Radiocarbon Dating The Shroud. A Critical Statistical Analysis.

Table 1 (Only for the Shroud)	d13C	Table 2
Ar. 591-+30 690-+35 606-+41 701-+33	-25	646-+31 689-+16*
Ox. 795-+65 730-+45 745-+55	-27	750-+30 691-+31'
Zu. 733-+61 722-+56 635-+57 639-+45 679-+51	-25.1	676-+24
* = Weighted mean ' = Unweighted mean.		X² = 6.4
		%SL. = 5

The reason for a statistical analysis of radiocarbon dates, is the FACT that, theoretically, no experimental result is absolute, and to some extent, subject to chance error. Even an EXACT measurement is still due, to some extent, to chance. Statistical analysis enables the analyst, to determine, with a pre-determined degree of confidence, (here claimed to be at least 95% ) the degree of variations which are due to chance and to known and unknown influences. When applying statistical tests, caution should be observed with data suspected of appreciably differing from a normal distribution.

Examples of distributions:
Normal = Z. Student = t. Chi^2 = X^2. Inverted Beta = F.

The British Museum, selected the Chi^2 to be the criterion for the assessment of the radiocarbon dating results for the Shroud. The MAXIMUM Chi^2 test value for 95% confidence and (3-1) degrees of freedom is 5.99. Theoretically, if the calculated Chi^2 test value could have occurred only by chance, with a probability LESS than that selected, then the set of data would be considered as being DIFFERENT. In practice : Any Chi^2 test value LARGER than 5.99, excludes the claimed 95 % confidence.

First I reworked the data given in Tables 1 & 2, using the classical statistical analyzing method, based on the "Central Limit Theorem." I used the method given in "Perry's Chemical Engineer's Handbook", my technical bible for many years. Note that I only verify the assessment of the radiocarbon dating results, NOT the conversion into calendar ages. All calculations are made on a Sharp PC 4700, using an Ability (Lotus) Spreadsheet. None of the results presented here are biased. Due to rounding up numbers, small differences may occur. For non-mathematical minds, all calculations are worked out completely. The use of complicated formulas is limited to the minimum.

Classical analysis radiocarbon dates for the Shroud of Turin.

Table 1 Means	Table 2
Ar. (591+690+606+701)/4 = 647	F-test = 4.7 > 4.26 = Reject.
Ox. (795+730+745)/3 = 757	(647+757+682)/3 = 695 (Un.Mean)
Zu. (733+722+639+635+679)/5= 682	(2588+2270+3408)/12 = 689 (W.Mean)

Table 2 Errors based on the scatter of results.

Ar.	{[(591-647)^2+...........+(701-647)^2]/(3x4)}^0.5 = 28
Ox.	{[(795-757)^2+(730-757)^2+(745-757)^2]/(2x3)}^0.5 = 20
Zu.	{[(733-682)^2+...........+(679-682)^2]/(4x5)}^0.5 = 20
UM.	{[(647-695)^2+(757-695)^2+(682-695)^2]/(2x3)}^0.5 = 32
WM.	{[(647-689)^2+(757-689)^2+(682-689)^2]/(2x3)}^0.5 = 33

Comparison Table 2 versus Computer.

	Arizona	Oxford	Zurich	U.Mean	W.Mean
*Nature*	646-+31	750-+30	676-+24	691-+31	689-+16
Computer	647-+28	757-+20	682-+20	695-+32	689-+33

Because the errors based on the scatter are NOT quoted errors, I did not use the X^2 test, but the more powerful F-test, based on the combination of the 3 means and 12 independent measurements. For 95% confidence, and (3-1) - (12-3) degrees of freedom the maximum F-test, value is 4.26. By chance alone, F should be -+ 1.

	Ari	Oxf	Zur
	701	795	733
	690	745	722
	606	730	679
	591		639
			635
Sums	2588	2270	3408	= 8266
	/4	/3	/5	/12
Mean	647	757	682	689

Between laboratories =2588^2/4+2270^2/3+3408^2/5-8266^2/12=21066
Residual = (701-647)^2+.............(635-682)^2=20167
Total = Sum (Between laboratories + Residual) =41233

Table of variances.

