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.)

Internet and Nobel Laureates Listed below are numbers of Internet users per 100 people and numbers of Nobel Laureates per 10 million people (from Data Set 16 “Nobel Laureates and Chocolate” in Appendix B) for different countries. Is there sufficient evidence to conclude that there is a linear correlation between Internet users and Nobel Laureates?

Internet Users	Nobel Laureates
79.5	5.5
79.6	9
56.8	3.3
67.6	1.7
77.9	10.8
38.3	0.1

The data is stated below:

A scatterplot visualizes paired data points for two data points corresponding to x and y axes.

Steps to sketch a scatterplot:

1. Sketch the x and y axes for the two variables.
2. Map each pair of values corresponding to the axes.
3. A scatter plot for the paired data is obtained.

The correlation coefficient is computedbelow:

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

The valuesare given in the table below:

Substitute the values in the formula:

\(\begin{aligned} r &= \frac{{6\left( {2301.16} \right) - \left( {399.7} \right)\left( {30.1} \right)}}{{\sqrt {6\left( {27987.71} \right) - {{\left( {399.7} \right)}^2}} \sqrt {6{{\left( {241.68} \right)}^2} - {{\left( {30.1} \right)}^2}} }}\\ &= 0.799\end{aligned}\)

Thus, the correlation coefficient is 0.799.

Let\(\rho \)be the true correlation coefficient.

For testing the claim, form the hypotheses as shown:

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

The samplesize is 6(n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.799}}{{\sqrt {\frac{{1 - {{0.799}^2}}}{{6 - 2}}} }}\\ &= 2.657\end{aligned}\)

Thus, the test statistic is 2.657.

The degree of freedom is computedbelow:

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

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

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > t} \right)\\ &= 2P\left( {T > 2.657} \right)\\ &= 2\left( {1 - P\left( {T < 2.657} \right)} \right)\\ &= 0.056\end{aligned}\)

Thus, the p-value is 0.056.

Since thep-value is greater than 0.05, the null hypothesis fails to be rejected.

Therefore, there is not enough evidence to conclude that variablesx and y have a linear correlation.