In this Exercise 14.58, we repeat the information from Exercises 14.22. Presuming that the assumptions for regression inferences are met, decide at the specified significance level whether the data provide sufficient evidence to conclude that the predictor variable is useful for predicting the response variable.

Following are the data on the percentage of investments in energy securities and tax efficiency from Exercise 14.22. Use $\alpha =0.05$.

$\alpha =0.05$

Decide the Null and Alternate hypothesis

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

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

We need to determine the significance level $\alpha$

The hypothesis test should be run at a significance threshold of $5\%$, or $\alpha =0.05$.

Computation table

$$\begin{gathered} S_{xx}=\sum x_i^2-{\left(\sum x_i\right)}^2/n \\ =365.05-(55.9{)}^2/10 \\ =365.05-3124.81/10 \\ =365.05-312.481 \\ =52.569 \end{gathered}$$

The entire amount of square SST is calculated as follows:

$$\begin{gathered} S_{yy}=\sum y_i^2-{\left(\sum y_i\right)}^2/n \\ =70838.49-(832.5{)}^2/10 \\ =70838.49-693056.25/10 \\ =70838.49-69305.625 \\ =1532.865 \end{gathered}$$

SSR regression sum of squares is calculated as follows:

$$\begin{gathered} SSR=\frac{S_{xy}^2}{S_{xx}} \\ =\frac{(-276.725{)}^2}{52.569} \\ =\frac{76576.72562}{52.569} \\ =1456.689791 \end{gathered}$$

$$\begin{gathered} SSE=SST-SSR \\ =1532.865-1456.689791 \\ =76.17520896 \end{gathered}$$

The slope of the regression line is calculated using the formula,

$$\begin{gathered} b_1=\frac{S_{xy}}{S_{xx}} \\ =\frac{-276.725}{52.569} \\ =-5.264033936 \end{gathered}$$

The standard error of the estimate is calculated using the formula:

$$\begin{gathered} s_e=\sqrt{\frac{SSE}{n-2}} \\ =\sqrt{\frac{76.17520896}{10-2}} \\ =3.085757787 \\ \approx 3.0857 \end{gathered}$$

We need to find the value of the test statistic

$$\begin{gathered} t=\frac{b_1}{s_e/\sqrt{S_{xx}}} \\ =\frac{-5.264033936}{3.085757787/\sqrt{52.569}} \\ =\frac{-5.264033936}{0.42559549} \\ =-12.36863304 \\ \approx -12.36 \end{gathered}$$

The test statistic's value is $t=-12.36$, as shown above. The $p$-value represents the likelihood of seeing a value of $t$of $-12.36$or larger in magnitude if the null hypothesis is true, because the test is two-tailed. The $P=0.00000170184$text is obtained using technology.

Since the $p$-value$=0.00000170184>\alpha =0.05$. We do not reject our null hypothesis $H_0$.