Question

problem 8
the table below shows the height of a tree from 1995 to 2001

years since 1995	0	1	3	5	6

regression equation:

using your regression equation approximately how tall was the tree in 2003?

using your regression equation approximately how tall will the tree be in 2030?

Explanation:

Step1: Assume linear regression model

Let the linear - regression equation be $y = ax + b$, where $x$ is the number of years since 1995 and $y$ is the height of the tree. We use the least - squares method. The formulas for $a$ and $b$ are:
$a=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}$ and $b=\bar{y}-a\bar{x}$, where $n$ is the number of data points.
We have $n = 5$, $\sum_{i = 1}^{5}x_i=0 + 1+3 + 5+6=15$, $\sum_{i = 1}^{5}y_i=5.1 + 6.4+9 + 11.6+12.9 = 45$, $\sum_{i = 1}^{5}x_i^{2}=0^{2}+1^{2}+3^{2}+5^{2}+6^{2}=0 + 1+9 + 25+36 = 71$, $\sum_{i = 1}^{5}x_iy_i=0\times5.1+1\times6.4 + 3\times9+5\times11.6+6\times12.9=0 + 6.4+27+58+77.4 = 168.8$.
$\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{15}{5}=3$, $\bar{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{45}{5}=9$.
$a=\frac{5\times168.8-15\times45}{5\times71 - 15^{2}}=\frac{844-675}{355 - 225}=\frac{169}{130}=1.3$.
$b=9-1.3\times3=9 - 3.9 = 5.1$.
So the regression equation is $y = 1.3x+5.1$.

Step2: Find height in 2003

The number of years since 1995 in 2003 is $x = 2003 - 1995=8$.
Substitute $x = 8$ into the regression equation: $y=1.3\times8 + 5.1=10.4+5.1 = 15.5$ feet.

Step3: Find height in 2030

The number of years since 1995 in 2030 is $x = 2030 - 1995 = 35$.
Substitute $x = 35$ into the regression equation: $y=1.3\times35+5.1=45.5+5.1 = 50.6$ feet.

Answer:

Regression Equation: $y = 1.3x+5.1$
Height in 2003: 15.5 feet
Height in 2030: 50.6 feet