Explore! Exercises 9 and 10 provide two data sets from “Graphs in Statistical Analysis,” by F. J. Anscombe, the American Statistician, Vol. 27. For each exercise,

a. Construct a scatterplot.

b. Find the value of the linear correlation coefficient r, then determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables.

c. Identify the feature of the data that would be missed if part (b) was completed without constructing the scatterplot.

x	10	8	13	9	11	14	6	4	12	7	5
y	9.14	8.14	8.74	8.77	9.26	8.10	6.13	3.10	9.13	7.26	4.74

The samples size is11(n).

The data for two variables is shown.

a.

When the data is visualized on a graph in paired form, it is referred to as a scatterplot.Here, one axis represents the values of x, and the other axis represents the values of y.

Steps to sketch a scatterplot:

1. Make two axes, x and y, for each of the two variables.
2. Map each paired value corresponding to the scale of the axes.
3. Thus, a scatter plot for the paired data is obtained.

b.

The formula for the correlation coefficient is given below:

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

The valuesare listed in the following table:

Substitute the values to obtain the value of r.

\(\begin{aligned} r &= \frac{{11\left( {797.47} \right) - \left( {99} \right)\left( {82.50} \right)}}{{\sqrt {11\left( {1001} \right) - {{\left( {99} \right)}^2}} \sqrt {11{{\left( {659.9762} \right)}^2} - {{\left( {82.5} \right)}^2}} }}\\ &= 0.8163\end{aligned}\)

Thus, the correlation coefficient is 0.8163.

The statistical hypotheses are formulated below:

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

Here,\(\rho \)isthe actual measure ofthe correlation coefficientfor the variables.

Calculate the test statistic as shown below:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.8163}}{{\sqrt {\frac{{1 - {{0.8163}^2}}}{{11 - 2}}} }}\\ &= 4.239\end{aligned}\)

Thus, the value of the test statistic is 4.239.

The degree of freedom is computedbelow:

\(\begin{aligned} df &= n - 2\\ &= 11 - 2\\ &= 9\end{aligned}\)

The p-value is computedfrom the t-distribution table.

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > 4.239} \right)\\ &= 2\left( {1 - P\left( {T < 4.239} \right)} \right)\\ &= 0.002\end{aligned}\)

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

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

c.

The scatterplot for the data reveals that one point lies at an extreme,beyond the straight-line pattern established by other points in the data. Resultant ofthis, the correlation measure, which was expected to be close to 1, is lower.