Answer to Question #333011 in Statistics and Probability for mimi

Question

The following data show the age (week) and height (cm) of soybean 

plant. 

Age (week) 1 2 3 4 5 6 7

Height (cm) 5 13 16 23 33 38 40

Find the least square regression line of height on age. 

Answer 1

Consider pairs of points $(x,y)$, where $x$ corresponds to the age and $y$ corresponds to the height. The least squares regression line is the line: $\hat{y}=m{x}+b$ that minimizes the following error: $E(m,b)=\sum_{i=1}^n(y_i-\hat{y})^2$. After substitution $\hat{y}=mx+b$ we receive: $E(m,b)=\sum_{i=1}^n(y_i-(mx_i+b))^2$. After taking the derivatives with respect to $m$ and $b$ we get: $E_m=2\sum_{i=1}^nx_i((mx_i+b)-y_i)=2(m\sum_{i=1}^nx_i^2+b\sum_{i=1}^nx_i-\sum_{i=1}^nx_iy_i)$ and $E_b=2\sum_{i=1}^n((mx_i+b)-y_i)=2(m\sum_{i=1}^nx_i+nb-\sum_{i=1}^ny_i)$. Denote $\alpha=\sum_{i=1}^nx_i^2$, $\beta=\sum_{i=1}^nx_iy_i$, $\gamma=\sum_{i=1}^nx_i$ and $\delta=\sum_{i=1}^ny_i$. Necessary conditions of extrema take the form: $E_m=0,E_b=0$. The latter is equivalent to: $m\alpha+b\gamma-\beta=0,$ $m\gamma+nb-\delta=0$. From the first equation we get: $m=\frac{\beta-b\gamma}{\alpha}$. We substitute it in the second equation and get: $\frac{\beta\gamma}{\alpha}-\frac{\gamma^2}{\alpha}b+nb-\delta=0$We receive: $b=\frac{\delta\alpha-\beta\gamma}{n\alpha-\gamma^2}$, $m=\frac{n\beta-\delta\gamma}{n\alpha-\gamma^2}$. Expressions for $m$ and $b$ as well as verifications of sufficient conditions of extrema can be also found in the respective literature about the least square regression line. Set $n=7$, take values from the tables and get: $\alpha=140$, $\beta=844$, $\delta=168$, $\gamma=28$. We receive: $b=\frac{168\cdot140-844\cdot28}{7\cdot 140-(28)^2}=-\frac47$, $m=\frac{7\cdot844-168\cdot28}{7\cdot 140-(28)^2}=\frac{43}{7}$.

Answer: the least square regression line has the form: $\hat{y}=\frac{43}{7}x-\frac47$. $x$ corresponds to the age and $\hat{y}$ corresponds to the height.