Correlation and RegressionMCQMTP June 22Question 3696 of 188
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For 10\displaystyle 10 pairs of observations, number of concurrent deviations was found to be 4\displaystyle 4. What is the value of the coefficient of concurrent deviation?

Options

A0.2\displaystyle \sqrt{0.2}
B1/3\displaystyle 1/3
C1/3\displaystyle -1/3
D0.2\displaystyle -\sqrt{0.2}
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Correct Answer

Option c1/3\displaystyle -1/3

All Options:

  • A0.2\displaystyle \sqrt{0.2}
  • B1/3\displaystyle 1/3
  • C1/3\displaystyle -1/3
  • D0.2\displaystyle -\sqrt{0.2}

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Detailed Solution & Explanation

The coefficient of concurrent deviation (rc\displaystyle r_c) is calculated using the formula: rc=±±2cmmr_c = \pm \sqrt{\pm \frac{2c - m}{m}} where: - n=10\displaystyle n = 10 is the number of pairs of observations. - m=n1=101=9\displaystyle m = n - 1 = 10 - 1 = 9 is the number of deviations. - c=4\displaystyle c = 4 is the number of concurrent deviations. First, let's calculate the value of 2cm\displaystyle 2c - m: 2cm=2(4)9=89=12c - m = 2(4) - 9 = 8 - 9 = -1 Since 2cm<0\displaystyle 2c - m < 0 (negative), we use the negative sign both inside and outside the square root to keep the value real and preserve the sign: rc=2cmmr_c = - \sqrt{- \frac{2c - m}{m}} rc=19=19=13r_c = - \sqrt{- \frac{-1}{9}} = - \sqrt{\frac{1}{9}} = -\frac{1}{3} *Note: The source file incorrectly lists Option B (1/3\displaystyle 1/3) as the correct answer. However, because 2cm=1\displaystyle 2c - m = -1 is negative, the coefficient of concurrent deviation must also be negative, which is -\frac{1}{3} (Option C).* Hence, **Option C** 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.

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