Consider the data given at the beginning of this section. [latex]\begin{array}{l|lll}x & 0 & 5 & 10 \\\hline y & 2 & 6 & 11\end{array}[/latex] We found that the best-fit line is y = (9/10)x + 11/6. Find the best-fit line that passes through the point x = 10, y = 11.

Subtracting 10 from all the x values and 11 from all the y values, we obtain a new set of data in terms of the new variables X = x − 10 and Y = y − 11. [latex]\begin{array}{l|lll}X & -10 & -5 & 0 \\\hline Y & -9 & -5 & 0\end{array}[/latex] Expressing (C.2.3) in terms of the new variables X and Y, we have [latex]\begin{aligned}m \sum_{i=1}^3 X_i^2 &=\sum_{i=1}^3 X_i Y_i \\\sum_{i=1}^3 X_i^2 &=(-10)^2+5^2+0=125 \\\sum_{i=1}^3 X_i Y_i &=(-10)(-9)+(-5)(-5)+0=115\end{aligned}[/latex] Thus, m=115∕125=23∕25 and the best-fit line is Y =(23/25)X. In terms of the original variables, this line is expressed as y − 11 = (23/25)(x − 10) or y = (23/25)x + 9/5.

Question C.2.2: Consider the data given at the beginning of this section.

System Dynamics

Consider the data given at the beginning of this section.

$\begin{array}{l|lll}x & 0 & 5 & 10 \\\hline y & 2 & 6 & 11\end{array}$

We found that the best-fit line is y = (9/10)x + 11/6. Find the best-fit line that passes through the point x = 10, y = 11.

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Subtracting 10 from all the x values and 11 from all the y values, we obtain a new set of data in terms of the new variables X = x − 10 and Y = y − 11.

$\begin{array}{l|lll}X & -10 & -5 & 0 \\\hline Y & -9 & -5 & 0\end{array}$

Expressing (C.2.3) in terms of the new variables X and Y, we have

$\begin{aligned}m \sum_{i=1}^3 X_i^2 &=\sum_{i=1}^3 X_i Y_i \\\sum_{i=1}^3 X_i^2 &=(-10)^2+5^2+0=125 \\\sum_{i=1}^3 X_i Y_i &=(-10)(-9)+(-5)(-5)+0=115\end{aligned}$

Thus, m=115∕125=23∕25 and the best-fit line is Y =(23/25)X. In terms of the original variables, this line is expressed as y − 11 = (23/25)(x − 10) or y = (23/25)x + 9/5.