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.)

Tips Listed below are amounts of bills for dinner and the amounts of the tips that were left. The data were collected by students of the author. Is there sufficient evidence to conclude that there is a linear correlation between the bill amounts and the tip amounts? If everyone were to tip with the same percentage, what should be the value of r?

Bill(dollars)	33.46	50.68	87.92	98.84	63.6	107.34
Tip(dollars)	5.5	5	8.08	17	12	16

The data is listedfor the amounts of bills and the tips given.

A scatterplot is a two-dimensional graph with dots marked for each variable in the paired form.

Steps to sketch a scatterplot:

1. Describe theaxes for bill and tip.
2. Mark the paired observations on the graph.

The scatterplotis 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 variable x be bill amounts, and y be the amount of tips.

The valuesare listed in the table below:

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{6\left( {5308.744} \right) - \left( {441.84} \right)\left( {63.58} \right)}}{{\sqrt {6\left( {36754.14} \right) - {{\left( {441.84} \right)}^2}} \sqrt {6\left( {809.5364} \right) - {{\left( {63.58} \right)}^2}} }}\\ &= 0.828\end{aligned}\)

Thus, the correlation coefficient is 0.828.

Definethe measure\(\rho \)as the true correlation measure for the two variables.

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:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.828}}{{\sqrt {\frac{{1 - {{\left( {0.828} \right)}^2}}}{{6 - 2}}} }}\\ &= 2.953\end{aligned}\)

Thus, the test statistic is 2.953.

The degree of freedom is calculated below:

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

The p-value is computed from the t-distribution table.

\(\begin{aligned} p - value &= 2P\left( {T > 2.953} \right)\\ &= 0.042\end{aligned}\)

Thus, the p-value is 0.042.

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

Therefore, there is sufficient evidence to support alinear correlation between theamount of bill and the tip left.

Assume each person offers a tip with the same percentage (say c percent) of the bill amount.

Then,

\({\rm{Tips}}\left( y \right) = {\rm{c}}\% \;{\rm{of}}\;{\rm{bill}}\left( x \right) \Rightarrow y = \frac{c}{{100}} \times x\).

Therefore, the equation that best explains the relationship is of a straight line with slope\(\frac{c}{{100}}\). The general equation of the line is\(y = ax + b\)for slope a and intercept b.

Hence, all observations willlie on the line.

The value of rwill be one as all observations are collinear. It means, a perfect linear relationship is established between xand y.