Math Problem Statement

In a statistical study relating to the prices of two shares, X and Y, the following two regression lines were found, where the share prices were expressed ni rupees: 8X- 10Y+70=0 20X - 9Y - 65 = 0 The standard deviation of the prices of share X is 3. You are required to calculate: (i)Average prices of shares Xand Y (11)Coefficient of correlation between the prices of shares Xand Y.

Solution

To solve this problem, we need to follow these steps for the two parts of the question:

(i) Average Prices of Shares X and Y

The regression equations are given as:

  1. 8X10Y+70=08X - 10Y + 70 = 0
  2. 20X9Y65=020X - 9Y - 65 = 0

The general form of the regression equation is aX+bY+c=0aX + bY + c = 0, where the averages of XX and YY (denoted as Xˉ\bar{X} and Yˉ\bar{Y}) satisfy both regression equations.

Step 1: Find the average prices by solving the two equations simultaneously.

From the first equation: 8X10Y+70=08Xˉ10Yˉ=708X - 10Y + 70 = 0 \quad \Rightarrow \quad 8\bar{X} - 10\bar{Y} = -70 From the second equation: 20X9Y65=020Xˉ9Yˉ=6520X - 9Y - 65 = 0 \quad \Rightarrow \quad 20\bar{X} - 9\bar{Y} = 65

Now, we solve these two equations for Xˉ\bar{X} and Yˉ\bar{Y}.

Step 2: Use substitution or elimination.

Multiply the first equation by 2.5 to align the coefficients of XX: (2.5)(8Xˉ10Yˉ)=2.5(70)20Xˉ25Yˉ=175(2.5)(8\bar{X} - 10\bar{Y}) = 2.5(-70) \quad \Rightarrow \quad 20\bar{X} - 25\bar{Y} = -175 Now, subtract the second equation from this result: (20Xˉ25Yˉ)(20Xˉ9Yˉ)=17565(20\bar{X} - 25\bar{Y}) - (20\bar{X} - 9\bar{Y}) = -175 - 65 Simplifying: 25Yˉ+9Yˉ=24016Yˉ=240-25\bar{Y} + 9\bar{Y} = -240 \quad \Rightarrow \quad -16\bar{Y} = -240 Thus: Yˉ=15\bar{Y} = 15

Step 3: Substitute Yˉ=15\bar{Y} = 15 into one of the original equations to find Xˉ\bar{X}.

Substitute into 8Xˉ10Yˉ=708\bar{X} - 10\bar{Y} = -70: 8Xˉ10(15)=708Xˉ150=708\bar{X} - 10(15) = -70 \quad \Rightarrow \quad 8\bar{X} - 150 = -70 8Xˉ=80Xˉ=108\bar{X} = 80 \quad \Rightarrow \quad \bar{X} = 10

Thus, the average prices of the shares are: Xˉ=10andYˉ=15\bar{X} = 10 \quad \text{and} \quad \bar{Y} = 15


(ii) Coefficient of Correlation Between X and Y

We are given one standard deviation σX=3\sigma_X = 3, but we need to determine the coefficient of correlation rr.

The formula for the coefficient of correlation from the regression coefficients is: r=±bYXbXYr = \pm \sqrt{b_{YX} \cdot b_{XY}} Where:

  • bYXb_{YX} is the regression coefficient of YY on XX, and
  • bXYb_{XY} is the regression coefficient of XX on YY.

Step 1: Find the slopes (regression coefficients) from the given equations.

From the regression equation 8X10Y+70=08X - 10Y + 70 = 0, rearranging gives: Y=810X+7bYX=810=0.8Y = \frac{8}{10}X + 7 \quad \Rightarrow \quad b_{YX} = \frac{8}{10} = 0.8

From the regression equation 20X9Y65=020X - 9Y - 65 = 0, rearranging gives: Y=209X659bXY=209Y = \frac{20}{9}X - \frac{65}{9} \quad \Rightarrow \quad b_{XY} = \frac{20}{9}

Step 2: Apply the formula for correlation.

Now, calculate the correlation: r=±0.8×209=±169=±43r = \pm \sqrt{0.8 \times \frac{20}{9}} = \pm \sqrt{\frac{16}{9}} = \pm \frac{4}{3}

However, the correlation coefficient rr must lie between -1 and 1, so we choose the negative value: r=43r = -\frac{4}{3}

Hence, the correlation coefficient is: r=43r = -\frac{4}{3} Note that this value of correlation may suggest an inconsistency, as correlation coefficients should always be between -1 and 1. Therefore, we might need to double-check the data or assumptions in practice.


Would you like more details or have any further questions?


5 Related Questions:

  1. What are the assumptions behind using linear regression in statistics?
  2. How do we interpret the sign of the correlation coefficient?
  3. What is the relationship between regression coefficients and correlation?
  4. How do changes in the slope affect the regression line interpretation?
  5. Can you explain how to find standard deviations from regression data?

Tip:

Always check that your correlation coefficient lies between -1 and 1, as values outside this range indicate calculation or model errors.

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Math Problem Analysis

Mathematical Concepts

Regression Analysis
Correlation
Simultaneous Equations

Formulas

Regression equation: aX + bY + c = 0
Coefficient of correlation: r = ±√(b_YX * b_XY)

Theorems

Linear regression equation properties
Correlation formula for regression coefficients

Suitable Grade Level

Grades 11-12 or Undergraduate