Anderson-Darling Test Calculator
Test whether sample data follows a specified distribution with emphasis on the tails.
Sample Data
Anderson-Darling A*
0.2467
Fail to Reject H0
Sample Statistics
Formula
A^2 = -n - Sum[(2i-1)/n * (ln(Fi) + ln(1-F(n+1-i)))]
What Is the Anderson-Darling Test?
The Anderson-Darling Test is a statistical goodness-of-fit test that determines whether a sample of data comes from a specific probability distribution. Unlike simpler tests like the Kolmogorov-Smirnov test, the Anderson-Darling test gives more weight to the tails of the distribution, making it especially sensitive to departures from normality in the extreme values.
This makes the test particularly valuable in fields where tail behavior matters — such as risk management, quality control, and financial modeling, where rare but extreme events have outsized consequences. The test was developed by Theodore Anderson and Donald Darling in 1952 as an improvement over the Cramér-von Mises criterion.
Our Anderson-Darling Test Calculator supports three distributions: normal (Gaussian), exponential, and uniform. Enter your sample data and select a significance level to instantly determine whether your data likely follows the chosen distribution.
The Anderson-Darling Test Formula
The Anderson-Darling statistic is built on the empirical distribution function (EDF) and measures the squared distance between the hypothesized CDF and the observed sample, weighted by the variance at each point. The formula below shows how the statistic is computed.
Anderson-Darling Formula
Where:
- A²= Anderson-Darling test statistic (raw)
- n= Sample size (number of data points)
- F(x)= Cumulative distribution function under the null hypothesis
- x_i= i-th ordered value in the sorted sample (ascending)
Adjusted Statistic, Critical Values, and P-Value
When testing for normality with both the mean and variance estimated from the data, the raw A² statistic must be adjusted using A²* = A² × (1 + 0.75/n + 2.25/n²). This correction accounts for the loss of degrees of freedom from parameter estimation and gives a more accurate test.
The critical values for the Anderson-Darling test at common significance levels for a normal distribution are:
| Significance Level (α) | Critical Value |
|---|---|
| 0.15 | 0.576 |
| 0.10 | 0.656 |
| 0.05 | 0.787 |
| 0.025 | 0.918 |
| 0.01 | 1.092 |
If the test statistic A²* exceeds the critical value at your chosen α, you reject the null hypothesis and conclude the data does not follow the specified distribution. A small p-value (typically below 0.05) also indicates rejection.
Our calculator also reports skewness and kurtosis alongside the test results, helping you understand the shape characteristics of your data distribution at a glance.
How to Use This Calculator
Using the Anderson-Darling Test Calculator takes just a few steps:
- Enter your sample data: Type or paste your numeric values separated by commas. The calculator accepts decimals and integers. You need at least 3 data points for a valid test. Example default data is provided to get you started.
- Choose a distribution: Select Normal, Exponential, or Uniform — this is the distribution you hypothesize your data follows. For most applications, you will test for normality.
- Set the significance level: Choose α (commonly 0.05). This is the threshold probability for rejecting the null hypothesis — a lower α means you require stronger evidence to reject.
- Read the results: The calculator displays the adjusted A²* statistic, raw A², p-value, critical value, and a clear reject/fail-to-reject verdict. It also shows descriptive statistics including mean, standard deviation, skewness, and kurtosis.
Understanding the Results
Interpreting Anderson-Darling test results correctly requires understanding what the statistic and p-value tell you. The null hypothesis H₀ is that the sample does follow the specified distribution. The alternative hypothesis H₁ is that it does not.
| Outcome | Condition | Meaning |
|---|---|---|
| Reject H₀ | A²* > critical value or p < α | Data does NOT follow the specified distribution |
| Fail to Reject H₀ | A²* ≤ critical value or p ≥ α | Insufficient evidence to conclude non-normality |
Failing to reject does not prove the data is normal — it only means the test did not find strong evidence against normality. With small sample sizes, the test has limited power and may fail to detect even substantial departures. Conversely, with very large samples, the test may reject normality for practically insignificant deviations. Always complement formal tests with diagnostic plots like Q-Q plots and histograms.
Real-World Applications
The Anderson-Darling test is used across many disciplines where verifying distributional assumptions is critical. In financial risk management, analysts test whether asset returns follow a normal distribution before applying models like Value-at-Risk (VaR). Since financial returns often exhibit fat tails, the Anderson-Darling test's tail sensitivity makes it a preferred choice over the Kolmogorov-Smirnov test.
In manufacturing and quality control, engineers use the test to verify that process measurements are normally distributed, a prerequisite for control charts and process capability indices like Cp and Cpk. The exponential distribution variant is used to test failure times and reliability data in survival analysis and reliability engineering.
