2. The Kolmogorov-Smirnov Test (K-S Test) is used to test the goodnessof-fit of a theoretical frequency distribution, i.e., whether there is a
significant difference between an observed frequency distribution and a
given theoretical (expected) frequency distribution.
•Similar to what the Chi-Square test does, but the K-S test has several
advantages:
More powerful test.
Easier to compute and use, as no grouping of data is required.
The test statistic is independent of the expected frequency
distribution. It only depends on the sample size n.
THE HYPOTHESES:
H0:
The observed frequency distribution is consistent with the
theoretical frequency distribution (Good fit).
H1: The observed frequency distribution is not consistent with the
theoretical frequency distribution (Bad fit).
α = Level of significance of the test.
•Here we use the cumulative probability distribution (CDF) of observed and
theoretical frequencies.
3. The K-S Test Statistic:
Here,
Fe = the expected relative cumulative frequencies(CDF).
Fo = the observed relative cumulative frequencies(CDF).
•If the gap between Fe and Fo is large then Ho should be rejected.
•The value of the test statistic is obtained from the observed data listed in
the tabular form.
•A K-S test is a one tailed test.
•The critical values of Dn have been tabulated and can be found from the
K-S table for corresponding levels of significance and sample size n.
•The calculated value of Dn is compared with the critical value of Dn.
If the calculated value > critical value, then reject H0.
4. Example:
Pg # 834, Prob. # SC14-7.
Soln.:
H0: The distn. is normal with µ= 6.80, σ= 1.24.
H1: Above not true.
Value of the
variable
fo
Cumulat
ive
fo
Fo
(obs.
CDF)
Fe
(exp.
CDF)
|Fe – Fo|
≤ 4.009
13
13
0.0173
0.0122
0.0051
4.010-5.869
158
171
0.2280
0.2266
0.0014
5.870-7.729
437
608
0.8017
0.7734
0.0373
7.730-9.589
122
730
0.9733
0.9878
0.0145
>9.590
20
750
1.0000
1.0000
0.0000
We obtain Fe values from the normal table, z= (X- µ)/ σ.
The calculated value of Dn is the maximum value in the | Fe - Fo |
column. Thus, 0.0373.
For 0.15 level of significance, Dcritical = 1.14/√n = 1.14/√750 = 0.0416.
Dn < Dcritical , so accept H0 and conclude that it is a good fit.
