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Exercises · Q6
Q.

The following data relate to advertising expenditure (XX, in ₹ lakh) and sales (YY, in ₹ lakh) of a firm for 5 years:

X35689
Y23465

Find the regression equation of YY on XX and of XX on YY. Estimate the sales when advertising expenditure is ₹10 lakh.

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Step 1 — Basic sums. n=5n=5; ΣX=3+5+6+8+9=31⇒Xˉ=31/5=6.2\Sigma X = 3+5+6+8+9=31 \Rightarrow \bar X = 31/5 = 6.2; ΣY=2+3+4+6+5=20⇒Yˉ=4\Sigma Y = 2+3+4+6+5=20 \Rightarrow \bar Y = 4.

ΣXY=(3)(2)+(5)(3)+(6)(4)+(8)(6)+(9)(5)=6+15+24+48+45=138\Sigma XY = (3)(2)+(5)(3)+(6)(4)+(8)(6)+(9)(5) = 6+15+24+48+45 = 138.

ΣX2=9+25+36+64+81=215\Sigma X^{2} = 9+25+36+64+81 = 215.

ΣY2=4+9+16+36+25=90\Sigma Y^{2} = 4+9+16+36+25 = 90.

Step 2 — byxb_{yx}.

byx=5(138)−(31)(20)5(215)−(31)2=690−6201075−961=70114≈0.614b_{yx} = \frac{5(138)-(31)(20)}{5(215)-(31)^{2}} = \frac{690-620}{1075-961} = \frac{70}{114} \approx 0.614

Step 3 — bxyb_{xy}.

bxy=5(138)−(31)(20)5(90)−(20)2=690−620450−400=7050=1.4b_{xy} = \frac{5(138)-(31)(20)}{5(90)-(20)^{2}} = \frac{690-620}{450-400} = \frac{70}{50} = 1.4

Step 4 — Regression equations.

YY on XX: Y−4=0.614(X−6.2)=0.614X−3.807⇒Y=0.614X+0.193Y - 4 = 0.614(X-6.2) = 0.614X - 3.807 \Rightarrow Y = 0.614X + 0.193

XX on YY: X−6.2=1.4(Y−4)=1.4Y−5.6⇒X=1.4Y+0.6X - 6.2 = 1.4(Y-4) = 1.4Y - 5.6 \Rightarrow X = 1.4Y + 0.6

Step 5 — Independent check. byx⋅bxy=0.614×1.4=0.860≤1b_{yx}\cdot b_{xy} = 0.614\times1.4 = 0.860 \le 1 ✓ (valid); r=0.860≈0.927r=\sqrt{0.860}\approx0.927, a strong positive relationship, consistent with sales rising with advertising in the data.

Step 6 — Prediction. At X=10X=10: Y=0.614(10)+0.193=6.14+0.193=6.33Y = 0.614(10)+0.193 = 6.14+0.193 = 6.33 (₹ lakh, approx.).

✓Final answer

Y=0.614X+0.193Y = 0.614X+0.193; estimated sales at ₹10 lakh advertising ≈\approx ₹6.33 lakh.

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