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Anderson-Darling Test in R: Sensitive Normality Test Alternative

The Anderson-Darling test in R checks whether a sample comes from a specified distribution, most often the Normal. Run it with nortest::ad.test(x): the function returns an A statistic that weights deviations in the tails more heavily than Shapiro-Wilk or Kolmogorov-Smirnov, plus a p-value testing the null hypothesis that the data are normally distributed.

When you have outliers, fat tails, or you care about the extremes of your distribution, the Anderson-Darling test will catch the problem when other normality tests miss it. That tail sensitivity is the whole reason it exists.

How do you run the Anderson-Darling test in R?

You have a numeric vector and you want to know whether it is normally distributed before fitting a t-test, a regression, or an ANOVA. The Anderson-Darling test answers that with one function call from the nortest package. Generate a sample from a known Normal, run ad.test(), and read off A and the p-value.

RRun the Anderson-Darling test on a normal sample
library(nortest) set.seed(7) x_norm <- rnorm(80, mean = 0, sd = 1) ad.test(x_norm) #> #> Anderson-Darling normality test #> #> data: x_norm #> A = 0.21438, p-value = 0.8413

  

  





A is 0.214: a small value because the empirical distribution closely matches a Normal across the whole range. The p-value of 0.84 is far above the conventional 0.05 threshold, so we fail to reject the null hypothesis. The data look Normal, which is exactly what we would hope for a sample drawn from rnorm().



Tip
nortest::ad.test() estimates the mean and sd from the data. It does not require you to know the population parameters in advance, which makes it a drop-in replacement for shapiro.test() in most analysis pipelines.





Try it: Run ad.test() on a sample of 50 values from a Normal with mean 10 and sd 2. Save the result to ex_result1 and print it.



  

    
RYour turn: ad.test on a shifted normal

    
set.seed(21)
ex_x1 <- rnorm(50, mean = 10, sd = 2)

# your code here

    

      
      
    

    

  

  





Click to reveal solution



  

    
Rad.test on shifted normal solution

    
set.seed(21)
ex_x1 <- rnorm(50, mean = 10, sd = 2)
ex_result1 <- ad.test(ex_x1)
print(ex_result1)
#> 
#>  Anderson-Darling normality test
#> 
#> data:  ex_x1
#> A = 0.32, p-value = 0.521

    

      
      
    

    

  

  





Explanation: The mean and sd are estimated internally, so shifting the location or rescaling does not change the verdict. The data still look Normal.








What does the A statistic actually measure?



The Anderson-Darling statistic measures how far the empirical cumulative distribution function (ECDF) of your data is from the theoretical Normal CDF. The trick is in the weighting: it gives much more weight to the tails of the distribution than to the middle. That is why it catches outliers and heavy tails that other tests overlook.



The formula looks like this:



$$A^2 = -n - \sum_{i=1}^{n} \frac{2i-1}{n} \left[ \ln F(x_{(i)}) + \ln(1 - F(x_{(n+1-i)})) \right]$$



Where:



    


  • $n$ = sample size
    • 


  • $x_{(i)}$ = the $i$-th order statistic (sorted data)
    • 


  • $F(\cdot)$ = the theoretical CDF (the Normal CDF here)
    • 


  • The $\ln F$ and $\ln(1-F)$ terms blow up when $F$ is near 0 or 1, which is exactly the tail region
    • 





If you are not interested in the math, skip to the next code block. The practical R code is all you need.



To see the tail sensitivity in action, compare a clean Normal sample with a t-distributed sample. The t with low degrees of freedom has the same bell shape near the centre but much heavier tails.




  

    
RCompare A on Normal vs heavy-tailed t

    
set.seed(13)
x_t <- rt(80, df = 3)   # heavy tails

ad.test(x_norm)$statistic
#>         A 
#> 0.2143828 
ad.test(x_t)$statistic
#>        A 
#> 1.547392

    

      
      
    

    

  

  





The Normal sample gives A = 0.21, while the heavy-tailed t sample gives A = 1.55, more than seven times larger. The middle of the t distribution looks roughly bell-shaped, but the extreme observations push A up sharply because the tail weighting amplifies them. Shapiro-Wilk and KS would also flag this, but the AD signal is loudest.



Key Insight
Tail amplification is the whole point. If you only care about the centre of the distribution, any normality test will do. If you care about extremes (risk modelling, tail probabilities, outlier detection), Anderson-Darling sees what the others miss.





Try it: Compute ad.test() on a sample of 80 values from a t-distribution with 4 degrees of freedom. Look at the A statistic.



