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.
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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 MathematicsUsed in Radiation Protection
Session 14 Statistics - Tests
Session I.2.14
IAEA Post Graduate Educational CourseRadiation Protection and Safety of Radiation Sources
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
3/2003 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 =
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
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
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
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 =
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
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
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 =
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
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
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
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
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
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.
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.
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
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.
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
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.
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
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
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)
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
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
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
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).