Question

(Ans: r = +0.97
16. Calculate Karl Pearson's correlation coefficient between X and Y from the following data:
N = 11, Ex = 117, Ex? 1313, Exy = 2827, 2y = 260, 2y2 : 6580 (Ans: r = 0.356
17. Calculate the coefficient of correlation between X and Y series from the following data;​

Answer 1

We recall the concept of Karl Pearson's coefficient between $x,y$

Karl Pearson's coefficient: It is defined as a linear correlation coefficient

Given:

$N=11,summation x=117, summation x^{2} =1313,\\Summation xy=2827,Ex=117,Ey=260$

Karl Pearson's coefficient $=\frac{11*2827-117*260}{(\sqrt{11*1313-117^{2} } )*\sqrt{11*6580-260^{2} } }$

                                           $=0.356$

Answer 2

Answer:

The coefficient of correlation between X and Y is r = 0.356.

Explanation:

From the above question,

They have given :

To calculate Karl Pearson's correlation coefficient (r) between X and Y, we can use the formula:

r = (NΣXY - ΣXΣY) / sqrt[(NΣ$X^2$ - (ΣX)) (NΣY$.^2$ - (ΣY)$.^2$)]

where N is the variety of facts points, ΣX is the sum of the X values, ΣY is the sum of the Y values, ΣXY is the sum of the merchandise of X and Y, Σ$X^2$ is the sum of the squares of X, and Σ$Y^2$ is the sum of the squares of Y.

Using the given data:

N = 11

ΣX = 117

Σ$X^2$ = 1313

ΣXY = 2827

2ΣY = 260 (since 2y = 260)

2Σ$Y^2$ = 6580 (since 2$y^2$ = 6580)

First, we want to calculate the sum of Y and the sum of $Y^2$:

ΣY = 260 / 2 = 130

ΣY$.^2$ = 6580 / 2 = 3290

Now we can plug in the values into the formula:

r = (11 * 2827 - 117 * 130) / sqrt[(11 * 1313 - $117^2$)(11 * 3290 - $130^2$)]

r = 0.356

Therefore, the coefficient of correlation between X and Y is r = 0.356.

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