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.)

Revised mpg Ratings Listed below are combined city-highway fuel economy ratings (in mi>gal) for different cars. The old ratings are based on tests used before 2008 and the new ratings are based on tests that went into effect in 2008. Is there sufficient evidence to conclude that there is a linear correlation between the old ratings and the new ratings? What do the data suggest about the old ratings?

Old	16	27	17	33	28	24	18	22	20	29	21
New	15	24	15	29	25	22	16	20	18	26	19

Short answer

The scatter plot is shown below:

The value of the correlation coefficient is 0.998.

The p-value is 0.000.

There is enough evidence to support the claim that there existsa linear correlation between old and new ratings.

The old ratings were higher than the new ratings for each car.

Step-by-step solution

Given information

The data is recorded for the two variables, old and new ratings.

Old	New
16	15
27	24
17	15
33	29
28	25
24	22
18	16
22	20
20	18
29	26
21	19

Sketch a scatterplot

A plot described using a paired set of observations is known as a scatterplot.

It givesa tentative relationship between two variables.

Steps to sketch a scatterplot:

1. Mark the horizontal for old ratings and the vertical for new ratings.
2. Mark each point in a pair onto the curve.

The resultant scatterplot is shown below.

Compute the measure of the correlation coefficient

The correlation coefficient formula is

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

Define variable xas old ratings and variable y as new ratings.

The valuesare listedin the table below:

x	y	\({x^2}\)	\({y^2}\)	\(xy\)
16	15	256	225	240
27	24	729	576	648
17	15	289	225	255
33	29	1089	841	957
28	25	784	625	700
24	22	576	484	528
18	16	324	256	288
22	20	484	400	440
20	18	400	324	360
29	26	841	676	754
21	19	441	361	399
\(\sum x = 255\)	\(\sum y = 229\)	\(\sum {{x^2}} = 6213\)	\(\sum {{y^2} = } \;4993\)	\(\sum {xy\; = \;} 5569\)

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{11\left( {5569} \right) - \left( {225} \right)\left( {229} \right)}}{{\sqrt {11\left( {6213} \right) - {{\left( {255} \right)}^2}} \sqrt {11\left( {4993} \right) - {{\left( {229} \right)}^2}} }}\\ &= 0.998\end{aligned}\)

Thus, the correlation coefficient is 0.998.

Step 4:Conduct a hypothesis test for correlation

Definethe actual measure of the correlation coefficient as\(\rho \).

For testing the claim, form the hypotheses:

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

The samplesize is 11 (n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.998}}{{\sqrt {\frac{{1 - {{\left( {0.998} \right)}^2}}}{{11 - 2}}} }}\\ &= 47.363\end{aligned}\)

Thus, the test statistic is 47.363.

The degree of freedom is

\(\begin{aligned} df &= n - 2\\ &= 11 - 2\\ &= 9.\end{aligned}\)

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > 47.363} \right)\\ &= 0.000\end{aligned}\)

Thus, the p-value is 0.000.

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

Therefore, there is enough evidence to conclude that old ratings are linearly correlated with new ratings.

Discuss the old ratings

The data suggests that throughout the period, the old ratings were higher than the new rating for each car.