Kolmogorov-Smirnov Test
Some statistical tests ask you to assume that your data follows a normal distribution. The Kolmogorov-Smirnov Goodness of Fit Test compares your data to data from a normal distribution with the same mean and standard deviation as your results. If the test is NOT significant, then the data can be assumed to be normal.
- You want to test whether your data can be assumed to follow the normal distribution.
- You have at least 2,000 data points.
This would most likely only occur if you are using online data from a previous extensive study. - Your data may be categorical data (see Types of Data for more information).
NOTE: For smaller samples (n < 2,000) use the Shapiro-Wilk Test.
Some statistical tests ask you to assume that your data follows a normal distribution. The Kolmogorov-Smirnov Goodness of Fit Test compares your data to data from a normal distribution with the same mean and standard deviation as your sample. If the test is NOT significant, then the data can be assumed to be normal.
The hypotheses for the test are:
H0: The data comes from the specified distribution.
H1: At least one value does not match the specified distribution.
The easiest way to do this test is to use an online calculator. A good example can be found HERE.
Worked Example
NOTE: It is not practical to give an example with over 2,000 data points, so the test will be done on this small sample.
The growth of plant seedlings in two different types of soil was measured. Growth (in cm) over a year was measured for 10 plants in each type of soil and is shown below.
Soil Type 1: 3.2, 4.5, 3.8, 4.1, 3.6, 3.1, 2.8, 5.1, 6.8, 3.0
Soil Type 2: 4.5, 6.3, 5.7, 6.0, 7.2, 5.8, 4.9, 5.3, 7.1, 4.9
Using an online calculator, test each set of data for the assumption of normality.
Using an online calculator on the data for Soil Type 1, we get: The value of the K-S test statistic is 0.16859. The p-value is .89544. Your data does not differ significantly from that which is normally distributed. For Soil Type 2, we get: The value of the K-S test statistic is 0.14244. The p-value is .97009. Your data does not differ significantly from that which is normally distributed. Therefore, in both cases, the assumption of normality is justified. |
NOTE: It is not practical to give an example with over 2,000 data points, so the test will be done on a smaller sample.
You have measured the number of seedlings in quadrant grid squares at 15 different points in a forest. Your data is shown below:
9, 2, 6, 7, 15, 8, 4, 23, 4, 14, 8, 12, 8, 13, 17
You wish to carry out statistical tests that require the assumption of normality.
Using the online calculator HERE, enter the data above.
The results show ...
As the p-value (0.675) is greater than the significance level (5% or 0.05) we fail to reject our null hypothesis that the data follows a normal distribution... so it is reasonable to assume that our data follows a normal distribution.