Linear Regression (y = a + bx)

Intro: We fit $\hat{y}=a+bx$, report slope/intercept, and optionally predict $\hat{y}(x_0)$.

Worked example

• $x:1,2,3,4,5;\; y:2,3,5,6,8;\;$ predict at $x=6$



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• Goal: Fit the least-squares line $\hat{y}=a+bx$ and then compute $\hat{y}(6)$. We use $$\displaystyle b=\frac{\sum (x_i-\bar{x})(y_i-\bar{y})}{\sum (x_i-\bar{x})^2},\qquad a=\bar{y}-b\bar{x}.$$



• Means: $\displaystyle \bar{x}=\frac{1+2+3+4+5}{5}=3,\quad \bar{y}=\frac{2+3+5+6+8}{5}=\frac{24}{5}=4.8.$



• Deviations in $x$: $(x_i-\bar{x})=(-2,-1,0,1,2)$. Deviations in $y$: $(y_i-\bar{y})=(-2.8,-1.8,0.2,1.2,3.2)$.



• Compute $S_{xx}=\sum (x_i-\bar{x})^2 = (-2)^2+(-1)^2+0^2+1^2+2^2=4+1+0+1+4=10.$



• Compute $S_{xy}=\sum (x_i-\bar{x})(y_i-\bar{y}) = (-2)(-2.8)+(-1)(-1.8)+0(0.2)+1(1.2)+2(3.2) = 5.6+1.8+0+1.2+6.4=15.0.$



• Slope: $\displaystyle b=\frac{S_{xy}}{S_{xx}}=\frac{15}{10}=1.5.$



• Intercept: $\displaystyle a=\bar{y}-b\bar{x}=4.8-(1.5)(3)=4.8-4.5=0.3.$



• Model: $$\boxed{\hat{y}=0.3+1.5x}.$$



• Prediction at $x=6$: $\hat{y}(6)=0.3+1.5\cdot6=0.3+9=9.3.$



• Answer: $$\boxed{\hat{y}=0.3+1.5x,\quad \hat{y}(6)=9.3}.$$



• Check (optional via alternative formula): $$\displaystyle b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2} = \frac{5\cdot87-15\cdot24}{5\cdot55-15^2}=\frac{435-360}{275-225}=\frac{75}{50}=1.5,$$ which matches.

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How this calculator works

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Related

• Correlation r