91Ó°ÊÓ

$\mathrm{Î\pm }=0.05$

Decide the Null and Alternate hypothesis

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

$H_a:{\mathrm{β}}_1≠0$($x$is useful for predicting $y$)

We need to determine the significance level $\mathrm{Î\pm }$

The hypothesis test should be run at a significance threshold of $5\%$, or $\mathrm{Î\pm }=0.05$.

Computation table

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

The entire amount of square SST is calculated as follows:

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

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:

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

We need to find the value of the test statistic

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

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$=0.00000170184>\mathrm{Î\pm }=0.05$. We do not reject our null hypothesis $H_0$.