Chapter 10.1, Problem 26E

Drive safely: How often does the average driver have an accident? The Allstate Insurance Company determined the average number of years between accidents for drivers in a large number of U.S. cities. Following are the results for 32 cities east of the Mississippi River and 32 cities west of the Mississippi River.

1. Construct a 95% confidence interval for the difference in mean scores between western and eastern cities.
2. An insurance company executive claims that the mean number of years between accidents for western cities is 1.5 years greater than the mean for eastern cities. Does the confidence interval contradict this claim?

To determine

To find: The $95\%$ confidence interval for the difference in mean scores between western and eastern cities.

Answer to Problem 26E

The $95\%$ confidence interval for the difference in mean scores between western and eastern cities is $\left(-0.773,0.992\right)$ .

Explanation of Solution

Given information:

The data is,

East
10.2	9.5	11.2	1 1.4	9.8	10.0	11.6	1 0.7
9.5	9.0	7.2	11.5	7.7	9.0	8.9	8.4
6.8	5.3	8.0	8.8	9.0	10.0	9.7	11.6
7.9	9.9	10.5	11.9	10.9	8.6	6.7	9.6

West
9.5	7.9	13.4	12.6	9.7	8.2	14.0	7.1
10.2	9.4	10.8	9.5	7.5	9.1	8.4	7.6
7.4	7.7	7.1	11.9	10.2	11.9	7.7	7.0
10.0	8.1	11.5	8.4	9.6	9.5	9.9	11.5

Concept used:

Minitab is used.

Calculation:

The steps for the confidence interval are,

Import the data, select start and choose the basic statistics option then select the 2 sample t and enter summarized data.

Click option button choose confidence level, test difference and alternative hypothesis and then click ok.

The data is shown below.

  

Figure-1

Therefore, the $95\%$ confidence interval for the difference in mean scores between western and eastern cities is $\left(-0.773,0.992\right)$ .

To determine

To find: Whether the confident interval contradict the claim.

Answer to Problem 26E

The confident interval contradicts the claim.

Explanation of Solution

Given information:

The data is,

East
10.2	9.5	11.2	1 1.4	9.8	10.0	11.6	1 0.7
9.5	9.0	7.2	11.5	7.7	9.0	8.9	8.4
6.8	5.3	8.0	8.8	9.0	10.0	9.7	11.6
7.9	9.9	10.5	11.9	10.9	8.6	6.7	9.6

West
9.5	7.9	13.4	12.6	9.7	8.2	14.0	7.1
10.2	9.4	10.8	9.5	7.5	9.1	8.4	7.6
7.4	7.7	7.1	11.9	10.2	11.9	7.7	7.0
10.0	8.1	11.5	8.4	9.6	9.5	9.9	11.5

Concept used:

Minitab is used.

Calculation:

Since, the $95\%$ confidence interval for the difference in mean scores between western and eastern cities is $\left(-0.773,0.992\right)$ and it does not include $1.5$ .

Therefore, the confident interval contradicts the claim.