Question

You are given the following information about x and y.

Answer 1

Answer:

$b_0 = 16.471$

Step-by-step explanation:

Given

$\begin{array}{ccccc}x & {15} & {12} & {10} & {7} \ \\ y & {5} & {7} & {9} & {11} \ \end{array}$

Required

The least square estimate $b_0$

Calculate the mean of x

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{15+12+10+7}{4} =\frac{44}{4} = 11$

Calculate the mean of y

$\bar y = \frac{\sum y}{n}$

$\bar y = \frac{5+7+9+11}{4} =\frac{32}{4} = 8$

Calculate $\sum(x - \bar x) * (y - \bar y)$

$\sum(x - \bar x) = (15 - 11) * (5 - 8)+ (12 - 11) * (7 - 8) + (10 - 11) * (9 - 8)+ (7 - 11) * (11 - 8)$

$\sum(x - \bar x) = -26$

Calculate $\sum(x - \bar x)^2$

$\sum(x - \bar x)^2 = (15 - 11)^2 + (12 - 11)^2 + (10 - 11)^2 + (7 - 11)^2$

$\sum(x - \bar x)^2 = 34$

So:

$b = \frac{\sum(x - \bar x) * (y - \bar y)}{\sum(x - \bar x)^2}$

$b = \frac{-26}{34}$

$b_0 = y - bx$

$b_0 = 5 - \frac{-26}{34}*15$

$b_0 = 5 + \frac{26*15}{34}$

$b_0 = 5 + \frac{390}{34}$

Take LCM

$b_0 = \frac{34*5+ 390}{34}$

$b_0 = \frac{560}{34}$

$b_0 = 16.471$