In Exercises 14.48, we repeat the information from Exercises 14.12.

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.

$\begin{array}{llll}x& 2& 4& 3\\ y& 3& 5& 7\end{array}$ role="math" localid="1652276835214" $\hat{y}=2+x$

$\begin{array}{|llll|}\hline x& 2& 4& 3\\ y& 3& 5& 7\\ \hline\end{array}$

$\hat{y}=2+x$

Define 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 $y$) .

Computation table

$\begin{array}{|ccccc|}\hline x& y& xy& x^2& y^2\\ 2& 3& 6& 4& 9\\ 4& 5& 20& 16& 25\\ 3& 7& 21& 9& 49\\ \sum x_i=9& \sum y_i=15& \sum x_i{y}_i=47& \sum x_i^2=29& \sum y_i^2=83\\ \hline\end{array}$

$$\begin{gathered} S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n \\ =47-\left(9\right)\left(15\right)/3 \\ =47-135/3 \\ =47-45 \\ =2 \end{gathered}$$

The entire amount of square SST is calculated as follows:

$$\begin{gathered} S_{yy}=\sum y_i^2-{\left(\sum y_i\right)}^2/n \\ =83-(15{)}^2/3 \\ =83-225/3 \\ =83-75 \\ =8 \end{gathered}$$

The regression sum of square SSR is calculated as follows:

$$\begin{gathered} SSR=\frac{{S_{xy}}^2}{S_{xx}} \\ =\frac{(2{)}^2}{2} \\ =\frac{4}{2} \\ =2 \end{gathered}$$

$$\begin{gathered} SSE=SST-SSR \\ =8-2 \\ =6 \end{gathered}$$

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

$$\begin{gathered} b_1=\frac{S_{xy}}{S_{xx}} \\ =\frac{2}{2} \\ =1 \end{gathered}$$

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

$$\begin{gathered} s_e=\sqrt{\frac{SSE}{n-2}} \\ =\frac{1}{2.45/\sqrt{2}} \\ =\frac{1}{1.732050808} \\ =0.577350269 \\ \approx 0.58 \end{gathered}$$

From above $\alpha =0.10.$For $n=3$

$$\begin{gathered} df=n-2 \\ =3-2 \\ =1 \end{gathered}$$

The critical values are role="math" localid="1652278581472" $$\begin{gathered} \pm t_{\alpha /2}=\pm t_{0.05} \\ =\pm 6.314 \end{gathered}$$, as determined by technology.

Because the test statistic's value is less than the critical value. Our null hypothesis $H_0$i.e $$\begin{gathered} t=0.58<t_{0.05,1} \\ =6.314 \end{gathered}$$. is not rejected.

$\begin{array}{|llll|}\hline x& 2& 4& 3\\ y& 3& 5& 7\\ \hline\end{array}$

$\hat{y}=2+x$

$\alpha =0.10$for a $90\%$confidence interval. Since $n=3$,

$$\begin{gathered} df=n-2 \\ =3-2 \\ =1 \end{gathered}$$

From technology, $$\begin{gathered} t_{\alpha /2}=t_{0.10/2} \\ =t_{0.05} \\ =6.314 \end{gathered}$$

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 $b_1=1$,

$s_e=2.45$

$S_{xx}=2$

So, role="math" $1\pm 6.314\times \frac{2.45}{\sqrt{2}}$

Or $1\pm 10.94$, or $-9.94\text{to}11.94$.