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

Mar 31, 2015

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Page 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.

3/2003 Rev 1 I.2.14 – slide 1 of 28

Part I Review of Fundamentals

Module 2 Basic Physics and MathematicsUsed in Radiation Protection

Session 14 Statistics - Tests

Session I.2.14

IAEA Post Graduate Educational CourseRadiation Protection and Safety of Radiation Sources

Page 2: 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.

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

Page 3: 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.

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

M1 – M2

r1

t1

r2

t2

+T =

Page 4: 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.

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

Page 5: 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.

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

M1 – M2

r1

t1

r2

t2

+T =

53 – 505310

5030

+ = = 1.13

Page 6: 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.

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

Page 7: 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.

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 Statistics

(n – 1)s2

o2

2 =

Page 8: 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.

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 Statistics

Page 9: 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.

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 Statistics

Page 10: 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.

3/2003 Rev 1 I.2.14 – slide 10 of 28

Degrees of Freedom (n-1)

2 Range @ 95% acceptance

19 8.91 – 32.85

24 12.40 – 39.36

29 16.05 – 45.72

50 32.36 – 71.42

2 Statistics

(Xi – Xave)2

Xave 2 =

Page 11: 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.

3/2003 Rev 1 I.2.14 – slide 11 of 28

NIST/SEMATECH e-Handbook of Statistical Methods, http://www.itl.nist.gov/div898/handbook/

2 Statistics

Page 12: 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.

3/2003 Rev 1 I.2.14 – slide 12 of 28

A long-lived radioactive sample is counted 20 times with the following results

526 542 539 502 540

535 514 519 555 527

522 533 518 512 520

526 525 516 541 516

Calculate the mean and standard deviation.

Sample Problem #1

Page 13: 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.

3/2003 Rev 1 I.2.14 – slide 13 of 28

Mean = 526.4 counts

Standard deviation = 13 counts

Calculate the 2 value for these data. Does the value indicate satisfactory operation?

Solution to Sample Problem #1

Page 14: 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.

3/2003 Rev 1 I.2.14 – slide 14 of 28

Solution to Sample Problem #1

(n – 1)s2

Xave 2 =

(19)(169)526.4

2 = = 6.1

Page 15: 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.

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 – 32.85 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

Page 16: 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.

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.

Page 17: 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.

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.

Page 18: 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.

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

Page 19: 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.

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.

Page 20: 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.

3/2003 Rev 1 I.2.14 – slide 20 of 28

Numberof data

LimitingRatio

2 1.15

3 1.38

4 1.54

5 1.68

6 1.73

7 1.79

8 1.86

9 1.92

10 1.96

12 2.03

Numberof data

LimitingRatio

15 2.13

19 2.22

20 2.24

25 2.33

30 2.39

35 2.45

40 2.50

50 2.58

75 2.71

100 2.80

Chauvenet’s Criterion

Page 21: 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.

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.

Page 22: 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.

3/2003 Rev 1 I.2.14 – slide 22 of 28

Observation Gross Counts

1 15

2 24

3 20

4 17

5 26

6 19

7 11

8 13

9 22

10 17

Sample Problem #1

Observation Gross Counts

11 19

12 20

13 29

14 22

15 18

16 28

17 23

18 23

19 32

20 20

Observation Gross Counts

21 18

22 19

23 14

24 30

25 24

Page 23: 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.

3/2003 Rev 1 I.2.14 – slide 23 of 28

Step 1 – Determine the mean, standard deviation, and number of degrees of freedom:

= 21 s2 = 28 N-1 = 24

Solution to Sample Problem #1

_x

Page 24: 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.

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)

Page 25: 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.

3/2003 Rev 1 I.2.14 – slide 25 of 28

Step 3 – Determine the Chauvenet’s Ratio for the data being evaluated:

CR32 = = 2.4

CR11 = = 2.2

Solution to Sample Problem #1

(32 – 21)

21

(21 - 11)

21

Page 26: 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.

3/2003 Rev 1 I.2.14 – slide 26 of 28

For 25 observations, the CR limit is 2.33 (see slide 20)

CR32 = 2.4 > 2.33, so the value of 32 should be rejected

CR11 = 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

Page 27: 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.

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

Page 28: 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.

3/2003 Rev 1 I.2.14 – slide 28 of 28

Cember, H., Johnson, T. E., “Introduction to Health Physics,” 4th Edition, McGraw-Hill, New York (2008).

Martin, A., Harbison, S. A., Beach, K., Cole, P., An Introduction to Radiation Protection, 6th 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).

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