Study Correlation Coefficient

In this article we will discuss about study correlation coefficient. The correlation coefficient, an idea from statistics calculates of how well trends in the expected values follow trends in past real values. The correlation coefficient is a value between 0 and 1. If there is no association between the predicted values and the definite values the correlation coefficient is 0 or very low.

Formula for study correlation coefficient:

`"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`

Where
              N = Total number of values
              X = 1st achieve
              Y = 2nd achieve
            sum XY = Addition of the product of 1st and 2nd achieve
              sum X = Addition of 1st achieve
              sum y = Addition of 2nd achieve
              sum x² = Addition of square 1st achieve
              sum y² = Addition of square 2nd get achieve

Example problems for study correlation coefficient:

Study Correlation coefficient – Example 1:
Calculate the Correlation co-efficient of following table.

X	60	61	62	63	64
Y	3.1	3.6	3.8	4	4.1

Solution 1 for study correlation coefficient:
            Step 1:  Count the number of values.
                         N = 5
            Step 2:  Calculate XY, X², Y²
                         See the below table

X	Y	x*y	x*x = x^2	y*y = y^2
60	3.1	60*3.1=186	60*60=3600	3.1*.31=9.61
61	3.6	61*3.6=219.6	61*61=3721	3.6*3.6=12.96
62	3.8	62*3.8=235.6	62*62=3844	3.8*3.8=14.44
63	4	63*4   =252	63*63=3969	4*4 = 16
64	4.1	64*4.1=262.4	64*64=4096	4.1*4.1=16.81

       Step 3:

       Find sumX, sumy, sumxy, sumx², sumy².

        Sum x = 310
        sum y = 18.6
        sum xy = 1155.6
        sum x² = 19230
        sum y² = 69.82

      Step 4:

       Now, Substitute in the above formula given.

            `"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`

             = `[5(1155.6) - ((310)(18.6)) / ((sqrt([5(19230)-(310)^2][5(69.82)-(18.6)^2]))]]`
              = `5778 - 5766 / sqrt(([96150 - 96100] [349.1-345.96]))`
              = `12 / sqrt(50 x 3.14 )`
              = `12/ sqrt(157)`
            = `12/12.5299`
            = 0.9577
            Answer is 0.9577
Study Correlation coefficient – Example 2:
         Find the Correlation co-efficient of following table

X	Y
50	3
51	3
52	3
53	4
54	4

        


 Solution for study correlation coefficient:
        Step 1: Count the number of values.
            N = 5
        Step 2: Find XY, X², Y²
            See the below table

X value	Y value	x* y	x*x	y*y
70	3	70*3=210	70*70=4900	3*3=9
71	3	71*3=213	71*71=5041	3*3=9
72	3	72*3=216	72*72=5184	3*3=9
73	4	73*4=292	73*73=5329	4*4=16
74	4	74*4=296	74*74=5476	4*4=16

          Step 3: Find sum X, sum y, sum xy, sum x², sum y².
            sum x = 360  
            sum y = 17
            sum xy = 1227
            sum x² = 25930
            sum y² = 59
           Step 4: Now, Substitute in the above formula given.
            `"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`
              = `[5(1227) - ((360)(17)) / ((sqrt([5(25930)-(360)^2][5(59)-(17)^2]))]]`                
              = `(6135 - 6120)/((sqrt([129650 -129600] [295-289])))`
              = 1`5/sqrt(50 xx6)`
              = `15/sqrt(300) `
              = `15/17.32`
              = 0.8660
              Answer is: 0.8660