examine the data below for a stalk of corn.\n| day, x | 9 | 12 | 22 | 40 |\n| height, y (in) | 5 | 17 | 45 |…

Explanation:

Step1: Set up the system

Let $X = \ln(x)$. We have the data points: For $x = 9$, $X_1=\ln(9)\approx2.197$ and $y_1 = 5$; For $x = 12$, $X_2=\ln(12)\approx2.485$ and $y_2 = 17$; For $x = 22$, $X_3=\ln(22)\approx3.091$ and $y_3 = 45$; For $x = 40$, $X_4=\ln(40)\approx3.689$ and $y_4 = 60$. The equation $y=a + bX$ gives the following system of linear - equations for the least - squares method. The normal equations for linear regression $y=a + bX$ are: $\sum_{i = 1}^{n}y_i=na + b\sum_{i = 1}^{n}X_i$ and $\sum_{i = 1}^{n}X_iy_i=a\sum_{i = 1}^{n}X_i + b\sum_{i = 1}^{n}X_i^{2}$. $n = 4$, $\sum_{i=1}^{4}y_i=5 + 17+45 + 60=127$, $\sum_{i = 1}^{4}X_i=2.197+2.485 + 3.091+3.689 = 11.462$, $\sum_{i = 1}^{4}X_i^{2}=2.197^{2}+2.485^{2}+3.091^{2}+3.689^{2}\approx4.827+6.175+9.554+13.609 = 34.165$, $\sum_{i = 1}^{4}X_iy_i=2.197\times5+2.485\times17+3.091\times45+3.689\times60=10.985 + 42.245+139.095+221.34=413.665$.

Step2: Solve the normal equations

The first normal equation is $127 = 4a+11.462b$. The second normal equation is $413.665 = 11.462a+34.165b$. From the first equation $a=\frac{127 - 11.462b}{4}$. Substitute $a$ into the second equation: [ \begin{align*} 413.665&=11.462\times\frac{127 - 11.462b}{4}+34.165b\ 413.665\times4&=11.462\times(127 - 11.462b)+34.165b\times4\ 1654.66&=11.462\times127-11.462\times11.462b + 136.66b\ 1654.66&=1455.674-(131.37 - 136.66)b\ 1654.66 - 1455.674&=- 5.29b\ 198.986&=- 5.29b\ b&\approx29.97 \end{align*} ] Substitute $b\approx29.97$ into $a=\frac{127 - 11.462b}{4}$: [ \begin{align*} a&=\frac{127-11.462\times29.97}{4}\ &=\frac{127 - 343.51}{4}\ &=\frac{- 216.51}{4}\ &\approx - 54.13 \end{align*} ]

Answer:

$a\approx - 54.13$ $b\approx29.97$