The following data relate to advertising expenditure (X, in ₹ lakh) and sales (Y, in ₹ lakh) of a firm for 5 years:
| X | 3 | 5 | 6 | 8 | 9 |
|---|---|---|---|---|---|
| Y | 2 | 3 | 4 | 6 | 5 |
Find the regression equation of Y on X and of X on Y. Estimate the sales when advertising expenditure is ₹10 lakh.
Concept understanding — Regression Lines (Y on X and X on Y)
There are two regression lines for any bivariate data set. The line of Y on X, (Y−Yˉ)=byx(X−Xˉ), minimizes vertical deviations and estimates Y from X. The line of X on Y, (X−Xˉ)=bxy(Y−Yˉ), minimizes horizontal deviations and estimates X from Y. Both pass through (Xˉ,Yˉ) and coincide only when r=±1.
With n=5, ΣX=31, ΣY=20, ΣXY=138, ΣX2=215, ΣY2=90, the direct-method formulas give byx≈0.614 and bxy=1.4.
Y on X: Y≈0.614X+0.193.
Estimated sales at X=10 lakh is about ₹6.33 lakh.
Step 1 — Basic sums. n=5; ΣX=3+5+6+8+9=31⇒Xˉ=31/5=6.2; ΣY=2+3+4+6+5=20⇒Yˉ=4.
ΣXY=(3)(2)+(5)(3)+(6)(4)+(8)(6)+(9)(5)=6+15+24+48+45=138.
ΣX2=9+25+36+64+81=215.
ΣY2=4+9+16+36+25=90.
Step 2 — byx.
byx=5(215)−(31)25(138)−(31)(20)=1075−961690−620=11470≈0.614
Step 3 — bxy.
bxy=5(90)−(20)25(138)−(31)(20)=450−400690−620=5070=1.4
Step 4 — Regression equations.
Y on X: Y−4=0.614(X−6.2)=0.614X−3.807⇒Y=0.614X+0.193
X on Y: X−6.2=1.4(Y−4)=1.4Y−5.6⇒X=1.4Y+0.6
Step 5 — Independent check. byx⋅bxy=0.614×1.4=0.860≤1 ✓ (valid); r=0.860≈0.927, a strong positive relationship, consistent with sales rising with advertising in the data.
Step 6 — Prediction. At X=10: Y=0.614(10)+0.193=6.14+0.193=6.33 (₹ lakh, approx.).
Y=0.614X+0.193; estimated sales at ₹10 lakh advertising ≈ ₹6.33 lakh.
Alternatively, compute deviations from the actual means (x=X−6.2, y=Y−4) and use byx=Σxy/Σx2 — arithmetically equivalent to the direct method above but requires decimal deviations since Xˉ=6.2 is not a whole number, which is exactly why the direct (raw-score) method is preferred here.
Rounding byx too early and carrying the rounding error through the equation; mixing up which sum of squares (ΣX2 or ΣY2) belongs in each coefficient's denominator.
- CBSE 2026Set MARCH1 markMCQQ.The best fitted line of regression can be obtained by which method?(a) Least Square Method(b) Karl Pearson's Method(c) Maximum Square Method(d) Bowley's Method
›Reveal solutionSolution
The best-fitted regression line is found by the method of least squares.
The regression line is chosen so that the sum of the squares of the deviations of the observed y from the estimated y^ is minimum, i.e. Σ(y−y^)2 is least. This is the least square method; Karl Pearson's method is for correlation, not for fitting a line.
✓Final answer(a) Least Square Method.
- CBSE 2026Set MARCH1 markMCQQ.The regression line of Y on X is y^=30−1.5x. What is the value of yˉ if xˉ=10?(a) 28.5(b) 20(c) 15(d) 45
›Reveal solutionSolution
The regression line passes through (xˉ,yˉ); putting xˉ=10 gives yˉ=15.
The regression line of Y on X always passes through the point of averages (xˉ,yˉ). Substituting xˉ=10 in y^=30−1.5x:
yˉ=30−1.5(10)=30−15=15.
✓Final answer(c) 15.
- CBSE 2025Set MARCH1 markMCQQ.The regression line always passes through which point?(a) (xˉ,yˉ)(b) (0,yˉ)(c) (xˉ,0)(d) (0,0)
›Reveal solutionSolution
A regression line always passes through the point of means (xˉ,yˉ) — option (a).
In the GSEB Class-12 Statistics Linear Regression chapter, the regression line of Y on X is written as:
y^−yˉ=byx(x−xˉ)
Substituting x=xˉ gives y^=yˉ, so the line passes through (xˉ,yˉ). The same is true for the line of X on Y; hence both lines intersect at the mean point.
✓Final answer(a) (xˉ,yˉ).
- CBSE 2022Set MARCH1 markMCQQ.The regression line always passes through which point?(a) (xˉ,yˉ)(b) (0,yˉ)(c) (xˉ,0)(d) (0,0)
›Reveal solutionSolution
Each regression line is constructed to pass through the point of averages (xˉ,yˉ), which is also where the two regression lines intersect.
Reasoning. The regression line of Y on X is y−yˉ=byx(x−xˉ) and of X on Y is x−xˉ=bxy(y−yˉ). Substituting x=xˉ,y=yˉ satisfies both equations, so both lines pass through (xˉ,yˉ).
✓Final answerOption (a) (xˉ,yˉ).
- CBSE 2022Set MARCH1 markQ.Give the name of a method to obtain the best fitted regression line.
›Reveal solutionSolution
The best-fitted regression line is obtained by the Method of Least Squares.
Explanation. The method of least squares chooses the line for which the sum of the squares of the deviations (errors) of the observed values from the estimated values is minimum. This criterion yields the best-fitted regression line of Y on X (and, similarly, of X on Y).
✓Final answerThe Method of Least Squares.
- CBSE 2020Set MARCH1 markMCQQ.What is the error e in estimation in case of regression line of Y on X?(a) y−y^(b) x^−y^(c) x−x^(d) y^−y
›Reveal solutionSolution
Error of estimation e=y−y^ (observed minus estimated) — option (a).
For the regression line of Y on X, we estimate Y as y^=a+bx. For a given observation the error (residual) of estimation is defined as the actual observed value of Y minus the value estimated by the line:
e=y−y^
The method of least squares chooses a and b so that ∑e2=∑(y−y^)2 is minimum. The other options either interchange the sign or use the X-on-Y residual.
✓Final answerThe error e=y−y^ — option (a).
- CBSE 2020Set MARCH1 markQ.What are the constants a and b in the regression line y^=a+bx?
›Reveal solutionSolution
a = intercept (y^ at x=0); b = regression coefficient byx = slope = change in y^ per unit change in x.
In the regression line of Y on X,
y^=a+bx,
- a is the intercept constant: the value the line predicts for y^ when x=0. It fixes the position of the line on the Y-axis.
- b is the regression coefficient of Y on X, denoted byx: it is the slope of the line, i.e. the amount by which the estimated y^ changes for a one-unit increase in x.
Both are determined by the method of least squares from the data, using
b=byx=∑(x−xˉ)2∑(x−xˉ)(y−yˉ),a=yˉ−bxˉ.
✓Final answera = intercept (value of y^ at x=0); b = slope / regression coefficient byx (change in y^ per unit change in x).
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