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42
Example: What is the probability of getting a count 15 in a measurement when the true
or average count is 10.
Using equation-1 above
Counting Statistics
eP
The probability is 0.0349 or 3.49% that a count 15 will be obtained when the average
count is 10.
In Poisson distribution the variance is equal to the mean. The variance is a number that
is equal to square of standard deviation.
2
Thus the standard deviation
In counting measurements the measured counts in a given time is assumed as mean. For
example if 100 counts are recorded in a given length of time in an experiment then 100 will
be taken as the mean counts. The standard deviation in this case will be 10. The observer
is 68% sure (explained in later part of this chapter) that the counts are within 10 of the
mean. If 1000 counts are acquired instead of 100 counts in the above example, observer will
be 68% sure that the data lie between 968 to 1032 with an error of 3.2%. Similarly a count
of 10,000 will have a standard deviation of 100. At the same confidence level of 68% the
data will lie between 9,900 to 10100 with an accuracy of 1%. It is therefore, important to
acquire sufficient number of counts to reduce error in the measurement.
μ)(
2
0349.010.49.3
)()(
meanVariance
121451510
)10.3.1/()10.999.9()10.54.4(!15/)10()10,15(
Gaussian distribution
When the number of measurement is large, the Poisson distribution is approximated by
normal (Gaussian) distribution. In normal distribution the area under 1 standard deviation
on either side of the mean is about 68.3% of the total area as shown in figure 1. Similarly
the lines drawn at 2 and 3 on either side will cover 95% and 99.7% respectively of the
total area under the normal distribution curve. It may be noted that the total area under the
frequency distribution curve is taken as unity. The equation for the Gaussian probability
(g) is
2
where and m have the same meaning as in Poisson distribution mentioned before.
Considering example given above, the probability of getting a count 15 (mean count = 10)
mmmg 2/)(exp)2(/1
(2)
Counting Statistics
r
r
43
can be estimated to be 0.036 or 3.6% with the help of equation-2 assuming that radioactive
disintegration follows Gaussian distribution rather than Poisson distribution.
Figure 1: Gaussian (normal) distribution
Binomial distribution
It refers to two alternatives (yes or no, head or tail etc) and can be used to determine
whether the results observed in a situation could have occurred randomly. If the number of
successes is r in n number of trials then the failure will be n-r. The bionomial probability of
r successes in n trials is given as:
rnr
ppC
(3)
The combination
n
n
is called binomial coefficient. Mathematically it can be calculated
C
)1.(.
as:
!
n
C
r
n
)!(!
rnr
As the value of n increases, so will be the range of r-values, the distribution becomes
bell shaped and can be approximated to a normal probability distribution.
Measures of central tendency
It is customary and convenient to summarize the results of a set of measurement with a
single value. The simplest value that represents the measurement is ‘mean’ or average value.
Simple mean is non-specific therefore three terms mean, mode and median are defined.
44
Counting Statistics
Mean
The mean of individual data points is simple to calculate. In case of grouped data the central
value of a group represents the class and the sum of its product with the frequency when
divided by the sum of frequency results in the mean of the data.
Mathematically,
NXMean
where Xi is the ith data and N is the total number of data points.
/
i
Mode
Other kind of central value of a data is the mode, which represents the most frequent value
in the data distribution.
Median
Median divides the data into two equal parts. Each part can further be divided into two more
parts. Thus there are three data points which divide the distribution into four parts. The first
data point is called first quartile or 25th(P25) percentile, second is median or 50th percentile
and the third is 3rd quartile or 75th(P75) percentile. Twenty five percent values lie below P25
(first quartile).
Skewness
When distribution curve is exactly symmetrical then mean, mode and median are equal.
The same is not true, whenever there is asymmetry in the distribution curve (skewness) as
shown in figure 2. For positive skewness in the distribution curve the mode is greater than
median and median is greater than mean. In case of negative skewness the mean is greater
than median and median is greater than mode. The skewness is defined in terms of its first
and second coefficient.
Counting Statistics
45
Figure 2: Symmetrical distribution (top), Asymmetrical distribution with positive skewness
(bottom left) and negative skewness (bottom right)
The first coefficient of skewness = (mean-mode)/standard deviation
The second coefficient = 3(mean-median)/standard deviation
Standard deviation
One of the most important statistical parameters which describes the variability in a set of
data is the standard deviation. The standard deviation for a series of measurements determines
the precision or reproducibility of the measured data. When data follows Poisson’s probability
distribution, the standard deviation equals the square root of the mean. In some situations,
where a single measurement is acquired as in digital imaging, the pixel counts is taken as
mean for all practical purposes (such as in estimating the noise). Estimation of mean with
single measurement is the only choice in such cases. For sample data the standard deviation
can be given by following equation
2
i
where xi is the ith data, x is the mean and N is the total number of data points in the
distribution.
