Answer:
[tex]b_0 = 16.471[/tex]
Step-by-step explanation:
Given
[tex]\begin{array}{ccccc}x & {15} & {12} & {10} & {7} \ \\ y & {5} & {7} & {9} & {11} \ \end{array}[/tex]
Required
The least square estimate [tex]b_0[/tex]
Calculate the mean of x
[tex]\bar x = \frac{\sum x}{n}[/tex]
[tex]\bar x = \frac{15+12+10+7}{4} =\frac{44}{4} = 11[/tex]
Calculate the mean of y
[tex]\bar y = \frac{\sum y}{n}[/tex]
[tex]\bar y = \frac{5+7+9+11}{4} =\frac{32}{4} = 8[/tex]
Calculate [tex]\sum(x - \bar x) * (y - \bar y)[/tex]
[tex]\sum(x - \bar x) = (15 - 11) * (5 - 8)+ (12 - 11) * (7 - 8) + (10 - 11) * (9 - 8)+ (7 - 11) * (11 - 8)[/tex]
[tex]\sum(x - \bar x) = -26[/tex]
Calculate [tex]\sum(x - \bar x)^2[/tex]
[tex]\sum(x - \bar x)^2 = (15 - 11)^2 + (12 - 11)^2 + (10 - 11)^2 + (7 - 11)^2[/tex]
[tex]\sum(x - \bar x)^2 = 34[/tex]
So:
[tex]b = \frac{\sum(x - \bar x) * (y - \bar y)}{\sum(x - \bar x)^2}[/tex]
[tex]b = \frac{-26}{34}[/tex]
[tex]b_0 = y - bx[/tex]
[tex]b_0 = 5 - \frac{-26}{34}*15[/tex]
[tex]b_0 = 5 + \frac{26*15}{34}[/tex]
[tex]b_0 = 5 + \frac{390}{34}[/tex]
Take LCM
[tex]b_0 = \frac{34*5+ 390}{34}[/tex]
[tex]b_0 = \frac{560}{34}[/tex]
[tex]b_0 = 16.471[/tex]
