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Fitting a Weibull Using the Anderson-Darling GoF Test

We now develop an example of testing for the Weibull assumption. We will use the data in Table 5, which will also be used for this same purpose in the implementation of the companion Kolmogorov-Smirnov GoF test [12]. The data consist of six measurements, drawn from the same Weibull (α = 10; β = 2) population. In our examples, however, the parameters are unknown and will be estimated from the data set.

Table 5. Data Set for Testing the Weibull Assumption
 11.7216 10.4286 8.0204 7.5778 1.4298 4.1154

We obtain the descriptive statistics (Table 6). Then, using graphical methods in [1], we get point estimations of the assumed Weibull parameters: shape β = 1.3 and scale α = 8.7. The parameters allow us to define the distribution hypothesis: Weibull (α = 8.7; β = 1.3).

Table 6. Descriptive Statistics
Variable
N Mean Median StDev Min Max Q1
Data Set 6 7.22 7.80 3.86 1.43 11.72 3.44

The Weibull version of the AD GoF test statistic is different from the one for Normality, given in the previous section. This Weibull version is explained in detail in [2, 5] and is defined by:

 (2)

where Z(i) = [x(i)/θ*]β* and where the asterisks (*) in the Weibull parameters denote the corresponding estimations. The OSL (observed significance level) probability (p-value) is now used for testing the Weibull assumption. If OSL < 0.05 then the Weibull assumption is rejected and the error committed is less than 5%. The OSL formula is given by:

OSL = 1/{1 + exp[-0.1 + 1.24 ln (AD*) + 4.48 (AD*)]}

To implement the AD GoF test, we first obtain the corresponding Weibull probabilities under the assumed distribution H0. For example, for the first data point (1.43):

Then, we use these values to work through formulas AD and AD* in (2). Intermediate results, for the small data set in Table 5, are given in Table 7.

Table 7. Intermediate Values for the AD GoF Test for the Weibull
Row DataSet Z(i) WeibProb Exp-Z(i) Ln(1-Ez) Zn-i+1 ith-term
1 1.430 0.09560 0.091176 0.908824 -2.39496 1.47336 0.64472
2 4.115 0.37789 0.314692 0.685308 -1.15616 1.26567 1.21092
3 7.578 0.83566 0.566413 0.433587 -0.56843 0.89967 1.22342
4 8.020 0.89967 0.593296 0.406704 -0.52206 0.83566 1.58401
5 10.429 1.26567 0.717949 0.282051 -0.33136 0.37789 1.06387
6 11.722 1.47336 0.770846 0.229154 -0.26027 0.09560 0.65242

The AD GoF test statistics (2) values are: AD = 0.3794 and AD* = 0.4104. The value corresponding to the OSL, or probability of rejecting the Weibull (8.7; 1.3) distribution erroneously with these results, is OSL = 0.3466 (much larger than the error = 0.05).

Hence, we accept the null hypothesis that the underlined distribution (of the population from where these data were obtained) is Weibull (α = 8.7; β = 1.3). Hence, the AD test was able to recognize that the data were actually Weibull. The GoF procedure for this case is summarized in Table 8.

Table 8. Step-by-Step Summary of the AD GoF Test for the Weibull
 Sort Original Sample (X) and Standardize: Z= [x(i)/θ*]β* (Cols & 2, Table 7) Establish the Null Hypothesis: assume Weibull distribution Obtain the distribution parameters: α = 8.7; β = 1.3 Obtain Weibull probability and Exp(-Z) (Cols. 3 & 4) Obtain the Logarithm of 1- Exp(-Z) (Col. 5) Sort the Z(i) in descending order (n-i+1) (Col. 6) Evaluate via (1): AD* = 0.4104 and OSL = 0.3466 Since OSL = 0.3466 > α = 0.05, assume Weibull (α = 8.7; β = 1.3) Software for this version of AD is not commonly available

Finally, recall that the Exponential distribution, with mean α, is only a special case of the Weibull (α; β) where the shape parameter β = 1. Therefore, if we are interested in using AD GoF test to assess Exponentiality, it is enough to estimate the sample mean (α) and then to implement the above Weibull procedure for this special case, using formula (2).

There are not, however, AD statistics (formulas) for all the distributions. Hence, if there is a need to fit other distributions than the four discussed in this START sheet, it is better to use the Kolmogorov Smirnov [12] or the Chi Square [11] GoF tests.