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F Ratio < 4.26
Between	21066	3-1	21066/2=10533	10533/2241 = 4.7
Residual	20167	12-3	20167/9= 2241	Conclusion :
Total	41233	11		4.7 > 4.26

Conclusion :
An F-test value of 4.7 > 4.26, states there is a SIGNIFICANT DIFFERENCE between the results given by the three laboratories. In regard to the tabulated data, given in Table 1 of Nature, a certain amount of unexplained variation of the individual runs within and between laboratories is indicated.

Student t distribution test.
The t test compares the sub mean -+error to the final mean. AMS date for 1 run are the mean of about 10 measurements and assuming that the quoted error is equal to the standard error, given in Nature. Because we have to deal with a two tail test, we have to take the critical t value 2.26 for 97.5% confidence and (10-1) degrees of freedom.

A : t-Test versus the sub-means t = < 2.26

Arizona : 646-+28	Oxford : 757-+20	Zurich : 682-+20
(701-646)/33 = 1.67	(795-757)/65 = 0.58	(733-682)/61 = 0.84
(690-646)/35 = 1.26	(757-745)/55 = 0.22	(722-682)/56 = 0.71
(646-606)/41 = 0.98	(757-730)/45 = 0.60	(682-679)/51 = 0.06
(646-591)/30 = 1.83		(682-639)/45 = 0.96
		(682-635)/57 = 0.82

B : t-Test versus the final Mean Final Error = 689-+31 t = < 2.26

Arizona	Oxford	Zurich
(701-689)/33 = 0.36	(795-689)/65 = 1.63	(733-689)/61 = 0.70
(690-689)/35 = 0.03	(745-689)/55 = 1.02	(722-689)/56 = 0.59
(689-606)/41 = 2.02	(730-689)/45 = 0.91	(689-679)/51 = 0.20
(689-591)/30 = 3.27		(689-639)/45 = 1.11
		(689-635)/57 = 0.95

This test indicates that the Arizona date 591-+30 is probably an OUTLIER. Later I tested the individual dates also by a NEW developed test, called the IEM-EEM criterion (Scott et al. 1990). Here the QUOTIENT of the difference between measurement and final mean is divided by the square root of the sum of BOTH the squared errors, it should be < 1.

Here all calculations are based on W & W.

IEM = < 1	EEM = < 1
Arizona 646-+17	Mean = 672-+13
(701-646)/(17^2+33^2)^0.5 = 1.49
(690-646)/(17^2+35^2)^0.5 = 1.13	(672-646)/(17^2+13^2)^0.5 = 1.21
(646-606)/(17^2+41^2)^0.5 = 0.91
(646-591)/(17^2+30^2)^0.5 = 1.59
Oxford 749-+31
(795-749)/(30^2+65^2)^0.5 = 0.64
(749-745)/(30^2+55^2)^0.5 = 0.06	(749-672)/(30^2+13^2)^0.5 = 2.35
(749-730)/(30^2+45^2)^0.5 = 0.41
Zurich 676-+24
(733-676)/(24^2+61^2)^0.5 = 0.87
(722-676)/(24^2+56^2)^0.5 = 0.75
(679-676)/(24^2+51^2)^0.5 = 0.05	(676-672)/(24^2+13^2)^0.5 = 0.15
(676-639)/(24^2+45^2)^0.5 = 0.73
(676-635)/(24^2+57^2)^0.5 = 0.66

Following the criterion, proposed by Dr. Scott et al, the dates for Arizona (3 x IEM + EEM ) and Oxford (EEM), are OUT of range. Upon meeting Dr. Scott at a Shroud Symposium in New York, I asked her in writing, to verify the Nature dates by the IEM-EEM criterion. But Dr. Scott refused a direct answer.......promising the reply by letter. I am still waiting.

I asked the British Museum for some explications. Dr. Morven Leese explained the LOW Arizona errors as follows : Arizona did not send in FOUR, but EIGHT "paired" dates, each pair measured the same day, with the same set of standards and blanks. The EIGHT "dependent" dates were combined in FOUR "independent dates, given in Table 1 (Nature). Each pair of DEPENDENT dates are combined into an INDEPENDENT date.

