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 paired data for two variables arerecorded.

a.

A scatterplot is a graph that represents observations for a paired set of data.

Steps to sketch a scatterplot:

1. Define thex and yaxes for each of the two variables. The horizontal axis is thex-axis, and the vertical axis is the y-axis.
2. Map each paired value corresponding to the axes.
3. Thus, a scatter plot for the paired data is obtained.

b.

The correlation coefficient is computed 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 listedin the table below:

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{11\left( {797.59} \right) - \left( {99} \right)\left( {82.51} \right)}}{{\sqrt {11\left( {1001} \right) - {{\left( {99} \right)}^2}} \sqrt {11{{\left( {660.1763} \right)}^2} - {{\left( {82.51} \right)}^2}} }}\\ &= 0.8162\end{aligned}\)

Thus, the correlation coefficient is 0.8162.

Let\(\rho \)be the true correlation coefficient measure for the paired variables.

For testing the claim, form the hypotheses as shown below:

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

The samples size is11(n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.8162}}{{\sqrt {\frac{{1 - {{0.8162}^2}}}{{11 - 2}}} }}\\ &= 4.238\end{aligned}\)

Thus, the test statistic is 4.238.

The degree of freedom is computed below:

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

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > t} \right)\\ &= 2P\left( {T > 4.238} \right)\\ &= 2\left( {1 - P\left( {T < 4.238} \right)} \right)\\ &= 0.002\end{aligned}\)

Thus, the p-value is 0.002.

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

Therefore, there is sufficient evidence to conclude that variables x and y have a linear correlation between them.

c.

The scatterplot reveals that the data follows a strong non-linear pattern. It means that the observations do not align on a straight line.

The characteristic of the data would be missed in part (b) if the scatterplot was not sketched.