## Radiocarbon Dating The Shroud

## A Critical Statistical Analysis.

By Remi Van Haelst, BelgiumGraduated Industrial ChemistCopyright 1997All Rights ReservedReprinted by Permission

TheNaturereport.

The report "Radiocarbon dating the Turin Shroud", Damon et al, inNature, (Vol.337, Nr. 6200 pages 611-615. 16 Feb. 1989) concludes:

Table 1 (Only for the Shroud)d13CTable 2Ar. 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. X ^{2}= 6.4%SL. = 5 The reason for a statistical analysis of radiocarbon dates, is the

FACTthat, theoretically, no experimental result is absolute, and to some extent, subject to chance error. Even anEXACTmeasurement 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

LESSthan that selected, then the set of data would be considered as beingDIFFERENT. In practice : Any Chi^2 test valueLARGERthan 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,

NOTthe 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 MeansTable 2Ar. (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.

ArizonaOxfordZurichU.MeanW.MeanNature646-+31 750-+30 676-+24 691-+31 689-+16 Computer647-+28 757-+20 682-+20 695-+32 689-+33 Because the errors based on the scatter are

NOTquoted 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.F-test :

Ari Oxf Zur 701 795 733 690 745 722 606 730 679 591 639 635 Sums 2588 2270 3408 = 8266 /4 /3 /5 /12 Mean647 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 VariationSum of SquaresDegrees of FreedomMean SquareF Ratio < 4.26Between 21066 3-1 21066/2=10533 10533/2241 = 4.7 Residual 20167 12-3 20167/9= 2241 Conclusion :Total41233114.7 > 4.26

Conclusion :

An F-test value of 4.7 > 4.26, states there is aSIGNIFICANT DIFFERENCEbetween the results given by the three laboratories. In regard to the tabulated data, given in Table 1 ofNature, 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 inNature. 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-+28Oxford : 757-+20Zurich : 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

ArizonaOxfordZurich(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 aNEWdeveloped test, called theIEM-EEMcriterion (Scott et al. 1990). Here theQUOTIENTof the difference between measurement and final mean is divided by the square root of the sum ofBOTHthe squared errors, it should be < 1.Here all calculations are based on W & W.

IEM = < 1EEM = < 1Arizona 646-+17Mean = 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

OUTof range. Upon meeting Dr. Scott at a Shroud Symposium in New York, I asked her in writing, to verify theNaturedates 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

LOWArizona errors as follows : Arizona did not send inFOUR, butEIGHT"paired" dates, each pair measured the same day, with the same set of standards and blanks. TheEIGHT"dependent" dates were combined inFOUR"independent dates, given in Table 1 (Nature). Each pair ofDEPENDENTdates are combined into anINDEPENDENTdate.

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

TWOyears 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 aNEWmethod, 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 theQUOTEDerror 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, theLARGERthe errors, theBETTERthe 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 :Error = [1/0.0010609]^0.5 = 942.6^2 = 30.70 = 31

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

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

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

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

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

Note :If we assume, as noted inNature, that all dates have the same weight, then we can avoid rounding off and calculate the mean -+error as follows :Error = [1/0.0062747]^0.5 = 159.37^0.5 = 12.62 = 13

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

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, withNO ROUNDING OFFof the values, will give X^2 = 8.76

Conclusion :

An X^2 test value 8.43 > 5.99 states that there is aSIGNIFICANT DIFFERENCEbetween 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 theSMALLsamples, taken at the same place, do not have the same radioactivity and are notREPRESENTATIVEfor the Shroud.

Comparison Table 2 versus the computer.

ArizonaOxfordZurichMeanChi^2S.L.%Nature646-+31 750-+30 676-+24 689-+16 6.4 5 Computer646-+17 749-+31 676-+24 672-+13 8.4 1.3 Following the

Naturereport :

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 isUNLIKELY, that the quoted errorsFULLYreflect the overall scatter. Not noted is, that a Chi^2 test value > 5.99DOES EXCLUDEthe 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

INCONSISTENCYfor Arizona and the Mean. But the British Museum, on finding"INCONSISTENCY"decidedNOTto check for outliers, but toBIASthe results for Arizona by enlarging the error from 17 to 31. This shifts theUNBIASEDmean 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 toREPLACEthe weighted mean 689-+16 by theUNWEIGHTEDmean 691-+31 and toENLARGEthe 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 wasESTIMATEDat 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 Multiplying the factor for d=14 and 97.5 % confidence = 2.15.