A: 606-+41 574-+45 >>>> [606/41^2 + 574/45^2]/[1/41^2 + 1/45^2] = 591

Error = [1/(1/41^2 + 1/45^2)]^0.5 = 30

B: 753-+51 632-+49 >>>> [753/51^2 + 632/49^2]/[1/51^2 + 1/49^2] = 690

Error = [1/(1/49^2 + 1/51^2)]^0.5 = 35

C: 540-+57 676-+59 >>>> [540/57^2 + 676/59^2]/[1/57^2 + 1/59^2] = 606

Error = [1/(1/57^2 + 1/59^2)]^0.5 = 41

D: 701-+47 701-+49 >>>> [701/47^2 + 701/47^2]/[1/47^2 + 1/47^2] = 701

Error = [1/(1/47^2 + 1/47^2)]^0.5 = 33

Yet mathematically correct, this "re-calculation" should have been notified by the authors of the report. At first, the "combination" was silently denied. It took more than TWO years before Arizona confirmed the combination made at the request of the British Museum. The arbitrary "enlarging" of the error from 17 to 31 was never solved. The best results were obtained by averaging the paired errors.
[41 + 45]/2=43 [51 + 49]/2=50 [57 + 59]/2=58 [47 + 47]/2=47
Error on the mean : [1/(1/43^2 + 1/50^2 + 1/58^2 + 1/47^2)]^0.5 = 24.

Based on the scatter, the standard error is 28. At first, I believed that the error was due to the fact, that Arizona included the error in delta 13C at a later stage. The D13C error can be estimated : 31 = [28^2 + (D13C)^2]^0.5
D13C = [961 - 784]^0.5 = 13.

The value appears to be too large. The British Museum gave several versions, but no specific answer was given........Dr. Leese explained also that the British Museum did not use the classical method, but a NEW method, developed by the Australian scientists Drs. Wilson & Ward. (Archeometry 20. 1978). I am still grateful to Drs. Wilson & Ward for their precious help. The method is based on the QUOTED error for each measurement. Quoted errors do incorporate the statistical (counting) error, the error of the scatter of results for standard and blanks, and the (small) uncertainty in the delta 13C determination. Normal statistical and systematic errors are 0.25 %. The total error will be : [0.25^2 + 0.25^2]^0.5 = 0.356 %.

Note : It seems a little strange, that in the W & W method, the LARGER the errors, the BETTER the CHI^2 test will become.

Formula W & W :
Consider a number of measurements -+errors : A-+a........N-+n
Mean X = [A/a^2 + ... + N/n^2]/[1/a^2 + .... + 1/n^2]
Variance: 1/(1/a^2 + ... + 1/n^2)
Error x = [1/(1/a^2 + ...1/n^2)]^0.5
Chi^2 = (A-X)^2/a^2 + ....(N-x)^2/n^2

Calculations :
Arizona :

701/33^2 + 690/35^2 + 606/41^2 + 591/30^2	2.2241414
--------------------------------------------------	= --------------- = 646
1/33^2 + 1/35^2 + 1/41^2 + 1/30^2	0.0034404
Error = [1/0.0034404]^0.5 = 290.7^2 = 17.05 = 17

Oxford :

795/65^2 + 745/55^2 + 730/30^2	0.7949403
---------------------------------------	= ------------- = 749
1/65^2 + 1/55^2 + 1/45^2	0.0010609

Error = [1/0.0010609]^0.5 = 942.6^2 = 30.70 = 31

Zurich :

733/61^2 + 722/56^2 + 679/51^2 + 639/45^2 + 635^2	1.1992731
-------------------------------------------------------------	= ---------- = 676
1/61^2 + 1/56^2 + 1.51^2 + 1/45^2 + 1.57^2	0.0017734

Error = [1/0.0017734]^0.5 = 563.9^2 = 23.75 = 24

Weighted Mean :

646/17^2 + 749/31^2 + 676/24^2	4.1883016
--------------------------------------	= ----------- = 671.55 = 672
1/17^2 + 1/31^2 + 1/24^2	0.0062368

Error = [1/0.0062368]^0.5 = 160.3^2 = 12.66 = 13.

