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 Repeat the preceding exercise using diameters and volumes.

Refer to Exercise 27 for the data on diameter and volume.

A scatterplot is described for a paired set of data obtained from two variables.

Steps to sketch a scatterplot:

1. Drawx-axisas the diameter and y-axis as the volume data.
2. Mark allthe paired values.

The resultant scatterplot is given below.

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 diameter and yas the volume.

The valuesare tabulatedbelow:

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{8\left( {396776} \right) - \left( {99} \right)\left( {18297.6} \right)}}{{\sqrt {8\left( {1715.6} \right) - {{\left( {99} \right)}^2}} \sqrt {8\left( {103679501} \right) - {{\left( {18297.6} \right)}^2}} }}\\ &= 0.978\end{aligned}\)

Thus, the correlation coefficient is 0.978.

Define\(\rho \)asthe correlation measure for diameter and volumes.

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:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.978}}{{\sqrt {\frac{{1 - {{\left( {0.978} \right)}^2}}}{{8 - 2}}} }}\\ &= 11.533\end{aligned}\)

Thus, the test statistic is11.533.

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.

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > 11.533} \right)\\ &= 0.000\end{aligned}\)

Thus, the p-value is 0.000.

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

Therefore, there issufficient evidence to support the existence of alinear correlation between diameter and volume.

The scatterplot does not support the linear associationestablished in the results above. On the contrary, it suggests anon-linear pattern between the two variables.