1/)(..
Nxxds
(4)
46
Counting Statistics
Example: Calculate the mean, standard deviation, variance and coefficient of variation
in the following set of measurement.
S.No. Counts observed (xi) Deviation (xi – x–) (xi – x–)
1. 100 -5 25
2. 110 5 25
3. 97 -8 64
4. 98 -7 49
5. 115 10 100
6. 106 1 1
7. 95 -10 100
8. 120 15 225
9. 100 -5 25
10. 109 4 16
11. Sum = 1050 630
2
i
2
70.).(
dsVariance
10510/105010/1050/)( nxxMean
2
nxxids
36.8)9/630()1/()(.).(deviationStandard
Coefficient of variation (%) = (s.d./mean) 100 = 7.96%
In presence of outliers the measurement should be repeated. If it still persists then a
statistical test is available to determine whether to discard or include it in the distribution.
Graphical display
The graphical representation shows the correct picture of the data distribution. It is not
necessary that the two sets of data with the same mean and standard deviation will have the
same graphical display. There are several methods of graphical representation of the data
such as bar diagram, histogram, pie charts etc. Statistical parameters with same mean and
standard deviations can have different distribution
Standard error
The spread of observations in a given measurement is estimated by the standard deviation. If
such measurements are repeated several times, a mean of all means can be estimated. The
spread of means in such situations is measured in terms of standard error. Distribution of
such means is usually normal. The standard error is calculated by the formula:
Counting Statistics
Standard error = standard deviation/n
Where n is the number of individual measurements.
Example: The mean uptake value of 10 patients for 10 consecutive weeks is listed
below. Calculate the standard error in this set of data.
Week Mean percentage uptake
1. 30
2. 25
3. 35
4. 30
5. 30
6. 25
7. 35
8. 25
9. 30
10. 30
47
Mean = 29.5
Mean = 29.5
Standard deviation for this data = 3.68
Standard error = standard deviation / n
= standard deviation / 10
= 3.68 / 3.16
= 1.16
Mean thyroid uptake value = (29.5 1.16)%
Confidence limits are related to standard errors. The spread of mean values follows
normal distribution. Thus at 95% confidence limits the uptake values will range from 29.5(1.96×1.16) to 29.5+(1.96×1.16) i.e. from 27.23 to 31.77. Thus there is only 5% probability
that the uptake range from 27.23% to 31,77% excludes the population mean.
Propagation of error in counting measurements
Any parameter that is estimated from counting measurement is associated with some
degree of uncertainty. If the final result involves arithmetic operation of such parameters, as
48
1 2
( / / )
N N
Counting Statistics
is usually the case with nuclear medicine data, then the error gets propagated. The error
introduced in the result depends on the type of arithmetic operation performed and on the
value of uncertainty in the original data.
For sum and difference of two terms N1 and N2, the error will be propagated as:
)()()(
NNNNNN
212121
(5)
The error in case of multiplication and division is propagated as:
N1 × N2 or N1/ N2 = (N1 × N2) or N1/ N2 +
1 1
(6)
Example: A radioactive sample is measured in a counter. The sample holder gives
20,000 and 500 counts in 10 seconds with and without sample. Find true counts in the
sample.
Background counts =500; s.d = 500
Measured counts in the sample = 20,000; s.d. = 20,000
True sample counts = (20,000 - 500) (20,000 + 500)
= 19,500 143.18
Percentage uncertainty = (143.18/19,500) ×100 = 0.73%
If instead of 20,000 and 500 counts from sample and background, one observes 20,000
and 19,500 counts respectively for a given length of time (say 10 seconds) then the net
counts will be equal to 500 198.74 and the percentage uncertainty will be 39.74%. Therefore,
the actual counts should be much greater than the background counts to have better precision
in the final result.
Example: In a renogram study, the total counts between two cursors, placed on the
rising portion of parenchymal uptake phase of the curve, at 1 minute and 3 minutes show the
integrated counts for right and left kidney as 1500 and 1400 counts respectively. Calculate
the relative function of the kidneys.
Relative function of right kidney = integrated counts for right kidney (R)/Integrated
counts for right and left kidney for the same time interval (R+L).
