Question

Testing for a Linear Correlation. In Exercises 13–28, construct a scatterplot, and find the value of the linear correlation coefficient r. Also find the P-value or the critical values of r from Table A-6. Use a significance level of A = 0.05. Determine whether there is sufficient evidence to support a claim of a linear correlation between the two variables. (Save your work because the same data sets will be used in Section 10-2 exercises.)

Lemons and Car Crashes Listed below are annual data for various years. The data are weights (metric tons) of lemons imported from Mexico and U.S. car crash fatality rates per 100,000 population (based on data from “The Trouble with QSAR (or How I Learned to Stop Worrying and Embrace Fallacy),” by Stephen Johnson, Journal of Chemical Information and Modeling, Vol. 48, No. 1). Is there sufficient evidence to conclude that there is a linear correlation between weights of lemon imports from Mexico and U.S. car fatality rates? Do the results suggest that imported lemons cause car fatalities?

Lemon Imports	230	265	358	480	530
Crash Fatality Rate	15.9	15.7	15.4	15.3	14.9

Short answer

The scatter plot is shown below:

The value of the correlation coefficient is -0.959.

The p-value is 0.0010.

There is enough evidence to support the claim for a linear correlation between lemon imports and crash fatality rate.

Since correlation does not implycausation, lemon imports do not cause crash fatality.

Step-by-step solution

Given information

The data is recorded for the two variables, weights of lemon imports and crash fatality rate.

Lemon Imports	230	265	358	480	530
Crash Fatality Rate	15.9	15.7	15.4	15.3	14.9

Sketch a scatterplot

A scatterplot represents paired observations in the form of dots, marked on the x and y axes.

Steps to sketch a scatterplot:

1. Mark the horizontal axis for the weights of imports and the vertical for the crash fatality rate.
2. As per the data, mark the dots corresponding to the two axes.

The resultant scatterplot is shown below.

Compute the measure of the correlation coefficient

The correlation coefficient formulais

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

Let variable xbe the weights of lemon imports,and variable ybe the crash fatality rate.

The valuesare tabulated in the table below:

x	y	\({x^2}\)	\({y^2}\)	\(xy\)
230	15.9	52900	252.81	3657
265	15.7	70225	246.49	4160.5
358	15.4	128164	237.16	5513.2
480	15.3	230400	234.09	7344
530	14.9	280900	222.01	7897
\(\sum x = 1863\)	\(\sum y = 77.2\)	\(\sum {{x^2}} = 762589\)	\(\sum {{y^2} = } \;1192.56\)	\(\sum {xy\; = \;} 28571.7\)

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{5\left( {28571.7} \right) - \left( {1863} \right)\left( {77.2} \right)}}{{\sqrt {5\left( {762589} \right) - {{\left( {1863} \right)}^2}} \sqrt {5\left( {1192.56} \right) - {{\left( {77.2} \right)}^2}} }}\\ &= - 0.959\end{aligned}\)

Thus, the correlation coefficient is -0.959.

Step 4:Conduct a hypothesis test for correlation

Define\(\rho \)to be the actual value of the correlation coefficient for the two variables.

For testing the claim, form the hypotheses:

\(\begin{array}{l}{H_o}:\rho = 0\\{H_a}:\rho \ne 0\end{array}\)

The samplesize is 5 (n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{ - 0.959}}{{\sqrt {\frac{{1 - {{\left( { - 0.959} \right)}^2}}}{{5 - 2}}} }}\\ &= - 5.86\end{aligned}\)

Thus, the test statistic is -5.86.

The degree of freedom is

\(\begin{aligned} df &= n - 2\\ &= 5 - 2\\ &= 3.\end{aligned}\)

The p-value is computed from the t-distribution table.

\(\begin{aligned} p - value &= 2P\left( {T < - 5.86} \right)\\ &= 0.0010\end{aligned}\)

Thus, the p-value is 0.0010.

Since thep-value is less than 0.05, the null hypothesis isrejected.

Therefore, there is enough evidence to conclude that the weights of lemon import are linearly correlated to the fatality rate.

Discuss the cause of fatalities

If a correlation exists between two variables, it does not imply one causes the other to change.It is because correlation does not imply causation.

There are chances that a lurking variable determines the association between the variables. Thus, it can be stated that the weight of imported lemons does not cause car fatalities.