The mean time was $85$ minutes with a standard deviation of $3.5$ minutes.

Calculation:

The $95\%$ confidence interval formula is,

  $CI=\overline{X}\pm 1.96\frac{s}{\sqrt{n}}\mathrm{.....}(1)$

Substitute $\overline{X}=85\min ,s=3.5\min \&n=144$ in equation $(1)$ .

  $\begin{array}{l}CI=\overline{X}\pm 1.96\frac{s}{\sqrt{n}}\\ =85\pm 1.96\frac{3.5}{\sqrt{144}}\\ =85\pm 1.96\frac{3.5}{12}\\ =85\pm 1.96\times 0.2917\\ \approx 85\pm 0.5717\end{array}$

The lower bond of the interval is,

  $\begin{array}{l}l=85-0.5717\\ =84.4283\end{array}$

The upper bond of the interval is,

  $\begin{array}{l}u=85+0.5717\\ =85.5717\end{array}$

So, the $95\%$ confidence interval estimate is $84.4283\le \mu \le 85.5717$ .

Hence, the $95\%$ confidence interval estimate of the sample with $\overline{X}=85\min ,s=3.5\min \&n=144$ is $84.4283\le \mu \le 85.5717$ .