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

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.

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:

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.

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.