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Question 18 of 31

Q.For a bivariate data:
∑(x−xˉ)2=1200\sum(x - \bar{x})^2 = 1200, ∑(y−yˉ)2=300\sum(y - \bar{y})^2 = 300, ∑(x−xˉ)(y−yˉ)=−250\sum(x - \bar{x})(y - \bar{y}) = -250
Find:
byxb_{yx}
bxyb_{xy}
Correlation coefficient between x and y.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2023Subjective· 3mImportance★★★★★
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byx=∑(x−xˉ)(y−yˉ)∑(x−xˉ)2=−0.2083b_{yx} = \dfrac{\sum(x-\bar{x})(y-\bar{y})}{\sum(x-\bar{x})^2} = -0.2083, bxy=∑(x−xˉ)(y−yˉ)∑(y−yˉ)2=−0.8333b_{xy} = \dfrac{\sum(x-\bar{x})(y-\bar{y})}{\sum(y-\bar{y})^2} = -0.8333, and r=−byx bxy=−0.4167r = -\sqrt{b_{yx}\,b_{xy}} = -0.4167.

Given: ∑(x−xˉ)2=1200\sum(x-\bar{x})^2 = 1200, ∑(y−yˉ)2=300\sum(y-\bar{y})^2 = 300, ∑(x−xˉ)(y−yˉ)=−250\sum(x-\bar{x})(y-\bar{y}) = -250.

Step 1 — Regression coefficient of yy on xx:

byx=∑(x−xˉ)(y−yˉ)∑(x−xˉ)2=−2501200=−0.2083b_{yx} = \frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(x-\bar{x})^2} = \frac{-250}{1200} = -0.2083

Step 2 — Regression coefficient of xx on yy:

bxy=∑(x−xˉ)(y−yˉ)∑(y−yˉ)2=−250300=−0.8333b_{xy} = \frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(y-\bar{y})^2} = \frac{-250}{300} = -0.8333

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