Note : If we assume, as noted in Nature, that all dates have the same weight, then we can avoid rounding off and calculate the mean -+error as follows :

2.2241414 + 0.7949403 + 1.1992731	4.2183548
-----------------------------------------	= ------------- = 672
0.0034404 + 0.0010609 + 0.0017734	0.0062747

Error = [1/0.0062747]^0.5 = 159.37^0.5 = 12.62 = 13

X^2 test:
Maximum 5.99 for 95% confidence and (3-1) degrees of freedom.
[(646-672)^2/17^2]+[(749-672)^2/31^2]+(676-672)^2/24^2] = 8.43
The computer, with NO ROUNDING OFF of the values, will give X^2 = 8.76

Conclusion :
An X^2 test value 8.43 > 5.99 states that there is a SIGNIFICANT DIFFERENCE between the results of the 3 laboratories. From the X^2 test result, one can determinate the % significance level : 2.718^-(8.43/2) = 1.3 %. From this test, one may conclude, that the probability of obtaining, by chance alone, a scatter as high as that observed for the Shroud, is only 13 in 1000. Because we assume all radiocarbon dates to be correct, we must conclude, that the SMALL samples, taken at the same place, do not have the same radioactivity and are not REPRESENTATIVE for the Shroud.

Comparison Table 2 versus the computer.

	Arizona	Oxford	Zurich	Mean	Chi^2	S.L.%
*Nature*	646-+31	750-+30	676-+24	689-+16	6.4	5
Computer	646-+17	749-+31	676-+24	672-+13	8.4	1.3

Following the Nature report :
The spread of measurements for sample 1 (Shroud) is somewhat greater than would be expected from the errors quoted. The X^2 test shows that it is UNLIKELY, that the quoted errors FULLY reflect the overall scatter. Not noted is, that a Chi^2 test value > 5.99 DOES EXCLUDE the claimed 95 % confidence. In such cases, one should verify and reject possible "outliers." Possible "outliers" are the Arizona dates 591 and 606, and the Oxford date 795. One may wonder why, the British Museum, did not use the X^2 test for each laboratory.

X^2 test for Arizona (For 4-1 degrees of freedom X^2 = < 7.81)

(701-646)^2	(690-646)^2	(646-606)^2	(646-591)^2
--------------	+ -------------	+ ------------	+ ----------	= 8.33 > 7.81
33^2	35^2	41^2	30^2

X^2 test for Oxford (For 3-1 degrees of freedom X^2 = <5.99)

(795-749)2	(749-745)^2	(749-730)^2
-------------	+ ---------------	+ -------------	= 0.68 < 5.99
65^2	55^2	45^2

X^2 test for Zurich (For 5-1 degrees of freedom X^2. = <9.49)

(733-676)^2	(722-676)^2	(679-676)^2	(676-638)^2	(676-635)^2
--------------	+ ------------	+ ------------	+ ------------	+ --------------	= 2.69
61^2	56^2	51^2	45^2	57^2

X^2 for the Mean (For 3-1 degrees of freedom X^2 = <5.99)

(749-672)^2	(676-672)^2	(672-646)^2
--------------	+ -------------	+ -------------	= 8.54 > 5.99
31^2	24^2	17^2

The X^2 test indicates INCONSISTENCY for Arizona and the Mean. But the British Museum, on finding "INCONSISTENCY" decided NOT to check for outliers, but to BIAS the results for Arizona by enlarging the error from 17 to 31. This shifts the UNBIASED mean date 672-+13 to 689-+16 and consequently the Chi^2 test value from 8.54 to 6.4 and also the % significance level from 1.3 to 4.4. Finally, it was decided to REPLACE the weighted mean 689-+16 by the UNWEIGHTED mean 691-+31 and to ENLARGE the multiplying factor 1.96 to 2.6, based on an analysis of variance on the 12 individual measurements. The number of degrees of freedom (d), lying between (3-1) inter- and (12-3) intra- laboratory degrees of freedom was ESTIMATED at 5.

