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3/2003 Rev 1 I.2.14 – slide 1 of 28 Part IReview of Fundamentals Module 2Basic Physics and Mathematics Used in Radiation Protection Session 14Statistics.

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Presentation on theme: "3/2003 Rev 1 I.2.14 – slide 1 of 28 Part IReview of Fundamentals Module 2Basic Physics and Mathematics Used in Radiation Protection Session 14Statistics."— Presentation transcript:

1 3/2003 Rev 1 I.2.14 – slide 1 of 28 Part IReview of Fundamentals Module 2Basic Physics and Mathematics Used in Radiation Protection Session 14Statistics - Tests Session I.2.14 IAEA Post Graduate Educational Course Radiation Protection and Safety of Radiation Sources

2 3/2003 Rev 1 I.2.14 – slide 2 of 28  We will also discuss statistical tests including  Student T  Chi squared  Chauvenet’s Criteria Overview

3 3/2003 Rev 1 I.2.14 – slide 3 of 28 The “T” test is used to determine if two different sets of data have the same mean Student’s “T”- test M 1 – M 2 r1r1r1r1 t1t1t1t1 r2r2r2r2 t2t2t2t2 +  T =

4 3/2003 Rev 1 I.2.14 – slide 4 of 28 A sample of drinking water was counted for 10 minutes and resulted in 530 counts. A 30 minute background count gave a background count rate of 50 cpm. At the 95% confidence level, was there any radioactivity in the water? Sample Problem #1

5 3/2003 Rev 1 I.2.14 – slide 5 of 28 The results differ from zero by 1.13 standard deviations. This includes 74% of the area between  1.13 standard deviations. Solution to Sample Problem #1 M 1 – M 2 r1r1r1r1 t1t1t1t1 r2r2r2r2 t2t2t2t2 +  T = 53 –  = = 1.13

6 3/2003 Rev 1 I.2.14 – slide 6 of 28  We would expect a difference this great or greater 26 times out of 100 (100% - 74%) if the samples came from the same population  To be significant at the 95% confidence level, the difference between the two means would have to be great enough to be observed only 5 times or less out of 100.  The difference between the two samples is not significant. Solution to Sample Problem #1

7 3/2003 Rev 1 I.2.14 – slide 7 of 28 To test the null hypothesis that the normal population has a standard deviation  =  0 at the significance level  based on a sample size n, the chi-squared statistic (  2 ) is 2222Statistics (n – 1)s 2 o2o2o2o2 2 =2 =2 =2 = 

8 3/2003 Rev 1 I.2.14 – slide 8 of 28  Radioactive decay is a statistically random process  The chi-squared (  2 ) test compares the variability of results to the predicted variability to determine if the observed results are consistent with that expected for random radioactive decay. 2222Statistics

9 3/2003 Rev 1 I.2.14 – slide 9 of 28  2 is calculated as the sum of the squares of the difference of the observed count rate from the mean count rate, and is then divided by the mean count rate. 2222Statistics

10 3/2003 Rev 1 I.2.14 – slide 10 of 28 Degrees of Freedom (n-1)  2 95% acceptance – – – – 2222Statistics (X i – X ave ) 2 X ave 2 =2 =2 =2 =

11 3/2003 Rev 1 I.2.14 – slide 11 of 28 NIST/SEMATECH e-Handbook of Statistical Methods, 2222Statistics

12 3/2003 Rev 1 I.2.14 – slide 12 of 28 A long-lived radioactive sample is counted 20 times with the following results Calculate the mean and standard deviation. Sample Problem #1

13 3/2003 Rev 1 I.2.14 – slide 13 of 28 Mean = counts Standard deviation = 13 counts Calculate the  2 value for these data. Does the value indicate satisfactory operation? Solution to Sample Problem #1

14 3/2003 Rev 1 I.2.14 – slide 14 of 28 Solution to Sample Problem #1 (n – 1)s 2 X ave 2 =2 =2 =2 = (19)(169)526.4 2 =2 =2 =2 = = 6.1

15 3/2003 Rev 1 I.2.14 – slide 15 of 28  The  2 value of 6.1 is outside the acceptable range of 8.91 – for 19 degrees of freedom at the 95% confidence level  This indicates that the counting system does not have satisfactory performance  That is, the variation in the values observed cannot be accounted for by statistical variability in the number of counts alone Solution to Sample Problem #1

16 3/2003 Rev 1 I.2.14 – slide 16 of 28 Data Rejection  Sometimes there may be variability because of factors other than the random nature of radioactive decay. For example, an error might be made in preparing a sample for counting.  This could result in anomalous results which might provide a basis for data rejection. If the data is kept, the anomalous data may introduce a significant error in the overall average.

