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In order to determine which line of fit is better, the residuals for both lines will be calculated first. For a data point, its residual is the difference between the $y\text{-}$value of the data point and the $y\text{-}$value predicted by the line of fit.

$x$	$y$ (Actual)	$y$ Predicted by $y=\text{-}0.1x+10$	Residual for $y=\text{-}0.1x+10$
$1$	$9.87$	$y=\text{-}0.1(1)+10=9.9$	$9.87-9.9=\text{-}0.03$
$2$	$9.85$	$y=\text{-}0.1(2)+10=9.8$	$9.85-9.8=0.05$
$3$	$9.69$	$y=\text{-}0.1(3)+10=9.7$	$9.69-9.7=\text{-}0.01$
$4$	$9.63$	$y=\text{-}0.1(4)+10=9.6$	$9.63-9.6=0.03$
$5$	$9.81$	$y=\text{-}0.1(5)+10=9.5$	$9.81-9.5=0.31$
$6$	$9.80$	$y=\text{-}0.1(6)+10=9.4$	$9.80-9.4=0.40$

The values on the last column of the table will be squared and added. The sum of squared residuals for Equation I, $SS{R}_1,$ is $0.2605.$ Similarly, the residuals for Equation II will be calculated.

$x$	$y$ (Actual)	$y$ Predicted by $y=\text{-}0.05x+10$	Residual for $y=\text{-}0.05x+10$
$1$	$9.87$	$y=\text{-}0.05(1)+10=9.95$	$9.87-9.95=\text{-}0.08$
$2$	$9.85$	$y=\text{-}0.05(2)+10=9.9$	$9.85-9.9=\text{-}0.05$
$3$	$9.69$	$y=\text{-}0.05(3)+10=9.85$	$9.69-9.85=\text{-}0.16$
$4$	$9.63$	$y=\text{-}0.05(1)+10=9.8$	$9.63-9.8=\text{-}0.17$
$5$	$9.81$	$y=\text{-}0.05(5)+10=9.75$	$9.81-9.75=0.06$
$6$	$9.80$	$y=\text{-}0.05(6)+10=9.70$	$9.80-9.70=0.10$

Now that the residuals are found, they can be squared and added. The sum of squared residuals for Equation II, $SS{R}_2,$ is $0.077.$ As a result, Equation II is the better line of fit because it has a lesser sum of squared residuals.

$x$	Residual for $y=\text{-}0.1x+10$	Residual for $y=\text{-}0.05x+10$
$1$	$\text{-}0.03$	$\text{-}0.08$
$2$	$0.05$	$\text{-}0.05$
$3$	$\text{-}0.01$	$\text{-}0.16$
$4$	$0.03$	$\text{-}0.17$
$5$	$0.31$	$0.06$
$6$	$0.47$	$0.1$

The points $(x,\text{residual})$ for each equation will be graphed on a scatter plot.

As can be seen, the residuals for Equation II are close the $x\text{-}$axis and therefore the sum of their squares has a lower value.

First, the given data will be shown on a graph. Then, a line that passes close to the points will be drawn. It appears that the line passes through the points $(4,20)$ and $(16,16).$ Knowing two points on the line, the slope of the line and its equation can be found. To find the slope, the points need to be substituted into the Slope Formula. Next, using $m$ and either point, the equation of the line can be written in the point-slope form. Use $m=\text{-}\frac{1}{3}$ and $(x_1,y_1)=(4,20).$ To write the equation in slope-intercept form, $y$ will be isolated. Finally, the residuals for this equation can be calculated. To do so, the difference between the $y\text{-}$value of a data point and the corresponding $y\text{-}$value predicted by the line of fit will be calculated for each data point.

$x$	$y$ (Actual)	$y$ Predicted by the equation	Residual
$6$	$19$	$y=\text{-}\frac{1}{3}(6)+\frac{64}{3}=\frac{58}{3}$	$19-\frac{58}{3}=\text{-}\frac{1}{3}$
$12$	$15$	$y=\text{-}\frac{1}{3}(12)+\frac{64}{3}=\frac{52}{3}$	$15-\frac{52}{3}=\text{-}\frac{7}{3}$
$18$	$17$	$y=\text{-}\frac{1}{3}(18)+\frac{64}{3}=\frac{46}{3}$	$17-\frac{46}{3}=\frac{5}{3}$
$23$	$15$	$y=\text{-}\frac{1}{3}(23)+\frac{64}{3}=\frac{41}{3}$	$15-\frac{41}{3}=\frac{4}{3}$
$26$	$15$	$y=\text{-}\frac{1}{3}(26)+\frac{64}{3}=\frac{38}{3}$	$15-\frac{58}{3}=\frac{7}{3}$
$30$	$12$	$y=\text{-}\frac{1}{3}(30)+\frac{64}{3}=\frac{34}{3}$	$12-\frac{34}{3}=\frac{2}{3}$

Next, the sum of squared residuals can be found by adding the squares of the numbers in the last column of the table. The sum of squared residuals for the equation is $16.$ This equation does not satisfy the condition that Maya wants. Another equation can be found by slightly increasing the slope of the above equation and slightly decreasing its $y\text{-}$intercept. Use the applet below to find an equation. For example, $y=\text{-}0.25x+20$ can be a good candidate. The residuals for this equation can be found as follows.

$x$	$y$ (Actual)	$y$ Predicted by the equation	Residual
$6$	$19$	$y=\text{-}0.25(6)+20=18.5$	$19-18.5=0.5$
$12$	$15$	$y=\text{-}0.25(12)+20=17$	$15-17=\text{-}2$
$18$	$17$	$y=\text{-}0.25(18)+20=15.5$	$17-15.5=1.5$
$23$	$15$	$y=\text{-}0.25(23)+20=14.25$	$15-14.25=0.75$
$26$	$15$	$y=\text{-}0.25(26)+20=13.5$	$15-13.5=1.5$
$30$	$12$	$y=\text{-}0.25(30)+20=12.5$	$12-12.5=\text{-}0.5$

Now, the sum of squared residuals for this equation can be calculated. As a result, the sum of squared residuals for $y=\text{-}0.25x+20$ satisfies the condition. Note that there are numerous equations that satisfy Maya's condition. Here only one of them was shown. The following applet shows the equation of a line fit and the sum of squared residuals. Use it to see how the sum changes as the equation changes.