Chapter 12.6, Problem 2CYP

To determine

To test the hypothesis that the average length of time is greater than $2$ minutes.

Answer to Problem 2CYP

Rejected.

Explanation of Solution

Given:

The average time is $2$ minutes to evacuate the building during a fire drill. After collecting the data for $20$ fire drills, she arrives at a mean of $3.2$ minutes and a standard deviation of $0.9$ minute.

Calculation:

From the given information,

The values are as follows.

  $H_0=3.2,H_1>2,n=20,\overline{x}=2,s=0.9$

The confidence level is calculated below.

  $\begin{array}{l}CI=\overline{x}\pm 1.96\cdot \frac{s}{\sqrt{n}}\\ =2\pm 1.96\cdot \frac{0.9}{\sqrt{20}}\\ =2.39\text{ or }1.60\end{array}$

This means that the experience will produce a mean between $2.39$ and $1.60$ .

The confidence interval does not include $H_0$ , the null hypothesis will be rejected.