91Ó°ÊÓ

$\begin{array}{|llllll|}\hline x& 0& 4& 3& 1& 2\\ y& 1& 9& 8& 4& 3\\ \hline\end{array}$

$\hat{y}=1+2x$

Decide the null and alternate hypothesis

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

$H_{\mathrm{Î\pm }}:{\mathrm{β}}_1≠0$($x$is useful for predicting $y$),

Determine the significance level $\mathrm{Î\pm }$

The hypothesis test should be run at a significance level of $10\%$, or $\mathrm{Î\pm }=0.10$.

Computation table

$\begin{array}{|ccccc|}\hline x& y& xy& x^2& y^2\\ 0& 1& 0& 0& 1\\ 4& 9& 36& 16& 81\\ 3& 8& 24& 9& 64\\ 1& 4& 4& 1& 16\\ 2& 3& 6& 4& 9\\ ∑x_i=10& ∑y_i=25& ∑x_i{y}_i=70& ∑x_i^2=30& ∑y_i^2=171\\ \hline\end{array}$

$$\begin{gathered} S_{xy}=∑x_i{y}_i-\left(∑x_i\right)\left(∑y_i\right)/n \\ =70-\left(10\right)\left(25\right)/5 \\ =70-250/5 \\ =70-50 \\ =20 \end{gathered}$$

$$\begin{gathered} S_{xx}=∑x_i^2-{\left(∑x_i\right)}^2/n \\ =30-(10{)}^2/5 \\ =30-100/5 \\ =30-20 \\ =10 \end{gathered}$$

The entire amount of square SST is calculated as follows:

$$\begin{gathered} S_{yy}=∑y_i^2-{\left(∑y_i\right)}^2/n \\ =171-(25{)}^2/5 \\ =171-625/5 \\ =171-125 \\ =46 \end{gathered}$$

SSR regression sum of squares is calculated by:

$$\begin{gathered} SSR=\frac{{S_{xy}}^2}{S_{xx}} \\ =\frac{(20{)}^2}{10} \\ =\frac{400}{10} \\ =40 \end{gathered}$$

$$\begin{gathered} SSE=SST-SSR \\ =46-40 \\ =6 \end{gathered}$$

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

$$\begin{gathered} b_1=\frac{S_{xy}}{S_{xx}} \\ =\frac{20}{10} \\ =2 \end{gathered}$$

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

$$\begin{gathered} S_e=\sqrt{\frac{SSE}{n-2}} \\ =\sqrt{\frac{6}{5-2}} \\ =1.414213562 \\ ≈1.41 \end{gathered}$$

We need to find the value of test statistic

$$\begin{gathered} t=\frac{b_1}{s_e/\sqrt{S_{xx}}} \\ =\frac{2}{1.41/\sqrt{10}} \\ =\frac{2}{0.44588115} \\ =4.485500227 \\ ≈4.49 \end{gathered}$$

The test statistic's value is $t=4.49$, as shown above. The $p$-value is the probability of seeing a value of $t$of $4.49$or more in magnitude if the null hypothesis is true because the test is two-tailed. We get $\mathrm{P}=0.0103$by using technology.

Since the $p$-value $=0.0103<\mathrm{Î\pm }=0.10$. We reject our null hypothesis $H_0$.

$\begin{array}{|llllll|}\hline x& 0& 4& 3& 1& 2\\ y& 1& 9& 8& 4& 3\\ \hline\end{array}$

$\hat{y}=1+2x$

$\mathrm{Î\pm }=0.10$for a $90\%$confidence interval. Since $n=5$,

$$\begin{gathered} df=n-2 \\ =5-2 \\ =3 \end{gathered}$$

From technology $$\begin{gathered} t_{\mathrm{Î\pm }/2}=t_{0.10/2} \\ =t_{0.05} \\ =2.355 \end{gathered}$$

The formula for computing the confidence interval endpoints for ${\mathrm{β}}_1$is

$b_1Â\pm t_{\mathrm{Î\pm }/2}×\frac{s_e}{\sqrt{s_{xx}}}$

We have $b_1=2$,

$s_e=1.41$,

$S_{xx}=10$,

So, $2Â\pm 2.355×\frac{1.41}{\sqrt{10}}$

Or $2Â\pm 1.050050108$, or $0.949$to$3.050$