In this Exercise 14.51, we repeat the information from Exercise 14.15.

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.

role="math" localid="1652333468446" $\begin{array}{|ccccc|}\hline x& 3& 4& 1& 2\\ y& 4& 5& 0& -1\\ \hline\end{array}$ role="math" localid="1652333507650" $\hat{y}=-3+x$

$\begin{array}{|ccccc|}\hline x& 3& 4& 1& 2\\ y& 4& 5& 0& -1\\ \hline\end{array}$

$\hat{y}=-3+x$

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 not useful for predicting $y$)

Determine the significance level $\alpha$

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

Computation table

$\begin{array}{|ccccc|}\hline x& y& xy& x^2& y^2\\ 3& 4& 12& 9& 16\\ 4& 5& 20& 16& 25\\ 1& 0& 0& 1& 0\\ 2& -1& -2& 4& 1\\ \sum x_i=10& \sum y_i=8& \sum x_i{y}_i=30& \sum x_i^2=30& \sum y_i^2=42\\ \hline\end{array}$

$$\begin{gathered} S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n \\ =30-\left(10\right)\left(8\right)/4 \\ =30-80/4 \\ =30-20 \\ =10 \end{gathered}$$

role="math" localid="1652332784970" $$\begin{gathered} S_{xx}=\sum x_i^2-{\left(\sum x_i\right)}^2/n \\ =30-(10{)}^2/4 \\ =30-100/4 \\ =30-25 \\ =5 \end{gathered}$$

The entire amount of square SST is calculated as follows:

$$\begin{gathered} S_{yy}=\sum y_i^2-{\left(\sum y_i\right)}^2/n \\ =42-(8{)}^2/4 \\ =42-64/4 \\ =42-16 \\ =26 \end{gathered}$$

SSR regression sum of squares is calculated as:

$$\begin{gathered} SSR=\frac{S_{xy}^2}{S_{xx}} \\ =\frac{(10{)}^2}{5} \\ =\frac{100}{5} \\ =20 \end{gathered}$$

$$\begin{gathered} SSE=SST=SSR \\ =26-20 \\ =6 \end{gathered}$$

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

$$\begin{gathered} b_1=\frac{S_{xy}}{S_{xx}} \\ =\frac{10}{5} \\ =2 \end{gathered}$$

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

$$\begin{gathered} s_e=\sqrt{\frac{SSE}{n-2}} \\ =\sqrt{\frac{6}{4-2}} \\ =1.732050808 \\ \approx 1.73 \end{gathered}$$

We need to find the value of test statistic

$$\begin{gathered} t=\frac{b_1}{s_e/\sqrt{s_{xx}}} \\ =\frac{2}{1.73/\sqrt{5}} \\ =\frac{2}{0.77367952} \\ =2.585049685 \\ \approx 2.58 \end{gathered}$$

The test statistic's value is $t=2.58$, as shown above. The $p$-value is the likelihood of seeing a value of $t$of $2.58$or more in magnitude if the null hypothesis is true because the test is two-tailed. We get $P=0.1229$by using technology.

Since the $P$-value $=0.1229>\alpha =0.10$. We do not reject our null hypothesis $H_0$.

$\begin{array}{|ccccc|}\hline x& 3& 4& 1& 2\\ y& 4& 5& 0& -1\\ \hline\end{array}$

$\hat{y}=-3+x$

$\alpha =0.10$for a $90\%$confidence interval. Since $n=4$,

$$\begin{gathered} df=n-2 \\ =4-2 \\ =2 \end{gathered}$$

From technology $$\begin{gathered} t_{a/2}=t_{0.10/2} \\ =t_{0.05} \\ =2.920 \end{gathered}$$

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

role="math" localid="1652333749328" $b_1\pm t_{\alpha /2}\times \frac{s_e}{\sqrt{S_{xx}}}$

We have $b_1=2$

$s_e=1.73$,

$S_{xx}=5$,

So, $2\pm 2.920\times \frac{1.73}{\sqrt{5}}$

Or $2\pm 2.259144199$, or $-0.26$ to$4.26$