91影视

The paired data for the weights of paper and glass are recorded with the output of the correlation matrix. The correlation value between the two variables is 0.1174.

And the number of households (n) taken is62.

Let\(\rho \)be the true correlation coefficient measure between the weights of paper and glass.

To test the claim about the linear association, 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 samples number of households is62(n).

The test statistic is computed as follows:

\(\begin{aligned} t &= \frac{r}{{\sqrt {\frac{{1 - {r^2}}}{{n - 2}}} }}\\ &= \frac{{0.1174}}{{\sqrt {\frac{{1 - {{0.1174}^2}}}{{62 - 2}}} }}\\ &= 0.9157\end{aligned}\)

Thus, the test statistic is 0.9157.

The degree of freedom is computedbelow:

\(\begin{aligned} df &= n - 2\\ &= 62 - 2\\ &= 60\end{aligned}\)

The p-value is computed using thet-distribution table:

\(\begin{aligned} p{\rm{ - value}} &= 2P\left( {T > t} \right)\\ &= 2P\left( {T > 0.9157} \right)\\ &= 2\left( {1 - P\left( {T < 0.9157} \right)} \right)\\ &= 0.3635\end{aligned}\)

Thus, the p-value is 0.3635.

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

Thus, it can be concluded at the 0.05 level of significance,there is not enough evidence to support the existence ofa linear correlation between the weights of paper and glass, which were discarded.