Question

In this Exercise 14.49, we repeat the information from Exercises 14.13.

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.

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

Short answer

(a) The data do not give sufficient evidence to establish that the slope of the population regression line is not $0$, and so the variable $x$is not useful for predicting the variable $y$at the $10\%$significance level.

(b) The slope of the population regression line is somewhere between $8.94$and $-12.94$, and we can be $90\%$sure of that.

Step-by-step solution

Part(a) Step 1: Given Information

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

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

Part(a) Step 2: Explanation

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 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:

$\begin{array}{|ccccc|}\hline x& y& xy& x^2& y^2\\ 3& -4& -12& 9& 16\\ 1& 0& 0& 1& 0\\ 2& -5& -10& 4& 25\\ \sum x_i=6& \sum y_i=-9& \sum x_i{y}_i=-22& \sum x_i^2=14& \sum y_i^2=41\\ \hline\end{array}$

$$\begin{gathered} S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n \\ =-22-\left(6\right)(-9)/3 \\ =-22+54/3 \\ =-22+18 \\ =-4 \end{gathered}$$

$$\begin{gathered} S_{xx}=\sum x_i^2-{\left(\sum x_i\right)}^2/n \\ =14-(6{)}^2/3 \\ =14-36/3 \\ =14-12 \\ =2 \end{gathered}$$

Part(a) Step 3: Calculation

The entire amount of square SST is calculated as follows:

$$\begin{gathered} S_{yy}=\sum y_i^2-{\left(\sum y_i\right)}^2/n \\ =41-(-9{)}^2/3 \\ =41-81/3 \\ =41-27 \\ =14 \end{gathered}$$

The regression sum of square SSR is calculated as follows:

role="math" localid="1652281192077" $$\begin{gathered} SSR=\frac{S_{xy}^2}{S_{xx}} \\ =\frac{(-4{)}^2}{2} \\ =\frac{16}{2} \\ =8 \end{gathered}$$

$$\begin{gathered} SSE=SST-SSR \\ =14-8 \\ =6 \end{gathered}$$

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

$$\begin{gathered} b_1=\frac{S_{xy}}{S_{xx}} \\ =\frac{-4}{2} \\ =-2 \end{gathered}$$

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

$$\begin{gathered} S_e=\sqrt{\frac{SSE}{n-2}} \\ =\sqrt{\frac{6}{3-2}} \\ =2.449489743 \\ \approx 2.45 \end{gathered}$$

part(a) step 4: Final answer

Computing the value of test statistic

$$\begin{gathered} t=\frac{b_1}{s_e/\sqrt{S_{xx}}} \\ =\frac{-2}{2.45/\sqrt{2}} \\ =\frac{-2}{1.732050808} \\ =-1.154460051 \\ \approx -1.15 \end{gathered}$$

From above $\alpha =0.10$when $n=3$

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

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

Because the test statistic's value is less than the critical value. Our null hypothesis $H_0$, i.e.$$\begin{gathered} t=-1.15<t_{0.05,1} \\ =6.314 \end{gathered}$$, is not rejected.

Part(b) Step 1: Given Information

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

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

Part(b) Step 2: Explanation

$\alpha =0.10$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 end points for ${\beta }_1$is

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

$b_1=-2$,

$s_e=2.45$,

$S_{xx}=2$

So, $-2\pm 6.314\times \frac{2.45}{\sqrt{2}}$

Or $-2\pm 10.94$, or $8.94$ to$-12.94$