Analysis of variance :

Arizona	(701+690+606+591)/4 = 647
Var.	(701-647)^2+(690-647)^2+(647-606)^2+(647-591)^2 = 9582
Oxford	(795+745+730)/3 = 757
Var.	(795-757)^2+(757-745)^2+(757-730)^2 = 2317
Zurich	(733+722+679+639+635)/5 = 682
Var.	(733-682)62+(722-682)^2+(682-679)^2+(682-639)^2+ (682-635)^2 = 8261

[9582/4 + 2317/3 + 8261/5]^2

------------------------------------------------------------------ =

((9582/4)^2)/(4+1) + ((2317/3)^2)/(3+1) + ((8261/5)^2)/(5+1)

(2395.5 + 722.3 + 1652.2)^2	22752900
--------------------------------------	= ------------	= 13.13 = 13-3 = 10
1147684 + 130429.3 + 454960.8	1733074

The multiplying factor for d=10 and 97.5% confidence = 2.23. (This is for a two tail test and 1 -(2x0,25) = 95 % confidence). An analysis of variance based on 3 laboratories and 16 measurements will only differ for Arizona :

36587/8 = 4573.4 and 4573.4^2/9 = 2323973.1

[4573.4 + 722.3 + 1652.2]^2	48273314
-------------------------------------- =	------------	= 16.59 = 17-3 = 14
2323973.1 + 130429.3 + 454960.8	2909363

Note : Following the Wilson & Ward method the variance is the squared quoted error.

Strictly following W & W :

[17^2/4 + 30^2/3 + 24^2/5]^2
------------------------------------------------------	= 9.22 -3 = 6
[(17^2/4)^2]/5 + [(30^2/3)^2]/4 + [(24^2/5)^2]/6

With Arizona error 31 :

[31^2/4 + 30^2/3 + 24^2/5]^2
-----------------------------------------------------	= 11.85 - 3 = 9
[(31^2/4)^2]/5 + [(30^2/3)^2]/4 + [(24^2/5)^2]/6

These small differences may seem of little importance, but it is one more indication of a "biased" assessment, with the only purpose of a systematic enlargement of the error range. The same goes for the estimation of the error for the Shroud from the scatter. Here one "forgets" that quoted errors in AMS are deviated from NBS standards and blanks, measured during the SAME run with a series of targets, from the Shroud. Note that for samples 2, 3 and 4, the limits were obtained in the usual way.

This can be illustrated by the Chi^2 and the significance level, given in table 2 (Nature).

Sample	1	2	3	4
X^2 (2 d.f.)	6.4	0.1	1.2	2.4
Sig.Level %	5	90	50	30

Note : 5% means a 1 chance in 20 that the dates are consistent.

COMMENT :
The advice of Prof. Hoel (University of California) is : "In the case of a X^2 test value close to the limit, it is better not to use such dates in further calculations, but to ask for more and better dates." Despite the use of the NBS standard (Oxalic Acid Standard SRM 4990) the results for the radiocarbon dating of the Shroud show a SYSTEMATIC BIAS and UNEXPLAINED VARIABILITY. An indication of the inhomogeneity of the micro-samples, used in AMS.

Note : Theoretically, all statistical assessments should have been made BEFORE the dC13 correction to -25 o/oo has been applied. In this case the RAW Oxford date would have been -+ 40 years OLDER.

RAW RESULTS.

Mean :	(646/17^2 ) + (790/31^2) + (676/24^2)
	-----------------------------------------	= 678 -+ 13
	1/17^2 +1/31^2 + 1/24^2

Chi^2 :	(646-678)^2	(790-678)^2	(676-678)^2
	---------------	+ --------------	+ --------------	= 17.49 > 5.99
	17^2	30^2	24^2

Their can be no doubt, the RAW Oxford results are probably "OUTLIERS." Let us say, that Oxford was saved by the dC13 determination, made IN ANOTHER laboratory.

