A measure of the strength of linear association between two variables. Correlation will always between -1.0 and +1.0. If the correlation is positive, we have a positive relationship. If it is negative, the relationship is negative.
N = Number of values or elements X = First Score Y = Second Score ΣXY = Sum of the product of first and Second Scores ΣX = Sum of First Scores ΣY = Sum of Second Scores ΣX2 = Sum of square First Scores ΣY2 = Sum of square Second Scores
To find the Correlation of
X Values | Y Values |
---|---|
60 | 3.1 |
61 | 3.6 |
62 | 3.8 |
63 | 4 |
65 | 4.1 |
Count the number of values. N = 5
Find XY, X2, Y2 See the below table
X Value | Y Value | X*Y | X*X | Y*Y |
---|---|---|---|---|
60 | 3.1 | 60 * 3.1 = 186 | 60 * 60 = 3600 | 3.1 * 3.1 = 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 |
65 | 4.1 | 65 * 4.1 = 266.5 | 65 * 65 = 4225 | 4.1 * 4.1 = 16.81 |
Find ΣX, ΣY, ΣXY, ΣX2, ΣY2. ΣX = 311 ΣY = 18.6 ΣXY = 1159.7 ΣX2 = 19359 ΣY2 = 69.82
Now, Substitute in the above formula given. Correlation(r) =[ NΣXY - (ΣX)(ΣY) / Sqrt([NΣX2 - (ΣX)2][NΣY2 - (ΣY)2])] = ((5)*(1159.7)-(311)*(18.6))/sqrt([(5)*(19359)-(311)2]*[(5)*(69.82)-(18.6)2]) = (5798.5 - 5784.6)/sqrt([96795 - 96721]*[349.1 - 345.96]) = 13.9/sqrt(74*3.14) = 13.9/sqrt(232.36) = 13.9/15.24336 = 0.9119 This example will guide you to find the relationship between two variables by calculating the Correlation Co-efficient from the above steps.
This tutorial will help you dynamically to find the Correlation Co-efficient problems.