Chapter 14, Problem 34SGA

To determine

To find the interval about the sample mean that has $1\%$ level of confidence.

Answer to Problem 34SGA

  $\left[70.03,109.96\right]$

Explanation of Solution

Given:

  $\sigma =30,N=15,\overline{X}=90$ .

Calculation:

First determine the percentage using the table given on page no. 919.

The $1\%$ level of confidence correspond to $P=99\%$

Now, $P=99\%$ correspond to $t=2.58$ .

Now finding the standard error by using the following formula-

  $\begin{array}{l}{\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{N}}\\ {\sigma }_{\overline{X}}=\frac{30}{\sqrt{15}}\\ {\sigma }_{\overline{X}}\approx 7.75\end{array}$

Now, finally let us find the interval as follows-

  $\begin{array}{l}\left[\overline{X}-t\sigma ,\overline{X}+t\sigma \right]\\ =\left[90-2.58\times 7.74,90+2.58\times 7.74\right]\\ =\left[70.03,109.96\right]\end{array}$