Kolmogorov Smirnov test (K-S test) is a statistical analysis, used to evaluate probability of normal distribution for mean variables.
Calculate maximum of normal distribution mean variable for 6 rows from date set of population {120,123,110,100,200,210} using Kolmogorov Smirnov test method.
Row | DataSet |
---|---|
1 | 120 |
2 | 123 |
3 | 110 |
4 | 100 |
5 | 200 |
6 | 210 |
Substitute the given values in the formula to solve maximum of normal distribution,
First, calculate the value of Mean(μ)
Mean (μ) = ∑ X / n = ∑ (120+123+110+100+200+210) / 6 = 863 / 6 = 143.833Standard Deviation (σ) = √∑(X-μ)2 / n-1 = √∑(120-143.8)2 + (123-143.8)2 + (110-143.8)2 + (100-143.8)2 + (200-143.8)2 + (210-143.8)2 / 5 = √2320.166 = 48.168
F0(100) = Pμ=143.833, σ=14.85 (100) Pμ=143.833, σ=14.85 (100) = Normal[(100 - 143.833) / 48.168] = 0.181409 Fn(100) = 1/n = 1/6 = 0.167 Fn-1(100) = (i-1)/n = (1-1)/6 = 0
Follow the procedure of Step 3, upto N th value.
The following table shows all the values
Row | DataSet | F0 | Fn | Fn-1 | D+=Fn-F0 | D-=F0-Fn-1 |
---|---|---|---|---|---|---|
1 | 100 | 0.181409 | 0.166667 | 0 | -0.014742 | 0.181409 |
2 | 110 | 0.241215 | 0.333333 | 0.166667 | 0.092118 | 0.074548 |
3 | 120 | 0.310372 | 0.5 | 0.333333 | 0.189628 | -0.022961 |
4 | 123 | 0.332684 | 0.666667 | 0.5 | 0.333983 | -0.167316 |
5 | 200 | 0.878204 | 0.833333 | 0.666667 | -0.044871 | 0.211537 |
6 | 210 | 0.915227 | 1 | 0.833333 | 0.084773 | 0.081894 |
Max Value | 0.333983 | 0.211537 | ||||
KS statistic value : 0.333983 |