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Testing for a Linear Correlation. In Exercises 13鈥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.)

Sports Diameters (cm), circumferences (cm), and volumes (cm3) from balls used in different sports are listed in the table below. Is there sufficient evidence to conclude that there is a linear correlation between diameters and circumferences? Does the scatterplot confirm a linear association?


Diameter

Circumference

Volume

Baseball

7.4

23.2

212.2

Basketball

23.9

75.1

7148.1

Golf

4.3

13.5

41.6

Soccer

21.8

68.5

5424.6

Tennis

7

22

179.6

Ping-Pong

4

12.6

33.5

Volleyball

20.9

65.7

4780.1

Softball

9.7

30.5

477.9

Short Answer

Expert verified

The scatterplot is drawn below:

The value ofthe correlation coefficient is approximately1.000.

The p-value is 0.000.

There is sufficient evidence to support the existence of a linear correlation between the diameter and circumference.

Yes, the scatterplot reveals a straight-line pattern, which implies a strong linear association between the variables.

Step by step solution

01

Given information

The data for diameter and circumference is given.

Diameter

Circumference

7.4

23.2

23.9

75.1

4.3

13.5

21.8

68.5

7

22

4

12.6

20.9

65.7

9.7

30.5

02

Sketch a scatterplot

A scatterplot is used for determining a relationship between a pair of variables by tracing the paired values onto a graph.

Steps to sketch a scatterplot:

  1. Mark diameter on the x-axisand circumference on they-axis.
  2. Draw dots for each paired observation.

The resultantscatterplotis shown below.

03

Compute the measure of the correlation coefficient

The formula for correlation coefficient 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 diameter and y as the circumference.

The valuesare tabulatedbelow:

x

y

\({x^2}\)

\({y^2}\)

\(xy\)

7.4

23.2

54.76

538.24

171.68

23.9

75.1

571.21

5640.01

1794.89

4.3

13.5

18.49

182.25

58.05

21.8

68.5

475.24

4692.25

1493.3

7

22

49

484

154

4

12.6

16

158.76

50.4

20.9

65.7

436.81

4316.49

1373.13

9.7

30.5

94.09

930.25

295.85

\(\sum x = 99\)

\(\sum y = 311.1\)

\(\sum {{x^2}} = 1715.6\)

\(\sum {{y^2} = } \;16942.25\)

\(\sum {xy\; = \;} 5391.3\)

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{8\left( {5391.3} \right) - \left( {99} \right)\left( {311.1} \right)}}{{\sqrt {8\left( {1715.6} \right) - {{\left( {99} \right)}^2}} \sqrt {8\left( {16942.25} \right) - {{\left( {311.1} \right)}^2}} }}\\ &= 0.999999\\ &\approx 1.000\end{aligned}\)

Thus, the correlation coefficient is approximately 1.000.

04

Step 4:Conduct a hypothesis test for correlation

Define\(\rho \)asbe the measure of the linear correlation coefficient for diameter and circumference.

For testing the claim, form the hypotheses:

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

The samplesize is8(n).

The test statistic is calculated below:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.999999}}{{\sqrt {\frac{{1 - {{\left( {0.999999} \right)}^2}}}{{8 - 2}}} }}\\ &= 1732.05\end{aligned}\)

Thus, the test statistic is1732.05.

The degree of freedom iscalculated below:

\(\begin{aligned} df &= n - 2\\ &= 8 - 2\\ &= 6\end{aligned}\)

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > 21732.05} \right)\\ &= 0.000\end{aligned}\)

Thus, the p-value is 0.000.

Since the p-value is lesser than 0.05, the null hypothesis is rejected.

Therefore, there is sufficient evidence to support the existence of a linear correlation between diameter and circumference.

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