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Tamil Nadu Board of Secondary EducationHSC Commerce Class 11

The following data relate to advertisement expenditure (in lakh of rupees) and their corresponding sales (in crores of rupees) - Business Mathematics and Statistics

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Question

The following data relate to advertisement expenditure (in lakh of rupees) and their corresponding sales (in crores of rupees)

Advertisement expenditure 40 50 38 60 65 50 35
Sales 38 60 55 70 60 48 30

Estimate the sales corresponding to advertising expenditure of ₹ 30 lakh.

Sum
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Solution

X Y X2 Y2 XY
40 38 1600 1444 1520
50 60 2500 3600 3000
38 55 1444 3025 2090
60 70 3600 4900 4200
65 60 4225 3600 3900
50 48 2500 2304 2400
35 32 1225 900 1050
338 361 17094 19773 18160

N = 7, ΣX = 338, ΣY = 361, ΣX2 = 17094, ΣY2 = 19773, ΣXY = 18160.

`bar"X" = (sum"X")/"N" = 338/7` = 48.29

`bar"Y" = (sum"Y")/"N" = 361/7` = 51.57

byx = `("N"sum"XY" - (sum"X")(sum"Y"))/("N"sum"X"^2 - (sum"X")^2)`

= `(7(18160) - (388)(361))/(7(17094) - (338)^2)`

= `(127120 - 122018)/(119658 - 114244)`

= `5102/5414`

= 0.942

Regression equation of Y on X

`"Y" - bar"Y" = "b"_"yx"("X" - bar"X")`

Y – 51.57 = 0.942(X – 48.29)

Y – 51.57 = 0.942X – 0.942 × 48.29

Y – 51.57 = 0.942X – 45.48918

Y = 0.942X + 51.57 – 48.29

Y = 0.942X + 6.081

To find the sales, when the advertising is X = ₹ 30 lakh in the above equation we get,

Y = 0.942(30) + 6.081

= 28.26 + 6.081

= 34.341

= ₹ 34.34 crores

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Regression Analysis
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Chapter 9: Correlation and Regression Analysis - Exercise 9.2 [Page 227]

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Samacheer Kalvi Business Mathematics and Statistics [English] Class 11 TN Board
Chapter 9 Correlation and Regression Analysis
Exercise 9.2 | Q 6 | Page 227

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