CONCLUSION :
Facts :
The Arizona error was arbitrary enlarged from 17 to 31. The Wilson & Ward mean 689-+16 was replaced by the UNWEIGHTED mean 691-+31. The multiplying t-factor for 95% confidence was enlarged from 1.96 to 2.6. The claimed "at least 95 % confidence" for the medieval dating of the Shroud is NOT supported by statistical analysis. One may wonder, why these OBVIOUS facts, were not spotted by the "team of peers" who judge all papers before publication in Nature. Even stranger is the FACT, that Prof. Bray of the "Istuto di Metrologia" of Turin, confirmed that the results of the 3 labs were mutually compatible, and that, on the evidence submitted, none of the means WERE questionable. Prof. Bray declared not to be at liberty to answer any questions. His answer was : "On the evidence submitted, no averaged results APPEAR questionable. The scatter for sample 1 is about equal to the limit." The only possible explanation is, that NOT all evidence was submitted to Prof. Bray. Prof. Bray refused to comment on the "combination from EIGHT to FOUR Arizona dates. I asked the editor of Nature, to compare my calculations with the results given by Damon et al. Following Dr. Laura Garwin (Physical Science Editor) : "You are asking me questions that are beyond my ability to answer. The Damon et al paper was refereed by qualified referees and no dissatisfaction was raised with the assignment or errors." I also asked the advice of Prof. Bene (University of Geneve). "I would like to congratulate you for the quality of your work. You established definitive evidence, that the measurements made on the linen of the Shroud are NOT homogeneous and that they should be rejected." Prof. Jouvenoux (University of Marseille-Aix) : "Van Haelst was probably the first to question the radiocarbon dating of the Shroud in a scientific way."

Note :
These calculations have been verified by : Dr. Leese (British Museum). She argues that the differences are caused by the use of different weighting systems! The French professional statisticians Bouclier de Carbon and Forestier, agreed with my results. Dr. Hedges (Oxford) replied : "I do not need statistics....." Dr. Woeffli (Zurich) and Dr. Damon (Arizona) did not reply. Prof. Evin (Lyon. France) : "Statistics will not change the facts...." This paper has been presented at the CIELT International Shroud Symposium. Rome 1993. (Actes CIELT Rome. ISBN 2.868.39.311X Ed. Fr.X. Guibert Paris. pages 207-218).

A REQUEST.
Please use your computer to verify the calculations made in this paper. The W & W method is easily to program on a Lotus Spreadsheet. Let your computer judge the Nature dates and this paper.

Example (Mean -+ error for Arizona):

	A	B	C	D	E
	Age	Error	(b-)^2	(a-)/(c-)	1/(c-)
1	701	33	1089	701/1089=0.6437098	1/1089=0.0009182
2	690	35	1225	690/1225=0.5632653	1/1225=0.0008153
3	606	41	1681	606/1681=0.3604997	1/1681=0.0005948
4	591	30	900	591/ 900=0.6566666	1/ 900=0.0011111
5	(d5)/(e5)	[1/(e5)]^0.5		@sum = 2.2241414	@sum = 0.0034394

a5 = Age = 2.2241414/0.0034294 = 646,6 = 647
b5 = Error = [1/0.0034394]^0.5 = 290.7^0.5 = 17.05 = 17

Repeat this for Oxford, Zurich and calculate the Weighted Mean-+Error from the sub-means-+errors.

The Chi^2 (X^2) test :	Example :
(Ar.Mean-W.Mean)^2/Ar.Error^2	(646-672)^2/17^2 = 2.5130111
(Ox.Mean-W.Mean)^2/Ox.Error^2	(749-672)^2/31^2 = 6.1696149
(Zu.Mean-W.Mean)^2/Zu.Error^2	(676-672)^2/24^2 = 0.0277777
	Chi^2 = 8.7104037

HOW TO BIAS THE COMPUTER RESULTS, TOWARDS 691-+31 ?
The computer results differ TWO times SIGNIFICANTLY from the Nature dates (Table 2).

Mean* = Result with Arizona 646-+17.

Mean$ = Result with Arizona 646-+31.

	Arizona	Mean*	Mean$
Computer	646-+17	672-+13	689-+16
*Nature*	646-+31	689-+16	691-+31
Factor	31/17	31/16

ESTIMATION OF INDIVIDUAL ERRORS (Table 1) IN FUNCTION OF 691-+31.
First the individual Arizona errors should be multiplied by a factor of 31/17.
Secondly, all individual errors should be multiplied by a factor of 31/16.

Table 1 Individual errors.

