Step One: Find [tex]$\dot{y}$[/tex] for the [tex]$x$[/tex]-values in the data set. (Round to the nearest tenth.)
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$\dot{y}$[/tex] \\
\hline
5 & 4.6 \\
\hline
7 & 5.3 \\
\hline
9 & 6.0 \\
\hline
11 & 6.6 \\
\hline
13 & 7.3 \\
\hline
15 & 8.0 \\
\hline
\end{tabular}

Step Two: Find the residual (difference between [tex]$y$[/tex] and [tex]$\dot{y}$[/tex]).

Instructions: Create a residual plot of the following data using the regression line
[tex]$
\begin{aligned}
y= & 2.9 + 0.34x \\
& \{(5, 5), (7, 3), (9, 9), (11, 7), (13, 5), (15, 9)\}
\end{aligned}
$[/tex]

What is the sum of the residuals? [tex]$\square$[/tex]

Check



Answer :

Let's walk through how we find the residuals and calculate their sum.

### Step Two: Find the Residuals

Residuals are the differences between the observed [tex]\( y \)[/tex]-values and the predicted [tex]\( \dot{y} \)[/tex]-values using the regression line [tex]\( y = 2.9 + 0.34x \)[/tex].

Given the data:
[tex]\[ \{(5,5),(7,3),(9,9),(11,7),(13,5),(15,9)\} \][/tex]

And the predicted [tex]\( \dot{y} \)[/tex] values calculated using the regression line:
[tex]\[ \begin{aligned} \dot{y} \text{ for } x=5 & = 4.6 \\ \dot{y} \text{ for } x=7 & = 5.3 \\ \dot{y} \text{ for } x=9 & = 6.0 \\ \dot{y} \text{ for } x=11 & = 6.6 \\ \dot{y} \text{ for } x=13 & = 7.3 \\ \dot{y} \text{ for } x=15 & = 8.0 \\ \end{aligned} \][/tex]

We can find the residuals by subtracting the predicted [tex]\( \dot{y} \)[/tex] from the observed [tex]\( y \)[/tex]:
[tex]\[ \begin{aligned} \text{Residual for } x=5 & : 5 - 4.6 = 0.4 \\ \text{Residual for } x=7 & : 3 - 5.3 = -2.3 \\ \text{Residual for } x=9 & : 9 - 6.0 = 3.0 \\ \text{Residual for } x=11 & : 7 - 6.6 = 0.4 \\ \text{Residual for } x=13 & : 5 - 7.3 = -2.3 \\ \text{Residual for } x=15 & : 9 - 8.0 = 1.0 \\ \end{aligned} \][/tex]

Now, listing these residuals:
[tex]\[ [0.4, -2.3, 3.0, 0.4, -2.3, 1.0] \][/tex]

### Sum of the Residuals

The sum of the residuals is calculated by summing up all the individual residuals:
[tex]\[ 0.4 + (-2.3) + 3.0 + 0.4 + (-2.3) + 1.0 = 0.2 \][/tex]

Thus, the sum of the residuals is [tex]\( 0.2 \)[/tex].