Introduction
This is a normal law match test. It was T.W. Anderson and D. A. Darling who developed this test in 1954. It will then be taken over and developed by Stephens in the years 70. The interest of this test is that it gives importance to the tails of distribution.
This test can be used to demonstrate adequacy with normal, Lognormal, exponential or Weibull law. The operation is the same, simply, it will have to change the table of the critical value as well as the calculation of the Fi. In this article, we will focus solely on the adequacy of the normal law.
The principle
The calculation principle is very similar to the Kolmogorov-Smirnov Test. The difference lies in the fact that the critical value is calculated from the normal law, as well as the Fi test variable.
Step 1: Assumptions
Anderson Darling’s test is a unilateral match test. The test hypotheses are:
- H0: The data follow the normal law
- H1: Data does not follow normal law
Step 2: Calculate the practical value
- Sort the data in ascending order.
- Deduct the average and standard deviation.
- Calculate reduced centered data according to the normal law
- Use the reduced centered normal law distribution function to get the frequencies.
- Calculate the logarithm logarithm of this value.
- In the same way, we Shape Fn-i + 1 and then deduce the natural logarithm.
- We then calculate the sum S:
8. It is inferred from the A2 statistic via the following calculation:
Step 3: Calculating the critical value
The critical value of the test is not calculated but was tabbed by the creators. These values depending on the adjustment law, and in the case of an adjustment to the normal law, the values are as follows:
|
Level of risk |
1% |
2,5% |
5% |
10% |
|
Critical value |
1,035 |
0,873 |
0,752 |
0,631 |
Step 4: Calculate the p-value
P-value is calculated from the A2 statistic by interpolation from a table described by Stephens (1986). The following table is used:
|
Value of a2 |
P-Value |
|
A2 < 0.2 |
1-e(-13.436 + 101.14 * A2-223,73 * A2 ^ 2 |
|
0.2 ≤ A2 ≤ 0.34 |
1-e(-8.318 + 42.796 * A2-59.938 * A2 ^ 2) |
|
0.34 ≤ A2 < 0.6 |
e(0,9177 – 4,279 * A2 – 1,38 * A2^2) |
|
0.6 ≤ A2 |
e(1.2937-5.709 * A2 + 0.0186 * A2 ^ 2) |
Step 5: Interpretation
| Result | Statistical conclusion | Practical conclusion |
|---|---|---|
| Practical value ≥ Critical value | We reject H0 | Our data do not follow the normal law at the level of risk α given. |
| Practical value < Critical value | We retain H0 | Our data follow the normal distribution at given level of risk. |
| Result | Statistical conclusion | Practical conclusion |
|---|---|---|
| p-value > α | We retain H0 | Our data follows a normal law with a risk of being wrong with p-value% |
| p-Value < α | We rejectH0 | Our data do not follow a normal law with a risk of being wrong with p-value% |
Source
Mr. A. Stephens (1974) – EDF statistics for Goodness of Fit and Some comparisons
M. A. Stephens (1986) – Tests based on EDF statistics