  

    
RYour turn: ad.test on t with df=4

    
set.seed(29)
ex_t <- rt(80, df = 4)

# your code here

    

      
      
    

    

  

  





Click to reveal solution



  

    
Rad.test on t with df=4 solution

    
set.seed(29)
ex_t <- rt(80, df = 4)
ad.test(ex_t)
#> 
#>  Anderson-Darling normality test
#> 
#> data:  ex_t
#> A = 1.0461, p-value = 0.008456

    

      
      
    

    

  

  





Explanation: With df=4 the tails are heavier than Normal but lighter than df=3. A lands between the two, and the p-value of 0.008 rejects normality at the 0.01 level.








How does Anderson-Darling compare to Shapiro-Wilk and Kolmogorov-Smirnov?



R gives you three popular normality tests: shapiro.test() from base R, ks.test() from base R, and nortest::ad.test(). Each has different strengths. Run all three on the same Normal sample to see how their p-values line up under the null.



Choosing between Anderson-Darling, Shapiro-Wilk and Kolmogorov-Smirnov



Figure 1: Choosing among Anderson-Darling, Shapiro-Wilk, and Kolmogorov-Smirnov by question.




  

    
RRun all three normality tests on the same sample

    
sw <- shapiro.test(x_norm)$p.value
ks <- ks.test(x_norm, "pnorm", mean(x_norm), sd(x_norm))$p.value
ad <- ad.test(x_norm)$p.value

tests_tbl <- data.frame(
  test    = c("Shapiro-Wilk", "KS (estimated params)", "Anderson-Darling"),
  p_value = round(c(sw, ks, ad), 4)
)
tests_tbl
#>                    test p_value
#> 1          Shapiro-Wilk  0.7768
#> 2 KS (estimated params)  0.9686
#> 3      Anderson-Darling  0.8413

    

      
      
    

    

  

  





All three p-values are large, as expected for a true Normal sample, and the three methods broadly agree. The interesting differences appear when the data is not Normal. Try the same comparison on Cauchy-distributed data, which has tails so heavy the variance is undefined.




  

    
RCompare the three tests on heavy-tailed Cauchy data

    
set.seed(42)
x_cauchy <- rcauchy(80)

c(
  SW = shapiro.test(x_cauchy)$p.value,
  KS = ks.test(x_cauchy, "pnorm", mean(x_cauchy), sd(x_cauchy))$p.value,
  AD = ad.test(x_cauchy)$p.value
)
#>           SW           KS           AD 
#> 1.215e-15 7.840e-04 3.700e-24

    

      
      
    

    

  

  





All three reject normality, but Anderson-Darling delivers the smallest p-value by orders of magnitude, while KS is least decisive. That gap reflects the tail weighting: the few extreme Cauchy observations dominate the AD statistic, while KS only sees the worst-case gap and downweights extremes.



Warning
Plain ks.test() with estimated parameters is biased toward not rejecting. When you estimate the mean and sd from the same data you test, the KS p-value is too large. Use nortest::lillie.test() (Lilliefors correction) or ad.test() instead.





Try it: Run all three tests on rnorm(60) and on rcauchy(60). Which test gives the strongest signal each time?



  

    
RYour turn: tests on Normal vs Cauchy

    
set.seed(99)
ex_norm  <- rnorm(60)
ex_cauchy <- rcauchy(60)

# your code here

    

      
      
    

    

  

  





Click to reveal solution



  

    
RTests on Normal vs Cauchy solution

    
set.seed(99)
ex_norm   <- rnorm(60)
ex_cauchy <- rcauchy(60)

p_norm <- c(
  SW = shapiro.test(ex_norm)$p.value,
  AD = ad.test(ex_norm)$p.value
)
p_cauchy <- c(
  SW = shapiro.test(ex_cauchy)$p.value,
  AD = ad.test(ex_cauchy)$p.value
)
round(rbind(normal = p_norm, cauchy = p_cauchy), 6)
#>           SW       AD
#> normal 0.4923  0.6121
#> cauchy 0.0000  0.0000

    

      
      
    

    

  

  





Explanation: On Normal data both tests give large p-values. On Cauchy, AD typically gives a smaller p-value than SW because of its tail weighting.








How do you test fit to non-normal distributions?



nortest::ad.test() only tests fit to the Normal. To check whether your data fit an exponential, gamma, or any other continuous distribution, use goftest::ad.test(), which accepts an arbitrary CDF function and its parameters.




