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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

Expert verified

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

01

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.

02

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.

03

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.

04

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.

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