Chapter 11, Problem 23SGR

To determine

To find: The $95\%$ confidence interval.

Answer to Problem 23SGR

The alternative hypothesis is $\mu >35$ ,the null hypothesis is $\mu \le 35$ (claim).

Explanation of Solution

Given:

The mean time is given $64.3$ minutes with a standard deviation of $7.3$ minutes.

Calculation:

In a $95\%$ confidence $63.5$ interval, $2.5\%$ of the area lies on each tail. The corresponding $z$ -value is $1.960$ .

The maximum error of estimate $E$ for a population mean is given by:

  $E=z\cdot \frac{s}{\sqrt{n}}$

The confidence interval for the population is given as:

  $CI=\overline{x}\pm z\cdot \frac{s}{\sqrt{n}}$

The calculation of confidence interval is given as:

  $\begin{array}{c}CI=64.3\pm \mathrm{1.960.}\frac{7.3}{\sqrt{300}}\\ CI\approx 64.3\pm 0.8\left[\text{Use a calculator}\right]\end{array}$

The $95\%$ confidence interval is $63.5<\mu <65.1$ .

Therefore we are $95\%$ confident that the population mean time is between $63.5$ minutes and $65.1$ minutes.