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.)

POTUS Media periodically discuss the issue of heights of winning presidential candidates and heights of their main opponents. Listed below are those heights (cm) from severalrecent presidential elections (from Data Set 15 “Presidents” in Appendix B). Is there sufficient evidence to conclude that there is a linear correlation between heights of winning presidential candidates and heights of their main opponents? Should there be such a correlation?

President

178

182

188

175

179

183

192

182

177

185

188

188

183

188

Opponent

180

180

182

173

178

182

180

180

183

177

173

188

185

175

Short Answer

Expert verified

The scatterplot is shown below:

The value of the correlation coefficient is 0.113.

The p-value is 0.700.

There is not sufficient evidence to support the existence of a linear correlation between the heights of the president and the opponent.

No, they are not expected to be correlated as height is not one of the reasons for competing in elections.

Step by step solution

01

Given information

The data is listedfor the heights of presidential candidates and their opponents.

President

Opponent

178

180

182

180

188

182

175

173

179

178

183

182

192

180

182

180

177

183

185

177

188

173

188

188

183

185

188

175

02

Sketch a scatterplot

A scatterplot describes a trend for two variables recorded in the paired form.

Steps to sketch a scatterplot:

  1. Describe theaxes for the height of presidents and opponents.
  2. Mark dots for paired observations.

The resultantscatterplotis 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 x as the president’s height and y as the opponent’s height.

The valuesare tabulatedbelow:

x

y

\({x^2}\)

\({y^2}\)

\(xy\)

178

180

31684

32400

32040

182

180

33124

32400

32760

188

182

35344

33124

34216

175

173

30625

29929

30275

179

178

32041

31684

31862

183

182

33489

33124

33306

192

180

36864

32400

34560

182

180

33124

32400

32760

177

183

31329

33489

32391

185

177

34225

31329

32745

188

173

35344

29929

32524

188

188

35344

35344

35344

183

185

33489

34225

33855

188

175

35344

30625

32900

\(\sum x = 2568\)

\(\sum y = 2516\)

\(\sum {{x^2}} = 471370\)

\(\sum {{y^2} = } \;452402\)

\(\sum {xy\; = \;} 461538\)

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{14\left( {461538} \right) - \left( {2568} \right)\left( {2516} \right)}}{{\sqrt {14\left( {471370} \right) - {{\left( {2568} \right)}^2}} \sqrt {14\left( {452402} \right) - {{\left( {2516} \right)}^2}} }}\\ &= 0.113\end{aligned}\)

Thus, the correlation coefficient is 0.113.

04

Step 4:Conduct a hypothesis test for correlation

Definethe measure\(\rho \)as the linear correlation between two variables:the height of the president and the opponent.

For testing the claim, form the hypotheses:

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

The samplesize is14(n).

The test statistic is calculated below:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.113}}{{\sqrt {\frac{{1 - {{\left( {0.113} \right)}^2}}}{{14 - 2}}} }}\\ &= 0.394\end{aligned}\)

Thus, the test statistic is 0.394.

The degree of freedom iscalculated below:

\(\begin{aligned} df &= n - 2\\ &= 14 - 2\\ &= 12\end{aligned}\)

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > 0.394} \right)\\ &= 0.700\end{aligned}\)

Thus, the p-value is 0.700.

Since thep-value is greater than 0.05, the null hypothesis fails to be rejected.

Therefore, there is not sufficient evidence to conclude the existence of alinear correlation between the president’s and the opponent’s height.

05

Discuss the expected correlation

The heights of presidents and opponents are expected to be uncorrelated. One possible reason is that the candidates who participate inelections are expected to compete for more important reasons than their height.

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