Answer to Question #300567 in Statistics and Probability for Vamsi

Question #300567

The equations of two Regressions lines are




7π‘₯ βˆ’ 16𝑦 + 9 = 0 and 5𝑦 βˆ’ 4π‘₯ βˆ’ 3 = 0.




Find the coefficient of correlation and the means of x and y

1
Expert's answer
2022-02-22T12:01:30-0500

Solution:

"7\ud835\udc65 \u2212 16\ud835\udc66 + 9 = 0 \\ ...(i)\n\n\\\\5\ud835\udc66 \u2212 4\ud835\udc65 \u2212 3 = 0\\ ...(ii)"

From (i)

"16y=7x+9\n\\\\\\Rightarrow y=\\dfrac7{16}x+\\dfrac9{16}" ...(iii)

From (ii)

"5y=4x+3\n\\\\\\Rightarrow y=\\dfrac45x+\\dfrac35" ...(iv)

On equating these values of y,

"\\dfrac7{16}x+\\dfrac9{16}=\\dfrac45x+\\dfrac35\n\\\\ \\Rightarrow x=-\\dfrac{3}{29}"

Put this in (iii), we get,

"y=\\dfrac{15}{29}"

So, mean of x "=-\\dfrac3{29}"

And mean of y "=\\dfrac{15}{29}"

Next, slope of 1st line "=b_{yx}=\\dfrac7{16}" [From (iii)]

And slope of 2nd line "=b_{xy}=\\dfrac4{5}" [From (iv)]

Then, "r^2=b_{yx}b_{xy}=\\dfrac7{16}\\times \\dfrac4{5}=\\dfrac7{20}"

So, "r=\\pm\\sqrt{\\dfrac7{20}}=\\pm0.592"

But r has same sign as "b_{yx},b_{xy}" , so "r=0.592"


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