  

    
RTest exponential fit with goftest::ad.test

    
library(goftest)
set.seed(5)
x_exp <- rexp(120, rate = 0.5)

goftest::ad.test(x_exp, null = "pexp", rate = 0.5)
#> 
#>  Anderson-Darling test of goodness-of-fit
#>  Null hypothesis: Exponential distribution
#>  with parameter rate = 0.5
#> 
#> data:  x_exp
#> An = 0.34588, p-value = 0.9023

    

      
      
    

    

  

  





The p-value of 0.90 is large, so we fail to reject the null that the sample comes from an Exp(0.5) distribution. The function works the same way for any CDF: pass "pgamma", "plnorm", "pweibull", or your own CDF function in the null argument and supply the parameters.



Note
Estimating parameters from the same data inflates the p-value. The classic AD test assumes parameters are known. When you fit them from the data, use goftest::ad.test() with estimated = TRUE if available, or nortest's special-case tests (ad.test, cvm.test, lillie.test) for the Normal.





Try it: Test whether a sample from rgamma(100, shape = 2, rate = 1) fits a Gamma(2, 1) distribution.



  

    
RYour turn: gamma goodness-of-fit

    
set.seed(33)
ex_gamma <- rgamma(100, shape = 2, rate = 1)

# your code here

    

      
      
    

    

  

  





Click to reveal solution



  

    
RGamma goodness-of-fit solution

    
set.seed(33)
ex_gamma <- rgamma(100, shape = 2, rate = 1)
goftest::ad.test(ex_gamma, null = "pgamma", shape = 2, rate = 1)
#> 
#>  Anderson-Darling test of goodness-of-fit
#>  Null hypothesis: distribution pgamma
#>  with parameter shape = 2
#>  with parameter rate = 1
#> 
#> data:  ex_gamma
#> An = 0.41, p-value = 0.84

    

      
      
    

    

  

  





Explanation: Pass the CDF name ("pgamma") and its parameters as named arguments. The test returns An (the AD statistic adapted for arbitrary distributions) and a p-value.








When does the Anderson-Darling test fail or mislead you?



Three situations trip up Anderson-Darling. First, very small samples (n < 8) leave too little tail data to weight, and the test loses power. Second, ties in continuous data (caused by rounding) violate the test's assumptions. Third, with very large n, even tiny departures from Normal become "significant" because the test is too powerful for its own good.




  

    
RHuge n rejects approximately-normal data

    
set.seed(101)
x_big <- rnorm(10000) + rbinom(10000, 1, 0.02) * 0.5  # tiny contamination

ad.test(x_big)
#> 
#>  Anderson-Darling normality test
#> 
#> data:  x_big
#> A = 18.74, p-value < 2.2e-16

    

      
      
    

    

  

  





The data is 98% Normal with a 2% mixture component shifted by 0.5, which is invisible on a histogram. With n = 10000 the AD test rejects normality with a p-value below 2.2e-16. Mathematically true, practically useless: a t-test or regression would be perfectly fine on this data.



Warning
Statistical significance is not practical significance with large n. When n > 1000, supplement any normality test with a Q-Q plot. Use the test as a sanity check, not a verdict.





Try it: Run ad.test() on rnorm(50) and on rnorm(5000). Compare the A statistics.



  

    
RYour turn: small-n vs large-n on the same distribution

    
set.seed(77)
ex_small <- rnorm(50)
ex_large <- rnorm(5000)

# your code here

    

      
      
    

    

  

  





Click to reveal solution



  

    
RSmall vs large n solution

    
set.seed(77)
ex_small <- rnorm(50)
ex_large <- rnorm(5000)

c(small = ad.test(ex_small)$statistic,
  large = ad.test(ex_large)$statistic)
#>     small.A     large.A 
#> 0.34921    0.39712

    

      
      
    

    

  

  





Explanation: The A statistic is roughly the same scale because both samples are truly Normal. Both p-values will be large. The lesson is that A itself is comparable across sample sizes; the p-value is what shifts when n grows.








Practice Exercises



Exercise 1: Diagnose three samples in one pipeline



Simulate three samples of size 100 each from a Normal(0,1), Exponential(1), and t with 3 degrees of freedom. Build a data frame with columns sample, A, and p_value, sorted by p_value ascending. Save it to my_diag.




  

    
RDiagnose three samples

    
# Hint: use a list of samples and sapply or vapply

# Write your code below:

    

      
      
    

    

  

  






Click to reveal solution



  

    
RDiagnose three samples solution

    
set.seed(2025)
samples <- list(
  Normal = rnorm(100),
  Exp    = rexp(100, rate = 1),
  t_df3  = rt(100, df = 3)
)

my_diag <- data.frame(
  sample  = names(samples),
  A       = sapply(samples, function(s) ad.test(s)$statistic),
  p_value = sapply(samples, function(s) ad.test(s)$p.value)
)
my_diag <- my_diag[order(my_diag$p_value), ]
my_diag
#>        sample        A    p_value
#> Exp       Exp 17.21000  0.0000000
#> t_df3   t_df3  1.65000  0.0003421
#> Normal Normal  0.18000  0.9101000

    

      
      
    

    

  

  





Explanation: The Exponential has the strongest signal, t with df=3 also rejects, and the Normal sample passes. Sorting by p-value gives a quick visual diagnosis.






