Question

In Exercises \(14.118-14.127\). we repeat the data from Exercises \(14.12-14.21\) and specify an alternative hypothesis for a correlation \(t-\)test. For each exercise, decide at the \(10%\) significance level. whether the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.

\(H_{a}:p<0\)

Short answer

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

Thus, the null hypothesis is not rejected.

Step-by-step solution

Step 1. Given information

The level of significance is \(0.1\) and the data is,

Step 2. Calculation

The hypothesis are,

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

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

The table is shown below.

The value of \(r\) is,

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

\(=\frac{-22-(6)\left ( \frac{-9}{3} \right )}{\sqrt{14-\frac{(6)^{2}}{3}}\sqrt{41-\frac{(-9)^{2}}{3}}}\)

\(=-0.7559\)

The value of test statistic is,

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

\(=\frac{-0.7559}{\sqrt{\frac{1-(-0.7559)^{2}}{3-2}}}\)

\(=-1.15\)

The degree of freedom is,

\(dof=n-2\)

\(=3-2\)

\(=1\)

The curve is shown below.

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

Thus, the null hypothesis is not rejected.