Question

lesson 11: equation of lines

20.2	89
18	94
11.8	82
19.5	84
15.1	86
22.7	103
23.7	96
24.8	91
12.5	74

a. determine a linear model for the cherry tree data (round values to three decimal places):
\\(\hat{y} = \square\\)

b. use the model to estimate the height of the tree when the diameter is 26.1 inches. (round result to 1 decimal place.)
height = \square feet

Explanation:

Step1: Define variables

Let $x$ = diameter (inches), $\hat{y}$ = height (feet)

Step2: Calculate $\bar{x}$ and $\bar{y}$

$\bar{x} = \frac{20.2+18+11.8+19.5+15.1+22.7+23.7+24.8+12.5}{9} = \frac{168.3}{9} = 18.7$
$\bar{y} = \frac{89+94+82+84+86+103+96+91+74}{9} = \frac{799}{9} \approx 88.7778$

Step3: Compute slope $b_1$

First calculate $\sum(x_i-\bar{x})(y_i-\bar{y})$ and $\sum(x_i-\bar{x})^2$:
$\sum(x_i-\bar{x})(y_i-\bar{y}) = (20.2-18.7)(89-88.7778)+(18-18.7)(94-88.7778)+(11.8-18.7)(82-88.7778)+(19.5-18.7)(84-88.7778)+(15.1-18.7)(86-88.7778)+(22.7-18.7)(103-88.7778)+(23.7-18.7)(96-88.7778)+(24.8-18.7)(91-88.7778)+(12.5-18.7)(74-88.7778)$
$= (1.5)(0.2222)+(-0.7)(5.2222)+(-6.9)(-6.7778)+(0.8)(-4.7778)+(-3.6)(-2.7778)+(4)(14.2222)+(5)(7.2222)+(6.1)(2.2222)+(-6.2)(-14.7778)$
$\approx 0.3333 - 3.6555 + 46.7668 - 3.8222 + 10.0001 + 56.8888 + 36.111 + 13.5554 + 91.6224 = 247.7991$

$\sum(x_i-\bar{x})^2 = (20.2-18.7)^2+(18-18.7)^2+(11.8-18.7)^2+(19.5-18.7)^2+(15.1-18.7)^2+(22.7-18.7)^2+(23.7-18.7)^2+(24.8-18.7)^2+(12.5-18.7)^2$
$= 2.25 + 0.49 + 47.61 + 0.64 + 12.96 + 16 + 25 + 37.21 + 38.44 = 180.6$

$b_1 = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} = \frac{247.7991}{180.6} \approx 1.372$

Step4: Compute intercept $b_0$

$b_0 = \bar{y} - b_1\bar{x} = 88.7778 - (1.372)(18.7) \approx 88.7778 - 25.6564 = 63.121$

Step5: Write linear model

$\hat{y} = 63.121 + 1.372x$

Step6: Estimate height for $x=26.1$

$\hat{y} = 63.121 + 1.372(26.1) = 63.121 + 35.8092 = 98.9302$
Round to 1 decimal place: $98.9$

Answer:

a. $\hat{y} = 63.121 + 1.372x$
b. height= $98.9$ feet