Question

In this Exercise 14.52, we repeat the information from Exercise 14.16.

a. Decide, at the $10\%$significance level, whether the data provide sufficient evidence to conclude that $x$is useful for predicting $y$:

b. Find a $90\%$confidence interval for the slope of the population regression line.

$\begin{array}{|ccccc|}\hline x& 1& 3& 4& 4\\ y& 13& -1& 3& 5\\ \hline\end{array}$ $\hat{y}=14-3x$

Short answer

(a) As a result, the data is insufficient to establish that $x$is beneficial for forecasting $y$.

(b) The $90\%$confidence interval for the slope of the regression line is $(-8.960,2.960)$.

Step-by-step solution

Part(a) Step 1: Given Information

$\begin{array}{|ccccc|}\hline x& 1& 3& 4& 4\\ y& 13& -1& 3& 5\\ \hline\end{array}$

$\hat{y}=14-3x$

Part(a) Step 2: Explanation

Decide the Null and Alternate hypothesis

$H_0:{\beta }_1=0$($x$is not useful for predicting $y$).

$H_{\alpha }:{\beta }_1\ne 0$($x$is useful for predicting $y$).

If $p$-value $\le \alpha (=0.10)$, then reject the null hypothesis $\left(H_0\right)$.

Using Excel, find the test statistic and $p$-value.

Part(a) Step 3: Procedure

Step 1: Select Data >Data Analysis > Regression from the drop-down menu.

Step 2: In the Input Y Range box, type $y$.

Step 3:In the Input X Range box, type $x$.

Step 4: Select Confidence Level$90$from the drop-down menu.

Step 5: Click the OK button.

OUTPUT

The test statistic's value is $-1.470$, and the $p$-value is $0.2794$, according to the Excel output.

Part(a) Step 4: Conclusion

Use the $\alpha =0.10$significance level.

The p-value is higher than the level of significance in this case.

To put it another way, $p$-value $(=0.2794)>\alpha (=0.10)$.

As a result of the rejection rule, it may be argued that at $\alpha =0.10$, there is no evidence to reject the null hypothesis ( $H_0$).

Part(b) Step 1: Given Information

$\begin{array}{|ccccc|}\hline x& 1& 3& 4& 4\\ y& 13& -1& 3& 5\\ \hline\end{array}$

$\hat{y}=14-3x$

Part(b) Step 2: Explanation

Using Excel, calculate the $90\%$confidence interval for the slope of the regression line.

From the Excel output in part (a) We get as $(-8.960,2.960)$