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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

Expert verified

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

01

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

02

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.

03

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.

04

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.

05

Discuss the old ratings

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

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Most popular questions from this chapter

In Exercises 5–8, use a significance level of A = 0.05 and refer to the

accompanying displays.

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Critical Thinking: Is the pain medicine Duragesic effective in reducing pain? Listed below are measures of pain intensity before and after using the drug Duragesic (fontanels) (based on data from Janssen Pharmaceutical Products, L.P.). The data are listed in order by row, and corresponding measures are from the same subject before and after treatment. For example, the first subject had a measure of 1.2 before treatment and a measure of 0.4 after treatment. Each pair of measurements is from one subject, and the intensity of pain was measured using the standard visual analog score. A higher score corresponds to higher pain intensity.

Pain intensity before Duragestic Treatment

1.2

1.3

1.5

1.6

8

3.4

3.5

2.8

2.6

2.2

3

7.1

2.3

2.1

3.4

6.4

5

4.2

2.8

3.9

5.2

6.9

6.9

5

5.5

6

5.5

8.6

9.4

10

7.6

Pain intensity after Duragestic Treatment

0.4

1.4

1.8

2.9

6.0

1.4

0.7

3.9

0.9

1.8

0.9

9.3

8.0

6.8

2.3

0.4

0.7

1.2

4.5

2.0

1.6

2.0

2.0

6.8

6.6

4.1

4.6

2.9

5.4

4.8

4.1

Regression:Use the given data to find the equation of the regression line. Let the response (y) variable be the pain intensity after treatment. What would be the equation of the regression line for a treatment having absolutely no effect?

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Pizza and Subways r = 0.992 (x = cost of a slice of pizza, y = subway fare in New York City

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