Question

4.55 The data points in Exercise 4.43

a. find the regression equation for the data points. Use the defining formations in Definition $4.4$to obtain $S_{xx}$and $S_{xy}$

b. Graph the regression connection and the data points.

Short answer

(a) The regression equation is $\hat{y}=1+2x$,

(b) The graph will be

Step-by-step solution

Part (a) Step 1: Given information

The data points in Exercise 4.43 is:

$x$	$1$	$2$	$3$
$y$	$4$	$3$	$8$

Part (a) Step 2: Explanation

The terms $b_0$and $b_1$are used to create the regression equation. For a set of$n$ data points, the regression equation is:
$\hat{y}=b_0+b_{1}x$
where,
$$\begin{gathered} b_1=\frac{S_{xy}}{S_{xx}} \\ {b}_0=\frac{1}{n}\left(\sum y_i-b_1\sum x_i\right) \\ {b}_0=\overline{y}-b_1\overline{x} \end{gathered}$$

The table is available in the following formats:

$x$	$y$	$\sum \left(x_i-\overline{x}\right)$	$\sum {\left(x_i-\overline{x}\right)}^2$	$(y_i-\overline{y})$	$\left(x_i-\overline{x}\right)(y_i-\overline{y})$
$1$	$4$	$-1$	$1$	$-1$	$1$
$2$	$3$	$0$	$0$	$-2$	$0$
$3$	$8$	$1$	$1$	$3$	$3$
$\overline{x}=2$	$\overline{y}=5$		$\sum {\left(x_i-\overline{x}\right)}^2=2$		$\sum \left(x_i-\overline{x}\right)(y_i-\overline{y})=4$

Hence, $S_{xx}=2$, and $S_{xy}=4$.

Part (a) Step 3: Claculation

Since,

$$\begin{gathered} b_1=\frac{S_{xy}}{S_{xx}} \\ {b}_1=\frac{4}{2} \\ {b}_1=2 \end{gathered}$$

Then,

$$\begin{gathered} b_0=\overline{y}-b_1\overline{x} \\ {b}_0=5-\left(2\right)2 \\ {b}_0=5-4 \\ {b}_0=1 \end{gathered}$$

As a result, the regression equation will be as follows:

$$\begin{gathered} \hat{y}=b_0+b_{1}x \\ \hat{y}=1+\left(2\right)x \\ \hat{y}=1+2x \end{gathered}$$

Part (b) Step 1: Given information

Given in the question that

Part (b) Step 2: Graphical representation

The graph for the regression equation $\widehat{y}=1+2x$is