Arizona	x 31/17	x 31/16	Oxford	x 31/16	Zurich	x 31/16
701-+33	60	117	795-+65	126	733-+61	118
690-+35	64	124	745-+55	107	722-+56	108
606-+41	75	145	730-+45	87	679-+51	99
591-+30	55	106			639-+45	87
					635-+57	110

Table 2 Errors on the mean.
Arizona error :
[1/(1/117^2 + 1/124^2 + 1/145^2 + 1/106^2)]^0.5 = 60

Mean age :

646/60^2 + 750/60^2 + 676/46^2	0.7072483
-------------------------------------	= --------------	= 688
1/60^2 + 1/60^2 1/46^2	0.0012080

X^2 test :
(689-646)^2/60^2 + (750-689)^2/60^2 + (689-676)^2/46^2 = 1.64
Significance level = 2.718^-(1.64/2) = 44 %

Comparison with X^2 and % Sig.Level.
C = Computer strictly W & W for sample 1.
CC = Computer with errors in function of final error 31.

Sample	C	1	4	CC	3	2
X^2	1.6	6.4	2.4	1.64	1.3	0.1
S.L.%	1.3	5	30	44	50	90

Conclusion : The computer estimated individual errors for the Shroud, in function of error 31 on the unweighted mean 691 are NOT abnormally high and NOT out of range for the claimed AMS accuracy.

Because AMS radiocarbon dating results are NOT absolute values, but the result of a number of X runs, with a frequency Y, all results are given with a certain error. The more measurements closer to the UNKNOWN true value and TRUE error, the closer the averaged and error mean will become. Theoretically, an INFINITE number of measurements will yield a result INFINITELY close to the TRUE value-+ error. Practically, one has to take into account, the precision of the measurement and the systematic errors of the method.

Following the Nature report, the radiocarbon age of the Shroud, with a 95 % confidence, is 691-+31 years BP. (Before Present = 1950). By backward calculation from this result, the INDIVIDUAL errors can be estimated.

Table 1 in Nature would look like this :

Arizona	x 1.82	591-+ 55	701-+ 60	690-+ 64	606-+ 75
All errors	x 1.94	591-+107	701-+116	690-+124	606-+145		(Ar)
		795-+126	745-+107	730-+ 87			(Ox)
		733-+118	722-+108	679-+ 99	639-+ 87	635-+110	(Zu)

Because THEORETICALLY, any measurement between the limits 606 - 145 = 416 and 795 + 126 = 921 can be the TRUE date, a good estimation of the 95% confidence range can obtained from a DOUBLE statistical analysis based on MINIMUM and MAXIMUM values.

Minimum				Maximum
	Ari.	Oxf.	Zur.	Ari.	Oxf.	Zur.
	484	669	615	698	921	851
	585	638	614	817	852	830
	566	643	580	814	817	778
	461		552	751		726
			525			745
Tot.	2096	1950	2887	3072	2590	3930
Mean	524	650	577	768	863	786
Er.	33	10	18	28	31	29
W.Mean.	578			799
Error	37			30
95 %	578-(2x37)=504			799+(2x30)=859

Any date between those limits 504-859 can be the TRUE date. Here, about 30 % of this age range will be YOUNGER than the HISTORICAL age of 1355 AD.

The Bienayme-Tchebychev formula.
In the case of non normal distributed dates, the error multiplying factors can be estimated, by the Bienayme-Tchebychec inequality.
Probability : {[X-M] < MF X SIGMA} > 1-1/MF^2.

Thus MF for:
   68 % = 1-1/1.77^2 = 1-1/3.31 = 1-0.32 = 0.68
   75 % = 1-1/2.00^2 = 1-1/4.00 = 1-0.25 = 0.75
   90 % = 1-1/3.16^2 = 1-1/10.0 = 1-0.10 = 0.90
   95 % = 1-1/4.47^2 = 1-1/20.0 = 1-0.05 = 0.95

The error range for 691-+31 at 95% confidence will be :
691-(4.47x31) = 552 < 691 < 830 = 691+(4.47X31)

Here about 21 % of this age range will be YOUNGER than the HISTORICAL age of 1355 AD. It is clear, that such large errors are unacceptable for the precision claimed for AMS age rc. determinations. (Source Prof. Jouvenoux. University of Marseille-Aix. France).