= R/(R+L) error = 1500/2900 (1/1500 + 1/2900)
= 1500/2900 0.031 = .517 0.031 = (51.7 3.1)%
Percentage right kidney function = 51.7 3.1
Similarly percentage left kidney function = 48.3 3.1
While using count rate instead of counts one can use the same formula (equation-5 and
Counting Statistics
equation-6) but the uncertainty will have to be estimated for the count rate ‘R’ instead of
counts ‘N’.
Count rate(C) = N/t
Where N is the number of counts in time ‘t’
Uncertainty in count rate will be
= uncertainty in counts (N)/time (t)
= N/t
= 1/t × (N/t)
= (C/t) (7)
Example: In a measurement with 1500 counts in 10 seconds the uncertainty in count
rate (150 counts/sec) = (150/10) = 15 = 3.87
Thus the count rate = 150 3.87 counts/sec
49
Percentage uncertainty
Uncertainty in gross sample count rate (Cg) will be g = (Cg/tg).
Similarly, uncertainty in background count rate
b= (Cb /tb)
Difference in uncertainty between sample and background can be expressed as:
g - b= [(Cg/tg) + (Cb /tb)] (see equation-5)
Percentage uncertainty in their difference
)/()/[(
tCtC
bbgg
100
CC
bg
(8)
Optimum time of measurement for maximum precision
For a given time of measurement (tg+tb), the maximum precision can be obtained with
following condition
)/(/
CCtt (9)
bgbg
Example: In an experiment the total time available for measurement is 5 minutes. Gross
counts from the sample and background for 10 sec are 5000 and 250 respectively. Calculate
the optimum time for the sample and background for maximum precision.
50
Solution:
Gross sample count rate = 5000/10 = 500
Background count rate = 250/10 = 25
tg / tb= (Cg /Cb) (see equation-9)
= (500/25)
= 4.47
Counting Statistics
tg= 4.47 t
b
Given, tg + tb= 5 min
4.47 tb + tb= 5 min
5.47tb= 5 min
tb= 0.91 min
therefore tg= 4.47 × 0.91 = 4.09 min
Example: The total time allotted for measuring a sample and background is 5 min.
Gross counts from the sample and background for 10 sec were 50,000 and 30,000 respectively.
Calculate the optimum time for the sample and background for maximum precision?
Solution:
Gross Count rate = 50,000/10 = 5000
Background count rate = 30,000/10 = 3000
tg / tb= (Cg /Cb)
= (5000/3000)
= 1.29
tg= 1.29 t
b
Given, tg + tb= 5
1.29 tb + tb= 5
2.29tb= 5
tb= 2.18 min
therefore tg= 2.82 min
Statistical Significance and Null Hypothesis
Null hypothesis starts with the assumption that the difference between the results obtained
Counting Statistics
51
for a parameter from two experiments is only by chance. The p value obtained from any
statistical test indicates the probability of getting a result by chance. If p value is small it
provides less evidence to support Null hypothesis. Decisions are taken to accept or reject
Null hypothesis at some significance level (risk). Generally 5% significance level is chosen
for accepting/rejecting null hypothesis.
The p value smaller than the significance level chosen provides less evidence to support
null hypothesis and greater evidence to support the Alternative. As p becomes smaller and
smaller and creeps beyond the significance level, the difference becomes more and more
apparent (statistically significant).
For 0.05 > p > 0.01 at 5% significance level the null hypothesis is rejected.
Student t-test
W.S. Gosset did the theoretical work on t-distribution in the early 1900. He was an employee
of the Guinness Brewery in Dublin, Ireland, which did not permit employees to publish
research findings under their own names. So Gosset adopted the pen name “student” and
published his findings under this name. Thereafter, the t-distribution is commonly known as
Student’s t-distribution or simply Student’s distribution.
The t-distribution is used when sample size is 30 or less and the population standard
deviation is unknown. It has been derived mathematically under the assumption of a normally
distributed population with the following form:
t
meanindifference
meansindifferencetheoferrorStandard
Applications of t-distribution
The following are some of the examples to illustrate the way in which the Student distribution
is generally used to test the significance of the various results obtained from small samples.
1. To test the significance of mean of a random sample
2. To test difference between means of two samples (independent samples)
3. To test difference between means of two samples (dependent samples)
To test the significance of mean of a random sample
Example: A student estimated the normal range of
70.The review of 20 cases by his instructor gives the mean uptake value as 60, with a
standard deviation of 11. The table value of t for degree of freedom = 19 at 1% level is
1.729. The instructor wants to test the hypothesis at 0.01 level of significance. In this case
131
I thyroid uptake value to be
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