In this Exercise 14.5, we repeat the information from Exercise 14.14.

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.

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

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

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

Decide the null and alternate hypothesis

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

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

Determine the significance level $\alpha$

The hypothesis test should be run at a significance level of $10\%$, or $\alpha =0.10$.

Computation table

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

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

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

The entire amount of square SST is calculated as follows:

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

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:

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

We need to find the value of test statistic

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

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 $=0.0103<\alpha =0.10$. We reject our null hypothesis $H_0$.

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

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

$\alpha =0.10$for a $90\%$confidence interval. Since $n=5$,

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

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

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

$b_1\pm t_{\alpha /2}\times \frac{s_e}{\sqrt{s_{xx}}}$

We have $b_1=2$,

$s_e=1.41$,

$S_{xx}=10$,

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

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