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Question 14 of 31

Q.Find the Regression lines X on Y from the following data : N=25N = 25, ∑X=125\sum X = 125, ∑Y=100\sum Y = 100, ∑X2=650\sum X^2 = 650, ∑Y2=436\sum Y^2 = 436, ∑XY=520\sum XY = 520.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2020Subjective· 2mImportance★★★★★
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Compute means (Xˉ=5, Yˉ=4\bar X=5,\ \bar Y=4) and bxy=59b_{xy}=\dfrac{5}{9}, then write X−Xˉ=bxy(Y−Yˉ)X-\bar X=b_{xy}(Y-\bar Y): X=0.556Y+2.778X=0.556Y+2.778.

This is a regression-line problem from the Correlation and Regression unit of the Tamil Nadu HSC Class-11 Business Statistics syllabus.

Step 1 — Means. Xˉ=∑XN=12525=5,Yˉ=∑YN=10025=4.\bar X=\dfrac{\sum X}{N}=\dfrac{125}{25}=5,\qquad \bar Y=\dfrac{\sum Y}{N}=\dfrac{100}{25}=4.

Step 2 — Regression coefficient of XX on YY.

bxy=N∑XY−∑X∑YN∑Y2−(∑Y)2=25(520)−125(100)25(436)−(100)2=13000−1250010900−10000=500900=59.b_{xy}=\frac{N\sum XY-\sum X\sum Y}{N\sum Y^2-(\sum Y)^2}=\frac{25(520)-125(100)}{25(436)-(100)^2}=\frac{13000-12500}{10900-10000}=\frac{500}{900}=\frac{5}{9}.

Step 3 — Equation of the line XX on YY. …

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