The height (x) in cm and weight (y) in kg are given for 5 persons in a community. X 160 162 165 161 162 : : : : : : Y 63 62 64 60 61 Find the correlation coefficient between X and Y and regression…

Solution

Step 1: Calculate the necessary sums

Given the data:

We compute the following sums:

1. Mean of $X$ and $Y$: $$\bar{X}=\frac{160+162+165+161+162}{5}=\frac{810}{5}=162$$ $$\bar{Y}=\frac{63+62+64+60+61}{5}=\frac{310}{5}=62$$

2. Sum of deviations and their products: $$\sum (X-\bar{X})=(160-162)+(162-162)+(165-162)+(161-162)+(162-162)=-2+0+3-1+0=0$$ $$\sum (Y-\bar{Y})=(63-62)+(62-62)+(64-62)+(60-62)+(61-62)=1+0+2-2-1=0$$ $$\sum (X-\bar{X})(Y-\bar{Y})=(-2)(1)+(0)(0)+(3)(2)+(-1)(-2)+(0)(-1)=-2+0+6+2+0=6$$ $$\sum (X-\bar{X}{)}^2=(-2{)}^2+(0{)}^2+(3{)}^2+(-1{)}^2+(0{)}^2=4+0+9+1+0=14$$ $$\sum (Y-\bar{Y}{)}^2=(1{)}^2+(0{)}^2+(2{)}^2+(-2{)}^2+(-1{)}^2=1+0+4+4+1=10$$

Step 2: Compute the correlation coefficient ($r$)

The formula for the correlation coefficient is: $$r=\frac{\sum (X-\bar{X})(Y-\bar{Y})}{\sqrt{\sum (X-\bar{X}{)}^2\sum (Y-\bar{Y}{)}^2}}$$ Substituting the values: $$r=\frac{6}{\sqrt{14\times 10}}=\frac{6}{\sqrt{140}}=\frac{6}{11.83}\approx 0.507$$

Step 3: Compute the regression coefficient ($b_{YX}$)

The formula for the regression coefficient of $Y$ on $X$ is: $$b_{YX}=\frac{\sum (X-\bar{X})(Y-\bar{Y})}{\sum (X-\bar{X}{)}^2}$$ Substituting the values: $$b_{YX}=\frac{6}{14}=0.4286$$

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Final Answers

• Correlation coefficient ($r$): $0.507$
• Regression coefficient ($b_{YX}$): $0.4286$