Question

Car Repair Costs Listed below are repair costs (in dollars) for cars crashed at 6 mi/h in full-front crash tests and the same cars crashed at 6 mi/h in full-rear crash tests (based on data from the Insurance Institute for Highway Safety). The cars are the Toyota Camry, Mazda 6, Volvo S40, Saturn Aura, Subaru Legacy, Hyundai Sonata, and Honda Accord. Is there sufficient evidence to conclude that there is a linear correlation between the repair costs from full-front crashes and full-rear crashes?

Front	936	978	2252	1032	3911	4312	3469
Rear	1480	1202	802	3191	1122	739	2767

Short answer

There is not enough evidence to conclude that there is a significant linear correlation between the repair costs from full-front car crashes and full rear car crashes.

Step-by-step solution

Given information

The repair costs are tabulated for cars crashed at 6 mi/h in full-front crashes and for cars crashed at 6mi/h in full rear crashes.

Hypotheses

The null hypothesis is as follows:

There is no linear correlation between the repair costs of two different crashes.

The alternative hypothesis is as follows:

There is a linear correlation between the repair costs of two different crashes.

Correlation coefficient

Let x denote the repair costs of cars crashed at six mi/h in full-front crashes.

Let y denote the repair costs of cars crashed at six mi/h in full rear crashes.

Here, n is equal to 7.

The formula for computing the correlation coefficient (r) is as follows:

\(r = \frac{{n\sum {xy} - \sum x \sum y }}{{\sqrt {n\sum {{x^2}} - {{\left( {\sum x } \right)}^2}} \sqrt {n\sum {{y^2}} - {{\left( {\sum y } \right)}^2}} }}\)

The following calculations are done to compute the value of r:

x	y	xy	\({x^2}\)	\({y^2}\)
936	1480	1385280	876096	2190400
978	1202	1175556	956484	1444804
2252	802	1806104	5071504	643204
1032	3191	3293112	1065024	10182481
3911	1122	4388142	15295921	1258884
4312	739	3186568	18593344	546121
3469	2767	9598723	12033961	7656289
\(\sum x \)=16890	\(\sum y \)=11303	\(\sum {xy} \)=24833485	\(\sum {{x^2}} \)=53892334	\(\sum {{y^2}} \)=23922183

Substituting the above values, the following value of r is obtained:

\[\begin{aligned}{c}r = \frac{{n\sum {xy} - \sum x \sum y }}{{\sqrt {n\sum {{x^2}} - {{\left( {\sum x } \right)}^2}} \sqrt {n\sum {{y^2}} - {{\left( {\sum y } \right)}^2}} }}\\ = \frac{{7\left( {24833485} \right) - \left( {16890} \right)\left( {11303} \right)}}{{\sqrt {7\left( {53892334} \right) - {{\left( {16890} \right)}^2}} \sqrt {7\left( {23922183} \right) - {{\left( {11303} \right)}^2}} }}\\ = - 0.283\end{aligned}\]

Therefore, the value of r is equal to -0.283.

Test statistic, critical value of r and conclusion of the test

The test statistic value is computed as follows:

\(\begin{aligned}{c}t = \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\;\;\; \sim {t_{\left( {n - 2} \right)}}\\ = \frac{{ - 0.283}}{{\sqrt {\frac{{1 - {{\left( { - 0.283} \right)}^2}}}{{7 - 2}}} }}\\ = - 0.659\end{aligned}\)

Here, n=7.

The degrees of freedom are computed as follows:

\(\begin{aligned}{c}df = n - 2\\ = 7 - 2\\ = 5\end{aligned}\)

The critical correlation value of r for\(\alpha = 0.05\)and degrees of freedom equal to 5 is equal to 0.754.

The p-value obtained dusing the test statistic is equal to 0.539.

Since the absolute value of the test statistic is less than the critical value and the p-value is greater than 0.05, so the null hypothesis is rejected.

Therefore, there is no linear correlation between the repair costs of the two types of car crashes.