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.)

Weighing Seals with a Camera Listed below are the overhead widths (cm) of seals

measured from photographs and the weights (kg) of the seals (based on “Mass Estimation of Weddell Seals Using Techniques of Photogrammetry,” by R. Garrott of Montana State University). The purpose of the study was to determine if weights of seals could be determined from overhead photographs. Is there sufficient evidence to conclude that there is a linear correlation between overhead widths of seals from photographs and the weights of the seals?

Overhead Width	7.2	7.4	9.8	9.4	8.8	8.4
Weight	116	154	245	202	200	191

Short answer

The scatterplot is shown below:

The value of the correlation coefficient is 0.948.

The p-value is 0.004.

There is enough evidence to support the claim that there is linear correlation between overhead width and weight.

Step-by-step solution

Given information

The data for overhead width and weights are recorded as shown below:

Overhead Width	Weight
7.2	116
7.4	154
9.8	245
9.4	202
8.8	200
8.4	191

Sketch a scatterplot

A graph thatdenotes a paired set of observations in a plotcan be used to analyze the trend between two variables.

Steps to sketch a scatterplot:

1. Formthe x and y axes for overhead width and weight, respectively.
2. Mark the points as coordinates on the graph.

The graph formed is shown below.

Compute the measure of the correlation coefficient

The correlation coefficient formula is

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

Let x be the overhead width and y be the weight.

The valuesare listed in the table below:

x	y	\({x^2}\)	\({y^2}\)	\(xy\)
7.2	116	51.84	13456	835.2
7.4	154	54.76	23716	1139.6
9.8	245	96.04	60025	2401
9.4	202	88.36	40804	1898.8
8.8	200	77.44	40000	1760
8.4	191	70.56	36481	1604.4
\(\sum x = 51\)	\(\sum y = 1108\)	\(\sum {{x^2}} = 439\)	\(\sum {{y^2} = } \;214482\)	\(\sum {xy\; = \;} 9639\)

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{6\left( {9639} \right) - \left( {51} \right)\left( {1108} \right)}}{{\sqrt {6\left( {439} \right) - {{\left( {51} \right)}^2}} \sqrt {6\left( {214482} \right) - {{\left( {1108} \right)}^2}} }}\\ &= 0.948\end{aligned}\)

Thus, the correlation coefficient is 0.948.

Step 4:Conduct a hypothesis test for correlation

Definethe true measure of the correlation coefficientas\(\rho \).

For testing the claim, form the hypotheses.

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

The samplesize is6(n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.948}}{{\sqrt {\frac{{1 - {{\left( {0.948} \right)}^2}}}{{6 - 2}}} }}\\ &= 5.957\end{aligned}\)

Thus, the test statistic is 5.957.

The degree of freedom is

\(\begin{aligned} df &= n - 2\\ &= 6 - 2\\ &= 4.\end{aligned}\)

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > 5.957} \right)\\ &= 0.0039\\ &\approx 0.004\end{aligned}\)

Thus, the p-value is 0.004.

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

Therefore, there is enough evidence to conclude the existence of a linear correlation between the two variables.