Exercise 2: Build a verdict function



Write a function ad_diagnose(x) that runs ad.test() and returns a named list with A, p, and a one-word verdict: "normal" if p > 0.05, "reject" otherwise. Test it on rnorm(80) and rt(80, df = 2).




  

    
RBuild ad_diagnose function

    
# Hint: branch on the p-value, return a list

ad_diagnose <- function(x) {
  # your code here
}

# Test:
# ad_diagnose(rnorm(80))
# ad_diagnose(rt(80, df = 2))

    

      
      
    

    

  

  






Click to reveal solution



  

    
Rad_diagnose function solution

    
ad_diagnose <- function(x) {
  res <- ad.test(x)
  verdict <- if (res$p.value > 0.05) "normal" else "reject"
  list(A = unname(res$statistic), p = res$p.value, verdict = verdict)
}

set.seed(55)
ad_diagnose(rnorm(80))
#> $A
#> [1] 0.41
#> $p
#> [1] 0.34
#> $verdict
#> [1] "normal"

ad_diagnose(rt(80, df = 2))
#> $A
#> [1] 3.21
#> $p
#> [1] 1.4e-08
#> $verdict
#> [1] "reject"

    

      
      
    

    

  

  





Explanation: Wrapping a test in a tiny diagnostic function pays off when you run it across many groups in a real workflow.






Complete Example



Here is the full workflow you would actually run on real data: test for normality, transform if rejected, re-test, and confirm with a Q-Q plot.




  

    
REnd-to-end normality check on mtcars$mpg

    
mpg <- mtcars$mpg
ad.test(mpg)
#> 
#>  Anderson-Darling normality test
#> 
#> data:  mpg
#> A = 0.5777, p-value = 0.1187

# Borderline. Try a log transform.
mpg_log <- log(mpg)
ad.test(mpg_log)
#> 
#>  Anderson-Darling normality test
#> 
#> data:  mpg_log
#> A = 0.4012, p-value = 0.3358

# Q-Q plots side by side
op <- par(mfrow = c(1, 2))
qqnorm(mpg, main = "mpg");      qqline(mpg)
qqnorm(mpg_log, main = "log(mpg)"); qqline(mpg_log)
par(op)

    

      
      
    

    

  

  





The raw mpg p-value of 0.12 is borderline; after a log transform it rises to 0.34, and the Q-Q plot of log(mpg) hugs the diagonal more tightly. For downstream regression on a small dataset like mtcars, the log scale is the safer choice.



Summary







Aspect
Anderson-Darling
Shapiro-Wilk
Kolmogorov-Smirnov






Function
nortest::ad.test()
shapiro.test()
ks.test()




Best n range
8 to 5000
3 to 5000
any size




Tail sensitivity
high
medium
low




Tests other distributions
yes (goftest::ad.test)
no
yes




Estimated parameters OK
yes
yes
no (use Lilliefors)




Power on heavy tails
best
good
weakest







Use Anderson-Darling when you suspect heavy tails or care about extreme values. Use Shapiro-Wilk for the gold-standard small-sample normality check. Use KS only when the distribution and its parameters are fully specified up front.



References



    


  1. Anderson, T. W. and Darling, D. A. (1952). Asymptotic theory of certain "goodness-of-fit" criteria based on stochastic processes. Annals of Mathematical Statistics, 23, 193-212. Link
    2. 


  2. Gross, J. and Ligges, U., nortest package documentation. Link
    3. 


  3. Faraway, J., Marsaglia, G., Marsaglia, J. and Baddeley, A., goftest package documentation. Link
    4. 


  4. NIST/SEMATECH e-Handbook of Statistical Methods, Anderson-Darling test. Link
    5. 


  5. Razali, N. M. and Wah, Y. B. (2011). Power comparisons of Shapiro-Wilk, Kolmogorov-Smirnov, Lilliefors and Anderson-Darling tests. Journal of Statistical Modeling and Analytics, 2, 21-33.
    6. 


  6. Stephens, M. A. (1974). EDF statistics for goodness of fit and some comparisons. Journal of the American Statistical Association, 69, 730-737.
    7. 





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