Weibull analysis: Median Rank

Weibull statistical analysis is used to project failure rates on a population of
identical parts based on a sample of failure times. Part of the calculation
involves something called the median rank. For a complete explanation see this
reference:

http://www.mathpages.com/home/kmath122/kmath122.htm

So I decided to develop my own implementation of the analysis, including the
calculation of median rank. I wanted to calculate the value from the definition
given in the above article, which involves numerical iteration.

The only on-line reference I could find for median rank tables was this page:

http://www.weibull.com/GPaper/hread.htm

(look only at the 50% column - that's the median rank). My calculations are
pretty close to those in the tables, In fact we agree up to sample size = 14.
But starting with sample size = 15 and higher, we start to diverge a little.

My question is, who is right? The paper at www.weibull.com, or me? I took
great care to use 64-bit double precision calculations to make everything as
precise as I could. What I am looking for is a tie-breaker. Does anyone have a
source for median rank tables that they could check to see who is right?

I won't bother posting my tables for sample sizes 2-14, since they agree with
the cited on-line table, but here are my findings for sample sizes 15-25:

Sample Size 15

1 0.0451584
2 0.1093961
3 0.1743207
4 0.2393934
5 0.3045198
6 0.3696705
7 0.4348328
8 0.5000000
9 0.5651672
10 0.6303295
11 0.6954802
12 0.7606066
13 0.8256793
14 0.8906039
15 0.9548416

Sample Size 16

1 0.0423967
2 0.1027030
3 0.1636542
4 0.2247447
5 0.2858859
6 0.3470504
7 0.4082265
8 0.4694083
9 0.5305917
10 0.5917735
11 0.6529496
12 0.7141141
13 0.7752553
14 0.8363458
15 0.8972970
16 0.9576033

Sample Size 17

1 0.0399533
2 0.0967816
3 0.1542176
4 0.2117850
5 0.2694005
6 0.3270381
7 0.3846872
8 0.4423423
9 0.5000000
10 0.5576577
11 0.6153128
12 0.6729619
13 0.7305995
14 0.7882150
15 0.8457824
16 0.9032184
17 0.9600467

Sample Size 18

1 0.0377762
2 0.0915057
3 0.1458097
4 0.2002382
5 0.2547122
6 0.3092075
7 0.3637138
8 0.4182263
9 0.4727418
10 0.5272582
11 0.5817737
12 0.6362862
13 0.6907925
14 0.7452878
15 0.7997618
16 0.8541903
17 0.9084943
18 0.9622238

Sample Size 19

1 0.0358240
2 0.0867752
3 0.1382712
4 0.1898852
5 0.2415426
6 0.2932202
7 0.3449086
8 0.3966030
9 0.4483008
10 0.5000000
11 0.5516992
12 0.6033970
13 0.6550914
14 0.7067798
15 0.7584574
16 0.8101148
17 0.8617288
18 0.9132248
19 0.9641760

Sample Size 20

1 0.0340637
2 0.0825097
3 0.1314737
4 0.1805500
5 0.2296676
6 0.2788046
7 0.3279519
8 0.3771052
9 0.4262619
10 0.4754205
11 0.5245795
12 0.5737381
13 0.6228948
14 0.6720481
15 0.7211954
16 0.7703324
17 0.8194500
18 0.8685263
19 0.9174903
20 0.9659363

Sample Size 21

1 0.0324682
2 0.0786438
3 0.1253131
4 0.1720895
5 0.2189054
6 0.2657397
7 0.3125841
8 0.3594343
9 0.4062879
10 0.4531435
11 0.5000000
12 0.5468565
13 0.5937121
14 0.6405657
15 0.6874159
16 0.7342603
17 0.7810946
18 0.8279105
19 0.8746869
20 0.9213562
21 0.9675318

Sample Size 22

1 0.0310155
2 0.0751240
3 0.1197041
4 0.1643864
5 0.2091065
6 0.2538444
7 0.2985919
8 0.3433450
9 0.3881016
10 0.4328602
11 0.4776200
12 0.5223800
13 0.5671398
14 0.6118984
15 0.6566550
16 0.7014081
17 0.7461556
18 0.7908935
19 0.8356136
20 0.8802959
21 0.9248760
22 0.9689845

Sample Size 23

1 0.0296872
2 0.0719057
3 0.1145755
4 0.1573433
5 0.2001472
6 0.2429682
7 0.2857985
8 0.3286342
9 0.3714733
10 0.4143145
11 0.4571570
12 0.5000000
13 0.5428430
14 0.5856855
15 0.6285267
16 0.6713658
17 0.7142015
18 0.7570318
19 0.7998528
20 0.8426567
21 0.8854245
22 0.9280943
23 0.9703128

Sample Size 24

1 0.0284681
2 0.0689518
3 0.1098684
4 0.1508789
5 0.1919241
6 0.2329856
7 0.2740562
8 0.3151320
9 0.3562111
10 0.3972924
11 0.4383750
12 0.4794582
13 0.5205418
14 0.5616250
15 0.6027076
16 0.6437889
17 0.6848680
18 0.7259438
19 0.7670144
20 0.8080759
21 0.8491211
22 0.8901316
23 0.9310482
24 0.9715319

Sample Size 25

1 0.0273451
2 0.0662310
3 0.1055327
4 0.1449246
5 0.1843499
6 0.2237909
7 0.2632406
8 0.3026954
9 0.3421534
10 0.3816136
11 0.4210750
12 0.4605373
13 0.5000000
14 0.5394627
15 0.5789250
16 0.6183864
17 0.6578466
18 0.6973046
19 0.7367594
20 0.7762091
21 0.8156501
22 0.8550754
23 0.8944673
24 0.9337690
25 0.9726549



Robert Scott
Ypsilanti, Michigan

 

I guess that this stuff might be helpful:
http://people.sc.fsu.edu/~burkardt/cpp_src/prob/prob.html

It has the following calculations:
WEIBULL_CDF evaluates the Weibull CDF.
WEIBULL_CDF_INV inverts the Weibull CDF.
WEIBULL_CDF_VALUES returns some values of the Weibull CDF.
WEIBULL_CHECK checks the parameters of the Weibull CDF.
WEIBULL_MEAN returns the mean of the Weibull PDF.
WEIBULL_PDF evaluates the Weibull PDF.
WEIBULL_SAMPLE samples the Weibull PDF.
WEIBULL_VARIANCE returns the variance of the Weibull PDF.
WEIBULL_DISCRETE_CDF evaluates the Discrete Weibull CDF.
WEIBULL_DISCRETE_CDF_INV inverts the Discrete Weibull CDF.
WEIBULL_DISCRETE_CHECK checks the parameters of the discrete Weibull
CDF.
WEIBULL_DISCRETE_PDF evaluates the discrete Weibull PDF.
WEIBULL_DISCRETE_SAMPLE samples the discrete Weibull PDF.

For parameter estimation, usually Newton's method, simulated
annealing, or Bayesian methods are used.
Here is a paper with C code at the bottom that does it:
http://www.ohiolink.edu/etd/send-pdf.cgi/Thomas Gina M.pdf?acc_num=ohiou1178901727

 

The package called dataplot (which includes source code if you want
it) computes Weibull median ranks among other things.
http://www.itl.nist.gov/div898/software/dataplot/

 

Robert,

For what it's worth... You're answers match Excel's to the 7 decimal
places you've given. You can check for yourself using the FINV
function.