17 3/2003 Rev 1 I.2.14 – slide 17 of 28 Chauvenet’s Criterion  Chauvenet’s criterion is a method of evaluating an observation relative to other data to determine if it is acceptable or not.  Chauvenet’s ratio, CR, is defined as the ratio of the suspect count’s deviation from the experimental mean to the square root of the experimental mean.

18 3/2003 Rev 1 I.2.14 – slide 18 of 28 Chauvenet’s Criterion If the calculated ratio is greater than that listed in the table for the specified number of observations, the datum may need to be rejected. (x - )  CR = _x _x

19 3/2003 Rev 1 I.2.14 – slide 19 of 28 Chauvenet’s Criterion  One investigator noted that this method may result in rejection of good observations about 40% of the time.  As an alternative, you may consider rejecting data that deviates from the mean by 2  or 3 .

20 3/2003 Rev 1 I.2.14 – slide 20 of 28 Number of data LimitingRatio Number LimitingRatio Chauvenet’s Criterion

21 3/2003 Rev 1 I.2.14 – slide 21 of 28 Sample Problem #1 Using the data in the following table, check for possible rejection. Try the highest and lowest values, 32 and 11, respectively.

22 3/2003 Rev 1 I.2.14 – slide 22 of 28 Observation Gross Counts Sample Problem #1 Observation Gross Counts Observation

23 3/2003 Rev 1 I.2.14 – slide 23 of 28 Step 1 – Determine the mean, standard deviation, and number of degrees of freedom: = 21s 2 = 28N-1 = 24 = 21s 2 = 28N-1 = 24 Solution to Sample Problem #1 _x

24 3/2003 Rev 1 I.2.14 – slide 24 of 28 Step 2 - Perform a  2 analysis on the data:  2 = = 32 The  2 value is within the acceptable range. Solution to Sample Problem #1 (24)(28)(21)

25 3/2003 Rev 1 I.2.14 – slide 25 of 28 Step 3 – Determine the Chauvenet’s Ratio for the data being evaluated: CR 32 = = 2.4 CR 11 = = 2.2 Solution to Sample Problem #1 (32 – 21)  21 ( )  21

26 3/2003 Rev 1 I.2.14 – slide 26 of 28  For 25 observations, the CR limit is 2.33 (see slide 20)  CR 32 = 2.4 > 2.33, so the value of 32 should be rejected  CR 11 = 2.2 < 2.33, so the value of 11 should be retained  After rejecting the value of 32, the mean, variance, and chi-square should be recalculated and Chauvenet’s ratio re-evaluated Solution to Sample Problem #1

27 3/2003 Rev 1 I.2.14 – slide 27 of 28  Dan Lurie and Roger H. Moore, “Applying Statistics,” NUREG-1475, U.S. Nuclear Regulatory Commission, (1994).  “Statistics Manual,” U.S. Naval Ordnance Test Station, NAVORD Report 3369, Dover Publication, Inc., (1960). Statistics – References

28 3/2003 Rev 1 I.2.14 – slide 28 of 28  Cember, H., Johnson, T. E., “Introduction to Health Physics,” 4 th Edition, McGraw- Hill, New York (2008).  Martin, A., Harbison, S. A., Beach, K., Cole, P., An Introduction to Radiation Protection, 6 th Edition, Hodder Arnold, London (2012).  C. H. Wang, D. L. Willis, and W. D. Loveland, “Radiotracer Methodology in the Biological, Environmental, and Physical Sciences,” Prentice-Hall, (1975). Where to Get More Information


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