$$\text { Find the equation } y = \beta _ { 0 } + \beta _ { 1 } x \text { of the least-squares line that best fits the given data points. }$$
-Data points: (2, 1), (3, 2), (7, 3), (8, 1)33) $$\mathrm { X } = \left[ \begin{array} { l l }
1 & 2 \
1 & 3 \
1 & 7 \
1 & 8
\end{array} \right] , \mathrm { y } = \left[ \begin{array} { l }
1 \
2 \
3 \
1
\end{array} \right]$$
A) $y = - \frac { 9 } { 52 } + \frac { 1 } { 13 } x$
B) $y = \frac { 71 } { 52 } - \frac { 1 } { 26 } x$
C) $y = \frac { 71 } { 52 } + \frac { 1 } { 13 } x$
D) $y = - \frac { 105 } { 52 } + \frac { 1 } { 13 } x$

Question

$$\text { Find the equation } y = \beta _ { 0 } + \beta _ { 1 } x \text { of the least-squares line that best fits the given data points. }$$
-Data points: (2, 1), (3, 2), (7, 3), (8, 1)33) $$\mathrm { X } = \left[ \begin{array} { l l } 
1 & 2 \
1 & 3 \
1 & 7 \
1 & 8
\end{array} \right] , \mathrm { y } = \left[ \begin{array} { l } 
1 \
2 \
3 \
1
\end{array} \right]$$
A) $y = - \frac { 9 } { 52 } + \frac { 1 } { 13 } x$
B) $y = \frac { 71 } { 52 } - \frac { 1 } { 26 } x$
C) $y = \frac { 71 } { 52 } + \frac { 1 } { 13 } x$
D) $y = - \frac { 105 } { 52 } + \frac { 1 } { 13 } x$

Quizplus · Accepted Answer

The least-squares line equation $y = \beta_0 + \beta_1 x$ is found by solving the normal equations $\mathbf{X}^T\mathbf{X}\boldsymbol{\beta} = \mathbf{X}^T\mathbf{y}$, where $\mathbf{X}$ is the matrix of input values (with a column of 1s for the intercept), $\mathbf{y}$ is the vector of output values, and $\boldsymbol{\beta}$ is the vector of coefficients $[\beta_0, \beta_1]^T$. Solving these equations with the given data points yields the coefficients that minimize the sum of squared residuals, leading to the correct equation $y = \frac{71}{52} + \frac{1}{13}x$.