Find the Linear Correlation Coefficient table[[x,y],[0,4],[1,12],[2,36],[3,108]]

Problem

r=(n*(∑_^)(x)*y−((∑_^)(x))*((∑_^)(y)))/√(,[n*(∑_^)(x^2)−((∑_^)(x))^2]*[n*(∑_^)(y^2)−((∑_^)(y))^2])

Solution

1. Identify the given data points (x,y) and the sample size n

(0,4),(1,12),(2,36),(3,108)

n=4

2. Calculate the sum of x and the sum of y

(∑_^)(x)=0+1+2+3=6

(∑_^)(y)=4+12+36+108=160

3. Calculate the sum of x^2 and the sum of y^2

(∑_^)(x^2)=0^2+1^2+2^2+3^2=14

(∑_^)(y^2)=4^2+12^2+36^2+108^2=13120

4. Calculate the sum of the products x*y

(∑_^)(x)*y=(0⋅4)+(1⋅12)+(2⋅36)+(3⋅108)=408

5. Substitute the values into the numerator of the correlation coefficient formula.

Numerator=4*(408)−(6)*(160)

Numerator=1632−960=672

6. Substitute the values into the denominator of the correlation coefficient formula.

Denominator=√(,[4*(14)−(6)^2]*[4*(13120)−(160)^2])

Denominator=√(,[56−36]*[52480−25600])

Denominator=√(,[20]*[26880])

Denominator=√(,537600)≈733.2121

7. Divide the numerator by the denominator to find r

r=672/√(,537600)

r≈0.9165

Final Answer

r=(7√(,105))/80≈0.9165

Want more problems? Check here!