In clinical research and biostatistics, researchers apply normality tests to determine whether parametric tests like t-tests and ANOVA are appropriate. The Anderson-Darling test is often recommended by regulatory guidelines because of its higher power compared to alternative tests. Environmental scientists also use it to test whether pollutant concentrations follow a log-normal or exponential distribution.
Worked Examples
Testing Stock Returns for Normality
Problem:
A financial analyst has 10 daily stock returns: 1.5%, 2.3%, 3.1%, 3.8%, 4.2%, 5.0%, 5.7%, 6.3%, 7.1%, 8.0%. She tests whether these returns follow a normal distribution at α = 0.05.
Solution Steps:
- 1Step 1: Enter the data as comma-separated values and select 'Normal' distribution with α = 0.05.
- 2Step 2: The calculator computes n = 10, mean ≈ 4.70, standard deviation ≈ 2.09, then sorts the data and applies the normal CDF.
- 3Step 3: The Anderson-Darling S-statistic is calculated: S = Σ(2(i+1)-1)(ln(F(x_i)) + ln(1-F(x_{n+1-i}))) over all observations.
- 4Step 4: Raw A² = -10 - S/10. The adjusted A²* = A² × (1 + 0.75/10 + 2.25/100). The p-value is computed and compared against α = 0.05.
Result:
The adjusted A²* statistic ≈ 0.28, with a p-value of approximately 0.60. Since A²* (0.28) < critical value (0.787) and p (0.60) > 0.05, the analyst fails to reject H₀ — there is no strong evidence that the returns deviate from a normal distribution.
Testing Product Lifetimes (Exponential)
Problem:
A reliability engineer records the lifetimes (in hours) of 8 electronic components: 120, 250, 340, 410, 520, 680, 790, 950. She wants to test if the failure times follow an exponential distribution at α = 0.05.
Solution Steps:
- 1Step 1: Enter the data and select 'Exponential' distribution. Set α = 0.05.
- 2Step 2: The calculator computes mean lifetime ≈ 507.5 hours, then λ = 1/mean ≈ 0.00197. The exponential CDF F(x) = 1 − e^(−λx) is applied at each sorted value.
- 3Step 3: The weighted sum S = Σ(2(i+1)-1)(ln(F_i) + ln(1-F_{n+1-i})) is accumulated and divided by n = 8, then negated and subtracted from -n to get A².
- 4Step 4: The computed A² is compared against critical values for the exponential distribution test.
Result:
The raw A² statistic ≈ 0.45. At α = 0.05, the critical value for the exponential test indicates whether to reject. If A² is below the threshold, the engineer fails to reject H₀, suggesting the exponential model is adequate for the failure time data.
Checking Uniformity of Random Samples
Problem:
A researcher uses random numbers from a uniform generator to assign treatment groups: 0.12, 0.37, 0.45, 0.58, 0.63, 0.71, 0.84, 0.91, 0.95, 0.98. She tests for uniformity at α = 0.10.
Solution Steps:
- 1Step 1: Input the 10 numbers and select 'Uniform' distribution with α = 0.10.
- 2Step 2: The calculator identifies min = 0.12 and max = 0.98, then applies the uniform CDF: F(x) = (x − min)/(max − min) for each ordered value.
- 3Step 3: The Anderson-Darling S-statistic is computed with the uniform CDF values, then A² = −n − S/n.
- 4Step 4: The result is compared to critical values for uniformity testing at α = 0.10.
Result:
Provided the random numbers are truly uniform, the A² statistic should fall below the critical value, and the researcher fails to reject the null hypothesis of uniformity — confirming the random number generator is working as expected.
Tips & Best Practices
- ✓Always plot a histogram or Q-Q plot of your data alongside running the Anderson-Darling test — visual checks catch patterns a single statistic might miss.
- ✓For financial return data, the Anderson-Darling test on the normal distribution is usually rejected; consider log-normal or Student's t distributions instead.
- ✓With sample sizes below 25, the Anderson-Darling test has limited statistical power — interpret negative results cautiously.
- ✓Use α = 0.05 as a standard significance threshold, but adjust it based on the cost of Type I versus Type II errors in your specific context.
- ✓If your data has ties (identical values), the Anderson-Darling test still works but may be slightly conservative in its p-value estimates.
- ✓The exponential distribution test is useful for failure time analysis — reject exponential if your failure rate changes significantly over time.
- ✓Run the test with multiple distributions to compare fit quality — the distribution with the smallest A² statistic and highest p-value is usually the best match.
Frequently Asked Questions
Sources & References
Last updated: 2026-06-06
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Editorial Note
MyCalcBuddy Editorial Team
This page is maintained as an educational calculator reference.
Formula Source: Standard Mathematical References
by Various