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

SAMErun 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).

Sample1234X^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 aSYSTEMATIC BIASandUNEXPLAINED VARIABILITY.An indication of the inhomogeneity of the micro-samples, used in AMS.

Note :Theoretically, all statistical assessments should have been madeBEFOREthe dC13 correction to -25 o/oo has been applied. In this case theRAWOxford date would have been -+ 40 yearsOLDER.

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

RAWOxford results are probably"OUTLIERS."Let us say, that Oxford was saved by the dC13 determination, madeIN ANOTHERlaboratory.

CONCLUSION :Facts :

The Arizona error was arbitrary enlarged from 17 to 31. The Wilson & Ward mean 689-+16 was replaced by theUNWEIGHTEDmean 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 isNOTsupported by statistical analysis. One may wonder, why theseOBVIOUSfacts, were not spotted by the "team of peers" who judge all papers before publication inNature. Even stranger is theFACT, 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 meansWEREquestionable. Prof. Bray declared not to be at liberty to answer any questions. His answer was : "On the evidence submitted, no averaged resultsAPPEARquestionable. The scatter for sample 1 is about equal to the limit." The only possible explanation is, thatNOTall evidence was submitted to Prof. Bray. Prof. Bray refused to comment on the "combination fromEIGHTtoFOURArizona dates. I asked the editor ofNature, 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 areNOThomogeneous 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 theNaturedates and this paper.

Example (Mean -+ error for Arizona):

ABCDEAgeError(b-)^2(a-)/(c-)1/(c-)1701 33 1089 701/1089=0.6437098 1/1089=0.0009182 2690 35 1225 690/1225=0.5632653 1/1225=0.0008153 3606 41 1681 606/1681=0.3604997 1/1681=0.0005948 4591 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 = 647b5 = Error = [1/0.0034394]^0.5 = 290.7^0.5 = 17.05 = 17Repeat 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

APPENDIX.

HOW TO BIAS THE COMPUTER RESULTS, TOWARDS 691-+31 ?

The computer results differTWOtimesSIGNIFICANTLYfrom theNaturedates (Table 2).Mean* = Result with Arizona 646-+17.Mean$ = Result with Arizona 646-+31.

ArizonaMean*Mean$Computer646-+17 672-+13 689-+16 Nature646-+31 689-+16 691-+31 Factor31/17 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.

Arizonax 31/17x 31/16Oxfordx 31/16Zurichx 31/16701-+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

Oxford error :

[1/(1/126^2 + 1/107^2 + 1/87^2)]^0.5 = 60

Zurich error :

[1/(1/118^2 + 1/108^2 + 1/99^2 + 1/87^2 + 1/110)]^0.5 = 46

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

Error : [1/(1/60^2 + 1/60^2 + 1/46^2)]^0.5 = 31

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.

SampleC14CC32X^21.6 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 areNOTabnormally high andNOTout of range for the claimed AMS accuracy.

A NEW STATISTICAL ASSESSMENT FOR AMS. RADIOCARBON DATES.Because AMS radiocarbon dating results are

NOTabsolute 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 theUNKNOWNtrue value andTRUEerror, the closer the averaged and error mean will become. Theoretically, anINFINITEnumber of measurements will yield a resultINFINITELYclose to theTRUEvalue-+ error. Practically, one has to take into account, the precision of the measurement and the systematic errors of the method.

Example :All Oxford sub-errors, lower than 40 have been rounded up to 40.Following the

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

Table 1 inNaturewould look like this :

Arizonax 1.82 591-+ 55 701-+ 60 690-+ 64 606-+ 75 All errorsx 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 limits606 - 145 = 416 and 795 + 126 = 921can be theTRUEdate, a good estimation of the 95% confidence range can obtained from aDOUBLEstatistical analysis based onMINIMUMandMAXIMUMvalues.

MinimumMaximumAri.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 Mean524 650 577 768 863 786 Er.33 10 18 28 31 29 W.Mean. 578 799 Error37 30 95 %578-(2x37)=504 799+(2x30)=859 Any date between those limits 504-859 can be the

TRUEdate. Here, about 30 % of this age range will beYOUNGERthan theHISTORICALage 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

YOUNGERthan theHISTORICALage 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).

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