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Question
Using the following information you are requested to
- obtain the linear regression of Y on X
- Estimate the level of defective parts delivered when inspection expenditure amounts to ₹ 82
∑X = 424, ∑Y = 363, ∑X2 = 21926, ∑Y2 = 15123, ∑XY = 12815, N = 10.
Here X is the expenditure on inspection, Y is the defective parts delivered.
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Solution
Given ∑X = 424, ∑Y = 363, ∑X2 = 21926, ∑Y2 = 15123, ∑XY = 12815, N = 10.
`bar"X" = (sum"X")/"N" = 424/10` = 42.4
`bar"Y" = (sum"Y")/"N" = 363/10` = 36.3
Regression coefficient of Y on X is
byx = `("N"sum"XY" - (sum"X")(sum"Y"))/("N"sum"X"^2 - (sum"X")^2)`
= `(10(12815) - (424)(363))/(10(21926) - (424)^2)`
= `(128150 - 153912)/(219260 - 179776)`
= `(-25762)/39484`
= − 0.652
∴ Regression line of Y on X is
`"Y" - bar"Y" = "b"_"yx"("X" - bar"X")`
Y − 36.3 = − 0.652 (X − 42.4)
Y − 36.3 = − 0.652X + 27.645
Y = − 0.652X + 63.945
When X = 82, Y = − 0.652(82) + 63.945
Y = − 53.464 + 63.945
Y = 10.481
Y ≈ 10
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