Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.)

Oscars Listed below are ages of Oscar winners matched by the years in which the awards were won (from Data Set 14 “Oscar Winner Age” in Appendix B). Is there sufficient evidence to conclude that there is a linear correlation between the ages of Best Actresses and Best Actors? Should we expect that there would be a correlation?

Actress

28

30

29

61

32

33

45

29

62

22

44

54

Actor

43

37

38

45

50

48

60

50

39

55

44

33

Short Answer

Expert verified

The scatter plot is shown below:

The value of the correlation coefficient is –0.288.

The p-value is 0.364.

There is not enough evidence to support the claim that there existsa linear correlation between the ages of actresses and actors.

No correlation is expected between the ages of best actressesand actors in Oscars.

Step by step solution

01

Given information

The data is recorded for the ages of best actresses and actorsin Oscars.

Best Actress

Best Actor

28

43

30

37

29

38

61

45

32

50

33

48

45

60

29

50

62

39

22

55

44

44

54

33

02

Sketch a scatterplot

A scatterplot is used to reveal the pattern of association between two variables.

Steps to sketch a scatterplot:

  1. Define two axesfor the ages of actors and actresseseach year.
  2. Mark each paired set of values on the graph.

The resultant scatterplot is shown below.

03

Compute the measure of the correlation coefficient

The correlation coefficient formula is

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

For the variable of age for best actress (x) and age of best actor (y), compute the correlation coefficient.

The valuesare listedin the table below:

x

y

\({x^2}\)

\({y^2}\)

\(xy\)

28

43

784

1849

1204

30

37

900

1369

1110

29

38

841

1444

1102

61

45

3721

2025

2745

32

50

1024

2500

1600

33

48

1089

2304

1584

45

60

2025

3600

2700

29

50

841

2500

1450

62

39

3844

1521

2418

22

55

484

3025

1210

44

44

1936

1936

1936

54

33

2916

1089

1782

\(\sum x = 469\)

\(\sum y = 542\)

\(\sum {{x^2}} = 20405\)

\(\sum {{y^2} = } \;25162\)

\(\sum {xy\; = \;} 20841\)

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{12\left( {20841} \right) - \left( {469} \right)\left( {542} \right)}}{{\sqrt {12\left( {20405} \right) - {{\left( {469} \right)}^2}} \sqrt {12\left( {25162} \right) - {{\left( {542} \right)}^2}} }}\\ &= - 0.288\end{aligned}\)

Thus, the correlation coefficient is –0.288.

04

Step 4:Conduct a hypothesis test for correlation

Definethe actual measure of the correlation coefficient between the age of best actress and actor in Oscars 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 12 (n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{ - 0.288}}{{\sqrt {\frac{{1 - {{\left( { - 0.288} \right)}^2}}}{{12 - 2}}} }}\\ &= - 0.951\end{aligned}\)

Thus, the test statistic is–0.951.

The degree of freedom is

\(\begin{aligned} df &= n - 2\\ &= 12 - 2\\ &= 10.\end{aligned}\)

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T < - 0.951} \right)\\ &= 0.364\end{aligned}\)

Thus, the p-value is 0.364.

Since the p-value is greater than 0.05, the null hypothes is fails to berejected.

Therefore, there is not enough evidence to conclude that there is no linear correlation between the age of actors and actresses.

05

Discuss if there exists a correlation between the ages of actors and actresses

It was not expected that actors and actresses have acorrelation between their ages.They can beparts of different movies.There is no clear logic for any association between their ages.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The following exercises are based on the following sample data consisting of numbers of enrolled students (in thousands) and numbers of burglaries for randomly selected large colleges in a recent year (based on data from the New York Times).

If you had computed the value of the linear correlation coefficient to be 1.500, what should you conclude?

Stocks and Sunspots. Listed below are annual high values of the Dow Jones Industrial Average (DJIA) and annual mean sunspot numbers for eight recent years. Use the data for Exercises 1–5. A sunspot number is a measure of sunspots or groups of sunspots on the surface of the sun. The DJIA is a commonly used index that is a weighted mean calculated from different stock values.

DJIA

14,198

13,338

10,606

11,625

12,929

13,589

16,577

18,054

Sunspot

Number

7.5

2.9

3.1

16.5

55.7

57.6

64.7

79.3

Confidence Interval Construct a 95% confidence interval estimate of the mean sunspot number. Write a brief statement interpreting the confidence interval.

Cigarette Nicotine and Carbon Monoxide Refer to the table of data given in Exercise 1 and use the amounts of nicotine and carbon monoxide (CO).

a. Construct a scatterplot using nicotine for the xscale, or horizontal axis. What does the scatterplot suggest about a linear correlation between amounts of nicotine and carbon monoxide?

b. Find the value of the linear correlation coefficient and determine whether there is sufficient evidence to support a claim of a linear correlation between amounts of nicotine and carbon monoxide.

c. Letting yrepresent the amount of carbon monoxide and letting xrepresent the amount of nicotine, find the regression equation.

d. The Raleigh brand king size cigarette is not included in the table, and it has 1.3 mg of nicotine. What is the best predicted amount of carbon monoxide?

Tar

25

27

20

24

20

20

21

24

CO

18

16

16

16

16

16

14

17

Nicotine

1.5

1.7

1.1

1.6

1.1

1.0

1.2

1.4

The following exercises are based on the following sample data consisting of numbers of enrolled students (in thousands) and numbers of burglaries for randomly selected large colleges in a recent year (based on data from the New York Times).

Repeat the preceding exercise, assuming that the linear correlation coefficient is r= 0.997.

Finding a Prediction Interval. In Exercises 13–16, use the paired data consisting of registered Florida boats (tens of thousands) and manatee fatalities from boat encounters listed in Data Set 10 “Manatee Deaths” in Appendix B. Let x represent number of registered boats and let y represent the corresponding number of manatee deaths. Use the given number of registered boats and the given confidence level to construct a prediction interval estimate of manatee deaths.

Boats Use x = 85 (for 850,000 registered boats) with a 99% confidence level.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free