For 10 observations on price (X) and supply (Y) the following data were
obtained (in appropriate units):
∑X = 130, ∑Y = 220, ∑X
2 = 2288, ∑Y
2 = 5506 and ∑XY = 3467.
Obtain the line of regression of Y on X and estimate the supply when the price is 16
units, and find out the standard error of the estimate.
The line of regression of Y on X is expressed as:
Y=a+bXY=a+bXY=a+bX
Where;
b=nΣXY−(ΣX)(ΣY)nΣX2−(ΣX)2b=\frac{n\Sigma XY-(\Sigma X)(\Sigma Y)}{n \Sigma X^2-(\Sigma X)^2}b=nΣX2−(ΣX)2nΣXY−(ΣX)(ΣY)
b=(10×3467)−(130)(220)10(2288)2−(130)2b=\frac{(10\times3467)-(130)(220)}{10 (2288)^2-(130)^2}b=10(2288)2−(130)2(10×3467)−(130)(220)
b=1.02b=1.02b=1.02
a=ΣY−bΣXna=\frac{\Sigma Y-b\Sigma X}{n}a=nΣY−bΣX
a=220−1.02×13010a=\frac{220-1.02\times130}{10}a=10220−1.02×130
a=8.74a=8.74a=8.74
Y=8.74+1.02XY=8.74+1.02XY=8.74+1.02X
If the price (X) is 16 units then supply (Y) is:
Y=8.74+1.02×16Y=8.74+1.02\times16Y=8.74+1.02×16
Y=25.06Y=25.06Y=25.06
The standard error of the estimate is:
s=ΣY2−aΣY−bΣXYn−2s=\sqrt{\frac{\Sigma Y^2-a\Sigma Y-b\Sigma XY}{n-2}}s=n−2ΣY2−aΣY−bΣXY
s=5506−(8.74×220)−(1.02×3467)10−2s=\sqrt{\frac{5506-(8.74\times220)-(1.02\times3467)}{10-2}}s=10−25506−(8.74×220)−(1.02×3467)
s=2.42s=2.42s=2.42
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