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.)

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

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

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

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.

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.

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.