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.)

CSI Statistics Police sometimes measure shoe prints at crime scenes so that they can learn something about criminals. Listed below are shoe print lengths, foot lengths, and heights of males (from Data Set 2 “Foot and Height” in Appendix B). Is there sufficient evidence to conclude that there is a linear correlation between shoe print lengths and heights of males? Based on these results, does it appear that police can use a shoe print length to estimate the height of a male?

Shoe print(cm)	29.7	29.7	31.4	31.8	27.6
Foot length(cm)	25.7	25.4	27.9	26.7	25.1
Height (cm)	175.3	177.8	185.4	175.3	172.7

The data for shoe print lengths, foot lengths, and heights of males is recorded.

A two-dimensional plot based on a paired data plotted with reference to two axes is called a scatterplot.

Steps to sketch a scatterplot:

1. Mark horizontal axis for shoe print lengthand vertical axis for the height of males.
2. Mark points for the paired observations corresponding to both axes.

The resultant graph is the scatterplot.

The formula for the correlation coefficient is

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

Let shoe print be defined by variable x and the height of males be defined by variable y.

The valuesare listed in the table below:

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{5\left( {26649.69} \right) - \left( {150.2} \right)\left( {886.5} \right)}}{{\sqrt {5\left( {157271.5} \right) - {{\left( {150.2} \right)}^2}} \sqrt {5\left( {4523.14} \right) - {{\left( {886.5} \right)}^2}} }}\\ &= 0.591\end{aligned}\)

Thus, the correlation coefficient is 0.591.

Define\(\rho \)as the actual value of the correlation coefficient for shoe print length and the height of males.

For testing the claim, form the hypotheses:

\(\begin{array}{l}{{\rm{H}}_{\rm{o}}}:\rho = 0\\{{\rm{{\rm H}}}_{\rm{a}}}:\rho \ne 0\end{array}\)

The samplesize is 5 (n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ = \frac{{0.591}}{{\sqrt {\frac{{1 - {{0.591}^2}}}{{5 - 2}}} }}\\ &= 1.27\end{aligned}\)

Thus, the test statistic is 1.270.

The degree of freedom is

\(\begin{aligned} df &= n - 2\\ &= 5 - 2\\ &= 3.\end{aligned}\)

Thep-value is computed from the t-distribution table.

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > t} \right)\\ &= 2P\left( {T > 1.27} \right)\\ &= 2\left( {1 - P\left( {T < 1.27} \right)} \right)\\ &= 0.294\end{aligned}\)

Thus, the p-value is 0.294.

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

Therefore, there is not enough evidence to conclude that the variables shoe print length and height have a linear correlation between them.

Since the two variables are not linearly correlated to one other, one variable cannot be used to estimate the other.

Also, the scatterplot reveals no specific pattern between shoe print length and height of men (linear or non-linear).Thus, the two variables are not associated with one other.