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 a way of estimating F(x).
The distribution of Dn can be calculated (see Kolmogorov Distribution), but for our purposes now 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 an x value less than 100, 25 elements have an x 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 standard deviation 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 (more precisely that there is no significant difference between the data and data which is normally distributed). Note that is not the same conclusion we reached from looking at the histogram and QQ plot.
Real Statistics Excel Function: The following functions are provided in the Real Statistics Resource Pack:
KSCRIT(n, α, tails, h) = the critical value of the Kolmogorov-Smirnov test for a sample of size n, for the given value of alpha (default = .05) and tails = 1 (one tail) or 2 (two tails, default), based on the KS Table. If h = TRUE (default) harmonic interpolation is used; otherwise linear interpolation is used.
KSPROB(x, n, tails, iter, h) = an approximate p-value for the KS test for the Dn value equal to x for a sample of size n and tails = 1 (one tail) or 2 (two tails, default) based on a linear interpolation (if h = FALSE) or harmonic interpolation (if h = TRUE, default) of the values in the Kolmogorov-Smirnov Table, using iter number of iterations (default = 40).
Note that the values for α in the Kolmogorov-Smirnov Table range from .001 to .2 (for tails = 2) and .0005 to .1 for tails = 1. If the p-value is less than .001 (tails = 2) or .0005 (tails = 1) then the p-value is given as 0 and if the p-value is greater than .2 (tails = 2) or .1 (tails = 1) then the p-value is given as 1.
For Example 2, KSCRIT(15, .05, 2) = .338 (the same as shown in cell H21 of Figure 4). Also note that the p-value = KSPROB(H20, B21) = KSPROB(0.184177, 15) = 1 (meaning that p-value > .2), and so once again we can’t reject the null hypothesis that the data is normally distributed.
If the value of Dn had been .35 in Example 2, then Dn = .35 > .338 = Dcrit, and so we would have rejected the null hypothesis that the data is normally distributed. In this case we would have seen that p-value = KSPROB(.35,15) = .0427, which once again leads us to reject the null hypothesis.
As referenced above, the Kolmogorov distribution can be useful in conducting the Kolmogorov-Smirnov test. Click here for more information about this distribution, including some useful functions provided by the Real Statistics Resource Pack.
When the population mean and standard deviation for the Kolmogorov-Smirnov Test is estimated from the sample mean and standard deviation, as was done in Example 1 and 2, then the Kolmogorov-Smirnov Table yields results that are too conservative. More accurate results can be derived from the Liiliefors Table as described in the Lilliefors Test for Normality.