Following are the age and price data for custom homes, use $\alpha =0.01$

presuming that the assumption for regression inference are met, decide at the specified significance level whether the data provide sufficient evidence to conclude that the predictor variable is useful for providing the response variable.

Given table is

and, $\alpha =0.01$. we have to find out whether the data provide sufficient evidence to conclude that the predictor variable is useful for providing the response variable.

STEP -1: The null and alternative hypothesis are

$H_0:{\beta }_1=0$

$H_{\alpha }:{\beta }_1\ne 0$

STEP 2: Determine the$\left(\alpha \right)$significance level.

The hypothesis test should be run at a significance level of $1$percent, or $\alpha =0.01$

Table of computations:

$$\begin{gathered} S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n \\ =169993-\left(270\right)\left(5552\right)/9 \\ =3433 \end{gathered}$$

$$\begin{gathered} S_{xx}=\sum x_i^2-{\left(\sum x_i\right)}^2/n \\ =8316-(270{)}^2/9 \\ =216 \end{gathered}$$

The total sum of square SST is calculated by

$$\begin{gathered} S_{yy}=\sum y_i^2-{\left(\sum y_i\right)}^2/n \\ =3504412-(5552{)}^2/9 \\ =3504412-30824704/9 \\ =79444.88889 \end{gathered}$$

The regression sum of square is calculated by

$$\begin{gathered} SSR=\frac{S_{xy}^2}{S_{xx}} \\ =\frac{(3433{)}^2}{216} \\ =54562.44907 \end{gathered}$$

$SSE=SST-SSR$

$$\begin{gathered} SSE=79444.88889-54562.44907 \\ =24882.43981 \end{gathered}$$

The formula of standard error is

$$\begin{gathered} s_e=\sqrt{\frac{SSE}{n-2}} \\ =\sqrt{\frac{24882.43981}{9-2}} \\ \approx 59.621 \end{gathered}$$

STEP-3 The formula of test stastic is

$$\begin{gathered} t=\frac{b_1}{s_e/\sqrt{S_{xx}}} \\ =\frac{15.89351852}{59.6207536/\sqrt{216}} \\ \approx 3.92 \end{gathered}$$

STEP-4

$$\begin{gathered} df=n-1 \\ =9-2 \\ =7 \end{gathered}$$

The critical values are $\pm t_{\alpha /2}=\pm t_{0.005}=\pm 3.500$, as determined by technology.

Or STEP 4: The test statistic's value is $t=3.92$as of Step 3. The P-value is the probability of seeing a value of$tof3.92$or larger in magnitude if the null hypothesis is true, because the test is two-tailed. We derive the $p=00.5765$using technology.

STEP 5: The test statistic's value is smaller than the critical value. Our null hypothesis$H_0$$t=3.92<0.005,7=3.500$, i.e. is rejected. Our null hypothesis$H_0$is not rejected.

STEP 6: The data give sufficient evidence at the 1% significance level to infer that the slope of the population regression line is not $0,$and so the variable $\left(x\right)$is significant.