Correlation and RegressionMCQPYQ Dec 22Question 3715 of 188
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The equations of the two lines of regression are 4x+3y+7=0\displaystyle 4x + 3y + 7 = 0 and 3x+4y+8=0\displaystyle 3x + 4y + 8 = 0. Find the correlation coefficient between x\displaystyle x and y\displaystyle y?

Options

A-0.75
B0.25
C-0.92
D1.25
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Correct Answer

✅ Option a — -0.75

All Options:

  • A-0.75
  • B0.25
  • C-0.92
  • D1.25

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Detailed Solution & Explanation

Given the two regression equations: 1) 4x+3y+7=0\displaystyle 4x + 3y + 7 = 0 2) 3x+4y+8=0\displaystyle 3x + 4y + 8 = 0 We need to determine the regression lines of y\displaystyle y on x\displaystyle x and x\displaystyle x on y\displaystyle y. First, let's assume equation (1) is the regression line of y\displaystyle y on x\displaystyle x: 4x+3y+7=0  ⟹  3y=−4x−7  ⟹  y=−43x−734x + 3y + 7 = 0 \implies 3y = -4x - 7 \implies y = -\frac{4}{3}x - \frac{7}{3} This gives the regression coefficient of y\displaystyle y on x\displaystyle x as: byx=−43≈−1.33b_{yx} = -\frac{4}{3} \approx -1.33 Now, let's assume equation (2) is the regression line of x\displaystyle x on y\displaystyle y: 3x+4y+8=0  ⟹  3x=−4y−8  ⟹  x=−43y−833x + 4y + 8 = 0 \implies 3x = -4y - 8 \implies x = -\frac{4}{3}y - \frac{8}{3} This gives the regression coefficient of x\displaystyle x on y\displaystyle y as: bxy=−43≈−1.33b_{xy} = -\frac{4}{3} \approx -1.33 Now, let's check the product of these regression coefficients, which is r2\displaystyle r^2: r2=byx×bxy=(−43)×(−43)=169≈1.78>1r^2 = b_{yx} \times b_{xy} = \left(-\frac{4}{3}\right) \times \left(-\frac{4}{3}\right) = \frac{16}{9} \approx 1.78 > 1 Since the coefficient of determination r2\displaystyle r^2 must lie between 0\displaystyle 0 and 1\displaystyle 1 (0≤r2≤1\displaystyle 0 \le r^2 \le 1), this assumption is invalid. Therefore, we reverse the assumptions: Assume equation (1) is the regression line of x\displaystyle x on y\displaystyle y: 4x+3y+7=0  ⟹  4x=−3y−7  ⟹  x=−34y−744x + 3y + 7 = 0 \implies 4x = -3y - 7 \implies x = -\frac{3}{4}y - \frac{7}{4} Thus: bxy=−34=−0.75b_{xy} = -\frac{3}{4} = -0.75 Assume equation (2) is the regression line of y\displaystyle y on x\displaystyle x: 3x+4y+8=0  ⟹  4y=−3x−8  ⟹  y=−34x−23x + 4y + 8 = 0 \implies 4y = -3x - 8 \implies y = -\frac{3}{4}x - 2 Thus: byx=−34=−0.75b_{yx} = -\frac{3}{4} = -0.75 Now, let's calculate r2\displaystyle r^2: r2=byx×bxy=(−0.75)×(−0.75)=0.5625r^2 = b_{yx} \times b_{xy} = (-0.75) \times (-0.75) = 0.5625 This is valid since 0.5625≤1\displaystyle 0.5625 \le 1. The correlation coefficient r\displaystyle r is: r=±r2=±0.5625=±0.75r = \pm \sqrt{r^2} = \pm \sqrt{0.5625} = \pm 0.75 Since both byx\displaystyle b_{yx} and bxy\displaystyle b_{xy} are negative, the correlation coefficient r\displaystyle r must also be negative. Therefore, r=−0.75\displaystyle r = -0.75. Hence, **Option A** is the correct answer.

About This Chapter: Correlation and Regression

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Correlation Coefficient, Regression Equations

This chapter covers Correlation Coefficient, Regression Equations and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

Key Concepts to Understand

Correlation

A statistical measure that expresses the extent to which two variables are linearly related.

Regression

A statistical method used for estimating the relationships between a dependent variable and one or more independent variables.

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Correlation vs Regression: Statistics for CA Foundation

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