Two lines of regression are 5x – 8y + 17 = 0, 2x – 5y + 14 = 0. If x = 16, then find: i] y ii] the coefficient of correlation.
The Correct Answer and Explanation is:
We are given two regression lines:
- 5x – 8y + 17 = 0 → Regression line of x on y
- 2x – 5y + 14 = 0 → Regression line of y on x
We are to find:
i) Value of y when x = 16
Since we are given the regression line of y on x as:2x−5y+14=02x – 5y + 14 = 02x−5y+14=0
Substitute x=16x = 16x=16 into the equation:2(16)−5y+14=0⇒32−5y+14=0⇒46−5y=0⇒5y=46⇒y=465=9.22(16) – 5y + 14 = 0 \Rightarrow 32 – 5y + 14 = 0 \Rightarrow 46 – 5y = 0 \Rightarrow 5y = 46 \Rightarrow y = \frac{46}{5} = 9.22(16)−5y+14=0⇒32−5y+14=0⇒46−5y=0⇒5y=46⇒y=546=9.2
ii) Coefficient of Correlation (r)
The two regression lines are:
- Regression of x on y: 5x−8y+17=05x – 8y + 17 = 05x−8y+17=0
- Regression of y on x: 2x−5y+14=02x – 5y + 14 = 02x−5y+14=0
Step 1: Express each regression line in slope form
Regression of x on y:5x=8y−17⇒x=85y−175⇒slope (bxy)=855x = 8y – 17 \Rightarrow x = \frac{8}{5}y – \frac{17}{5} \Rightarrow \text{slope } (b_{xy}) = \frac{8}{5}5x=8y−17⇒x=58y−517⇒slope (bxy)=58
Regression of y on x:2x−5y+14=0⇒5y=2x+14⇒y=25x+145⇒slope (byx)=252x – 5y + 14 = 0 \Rightarrow 5y = 2x + 14 \Rightarrow y = \frac{2}{5}x + \frac{14}{5} \Rightarrow \text{slope } (b_{yx}) = \frac{2}{5}2x−5y+14=0⇒5y=2x+14⇒y=52x+514⇒slope (byx)=52
Step 2: Use the formula for the correlation coefficient rrr:
r=bxy⋅byx=85⋅25=1625=45=0.8r = \sqrt{b_{xy} \cdot b_{yx}} = \sqrt{\frac{8}{5} \cdot \frac{2}{5}} = \sqrt{\frac{16}{25}} = \frac{4}{5} = 0.8r=bxy⋅byx=58⋅52=2516=54=0.8
Note: Since both regression coefficients are positive, the correlation coefficient is also positive.
✅ Final Answers:
i) y=9.2y = 9.2y=9.2
ii) Coefficient of correlation r=0.8r = 0.8r=0.8
📘 Textbook-Style Explanation
In statistics, the regression lines are used to estimate the relationship between two variables. Given two regression equations, one for xxx on yyy and the other for yyy on xxx, we can determine the correlation between the variables and also predict the value of one variable given the other.
The regression line of yyy on xxx is used to predict the value of yyy when xxx is known. Substituting x=16x = 16x=16 into the equation 2x−5y+14=02x – 5y + 14 = 02x−5y+14=0, we solve for yyy:2(16)−5y+14=0⇒46−5y=0⇒y=465=9.22(16) – 5y + 14 = 0 \Rightarrow 46 – 5y = 0 \Rightarrow y = \frac{46}{5} = 9.22(16)−5y+14=0⇒46−5y=0⇒y=546=9.2
This gives us the estimated value of yyy when x=16x = 16x=16.
To compute the coefficient of correlation rrr, we first identify the slopes of the regression lines. The regression line of xxx on yyy is rearranged as:x=85y−175⇒bxy=85x = \frac{8}{5}y – \frac{17}{5} \Rightarrow b_{xy} = \frac{8}{5}x=58y−517⇒bxy=58
Similarly, the regression line of yyy on xxx is:y=25x+145⇒byx=25y = \frac{2}{5}x + \frac{14}{5} \Rightarrow b_{yx} = \frac{2}{5}y=52x+514⇒byx=52
The correlation coefficient is given by:r=bxy⋅byx=85⋅25=1625=0.8r = \sqrt{b_{xy} \cdot b_{yx}} = \sqrt{\frac{8}{5} \cdot \frac{2}{5}} = \sqrt{\frac{16}{25}} = 0.8r=bxy⋅byx=58⋅52=2516=0.8
Since both slopes are positive, the correlation coefficient is also positive. Thus, the variables are moderately and positively correlated.
