Question

In Exercises \(14.128-14.133\). we repeat the information from Exercises \(14.22-14.27\) Presuming that the assumptions for regression inferences are met, perform the required correlation \(t-\)tests, using either the critical- value approach or the \(P-\)value approach.

Following are the data on plant weight and quantity of volatile emissions from Exercise \(14.25\).

Do the data suggest that, for the potato plant Solanum tuberosum, weight and quantity of volatile emissions are linearly correlated?

Use \(\alpha= 0.05\).

Short answer

The null hypothesis is not rejected and it cannot be conclude that the potato plant Solanum tuberosum weight and quantity of volatile emission are linearly correlated.

Step-by-step solution

Step 1. Given information

The level of significance is \(0.05\) and the data is,

Step 2. Calculation

The hypothesis are,

\(H_{0}:\rho=0\)

\(H_{a}:\rho \neq 0\)

The table is shown below.

The value of \(r\) is,

\(r=\frac{\sum x_{i}y_{i}-\sum x_{i}\sum \frac{y_{i}}{n}}{\sqrt{\sum x_{i}^{2}-(\sum x_{i}^{2})}\sqrt{\sum y_{i}^{2}-(\sum y_{i})^{2}}}\)

\(=\frac{10486-(723)\left ( \frac{156.5}{11} \right )}{\sqrt{48747-\frac{(723)^{2}}{11}}\sqrt{2523.25-\frac{(156.5)^{2}}{11}}}\)

\(=0.33106\)

The value of test statistic is,

\(t=\frac{r}{\sqrt{\frac{1-r^{2}}{n-2}}}\)

\(=\frac{0.33106}{\sqrt{\frac{1-(0.33106)^{2}}{11-2}}}\)

\(=1.053\)

The degree of freedom is,

\(dof=n-2\)

\(=11-2\)

\(=9\)

The curve is shown below.

Since, the value do not lie in the rejection region.

Thus, the null hypothesis is not rejected and it cannot be conclude that the potato plant Solanum tuberosum weight and quantity of volatile emission are linearly correlated.