Definition 1: Let x1,…,xn be an ordered sample with x1 ≤ … ≤ xn and define Sn(x) as follows:
Now suppose that the sample comes from a population with cumulative distribution function F(x) and define Dn as follows:
Observation: It can be shown that Dn doesn’t depend on F. Since Sn(x) depends on the sample chosen, Dn is a random variable. Our objective is to use Dn as way of estimating F(x).
The distribution of Dn can be calculated, but for our purposes the important aspect of this distribution are the critical values. These can be found in the Kolmogorov-Smirnov Table.
If Dn,α is the critical value from the table, then P(Dn ≤ Dn,α) = 1 – α. Dn can be used to test the hypothesis that a random sample came from a population with a specific distribution function F(x). If
then the sample data is a good fit with F(x).
Also from the definition of Dn given above, it follows that
Thus Sn(x) ± Dn,α provides a confidence interval for F(x)
Example 1: Determine whether the data represented in the following frequency table is normally distributed.
Figure 1 – Frequency table for Example 1
This means that 8 elements have value less than 100, 25 elements have value between 101 and 200, etc. We need to find the mean and standard deviation of this data. Since this is a frequency table, we can’t simply use Excel’s AVERAGE and STDEV functions. Instead we first use the midpoints of each interval and then use an approach similar to that described in Frequency Tables as follows:
Figure 2 – Calculating mean and std dev for data in frequency table
Thus, the mean is 481.4 and the standard deviation is 155.2. We can now build the table that allows us to carry out the KS test, namely:
Figure 3 – Kolmogorov-Smirnov test for Example 1
Columns A and B contain the data from the original frequency table. Column C contains the corresponding cumulative frequency values and column D simply divides these values by the sample size (n = 1000) to yield the cumulative distribution function Sn(x)
Column E uses the mean and standard deviation calculated previously to standardize the values of x from column A. E.g. the formula in cell E4 is =STANDARDIZE(A4,N$5,N$10), where cell N5 contains the mean and cell N10 contains the standard deviation. Column F uses these standardized values to calculate the cumulative distribution function values assuming that the original data is normally distributed. E.g. cell F4 contains the formula =NORMSDIST(E4). Finally column G contains the differences between the values in columns D and F. E.g. cell G4 contains the formula =ABS(F4—D4). If the original data is normally distributed these differences will be zero.
Now Dn = the largest value in column G, which in our case is 0.0117. If the data is normally distributed then the critical value Dn,α will be larger than Dn. From the Kolmogorov-Smirnov Table we see that
Dn,α = D1000,.05 = 1.36 / SQRT(1000) = 0.043007
Since Dn = 0.0117 < 0.043007 = Dn,α, we conclude that the data is a good fit with the normal distribution.
Example 2: Using the KS test, determine whether the data in Example 1 of Graphical Tests for Normality and Symmetry is normally distributed.
We follow the same procedure as in the previous example to obtain the following results. Since the frequencies are all 1, this example should be a bit easier to understand.
Figure 4 – KS test for data from Example 2
The Kolmogorov-Smirnov Table shows that the critical value Dn,α = D15,.05 = .338
Since Dn = 0.1874988 < 0.338 = Dn,α, we conclude that the data is a reasonably good fit with the normal distribution, which a bit different from what we concluded from looking at the histogram and QQ plot.
Real Statistics Excel Function: The following function is provided in the Real Statistics Resource Pack:
KSCRIT(n, α, t) = the critical value of the Kolmogorov-Smirnov test for a sample of size n, for the given value of alpha and t = 1 (one tail) or 2 (two tails).