Chapter 10.2, Problem 10.40E

To determine

To find: that to give 90% confidence interval for the mean of the company’s executive population and expectation about 128 to be inside these intervals.

Answer to Problem 10.40E

(123.1622, 129.0378)

Mean did not differ significantly from 128 at the 10% level

Explanation of Solution

Given:

  $\begin{array}{l}n=72\\ \overline{x}=126.1\\ s=15.2\\ c=90\%=0.90\end{array}$

Formula used:

  $E=z_{a/2}\times \frac{s}{\sqrt{n}}$

Confidence interval

  $\begin{array}{l}\overline{x}-E\\ \overline{x}+E\end{array}$

Calculation:

For confidence level $1-\alpha =0.90$

  $z_{a/2}=z_{0.005}$ using the normal probability table

  $z_{a/2}=1.64$

The margin of error is then:

  $E=z_{a/2}\times \frac{s}{\sqrt{n}}=1.64\times \frac{15.2}{\sqrt{72}}=2.9378$

Confidence intervals are

  $\begin{array}{l}\overline{x}-E=126.1-2.9378=123.1622\\ \overline{x}+E=126.1+2.9378=129.0378\end{array}$

It is expected that 128 to be inside this interval, because it is concluded that the mean did not differ significantly from 128 at the 10% level of significance and a 90% confidence interval associates with a significant test at the 10% level.