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It allows us to test whether our data follow a diverse law, especially the normal law.

## 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

1. Sort the data in ascending order.
2. Deduct the average and standard deviation.
3. Calculate reduced centered data according to the normal law
4. Use the reduced centered normal law distribution function to get the frequencies.
5. Calculate the logarithm logarithm of this value.
6. In the same way, we Shape Fn-i + 1 and then deduce the natural logarithm.
7. 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

ResultStatistical conclusionPractical conclusion
Practical value ≥ Critical valueWe reject H0Our data do not follow the normal law at the level of risk α given.
Practical value < Critical valueWe retain H0Our data follow the normal distribution at given level of risk.
ResultStatistical conclusionPractical conclusion
p-value > αWe retain H0Our data follows a normal law with a risk of being wrong with p-value%
p-Value < αWe rejectH0Our 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