91影视

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.