Statistical considerations
Alfredo García – Arieta, PhD
Training workshop: Training of BE assessors, Kiev, October 2009
Outline
 Basic statistical concepts on equivalence
 How to perform the statistical analysis of a
2x2 cross-over bioequivalence study
 How to calculate the sample size of a 2x2
cross-over bioequivalence study
 How to calculate the CV based on the 90% CI
of a BE study
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Training workshop: Training of BE assessors, Kiev, October 2009
Basic statistical concepts
3|
Training workshop: Training of BE assessors, Kiev, October 2009
Type of studies
 Superiority studies
– A is better than B (A = active and B = placebo or gold-standard)
– Conventional one-sided hypothesis test
 Equivalence studies
– A is more or less like B (A = active and B = standard)
– Two-sided interval hypothesis
 Non-inferiority studies
– A is not worse than B (A = active and B = standard with adverse
effects)
– One-sided interval hypothesis
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Training workshop: Training of BE assessors, Kiev, October 2009
Hypothesis test
 Conventional hypothesis test
 H0:  = 1
H1:   1
(in this case it is two-sided)
 If P<0,05 we can conclude that statistical significant difference exists
 If P≥0,05 we cannot conclude
– With the available potency we cannot detect a difference
– But it does not mean that the difference does not exist
– And it does not mean that they are equivalent or equal
 We only have certainty when we reject the null hypothesis
– In superiority trials: H1 is for existence of differences
 This conventional test is inadequate to conclude about “equalities”
– In fact, it is impossible to conclude “equality”
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Training workshop: Training of BE assessors, Kiev, October 2009
Null vs. Alternative hypothesis
 Fisher, R.A. The Design of Experiments, Oliver and Boyd,
London, 1935
 “The null hypothesis is never proved or established, but is
possibly disproved in the course of experimentation.
Every experiment may be said to exist only in order to
give the facts a chance of disproving the null hypothesis”
 Frequent mistake: The absence of statistical significance
has been interpreted incorrectly as absence of clinically
relevant differences.
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Training workshop: Training of BE assessors, Kiev, October 2009
Equivalence
 We are interested in verifying (instead of rejecting) the null
hypothesis of a conventional hypothesis test
 We have to redefine the alternative hypothesis as a range of values
with an equivalent effect
 The differences within this range are considered clinically irrelevant
 Problem: it is very difficult to define the maximum difference without
clinical relevance for the Cmax and AUC of each drug
 Solution: 20% based on a survey among physicians
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Training workshop: Training of BE assessors, Kiev, October 2009
Interval hypothesis or two one-sided tests
 Redefine the null hypothesis: How?
 Solution: It is like changing the null to the alternative hypothesis
and vice versa.
 Alternative hypothesis test: Schuirmann, 1981
– H01:  1
– H02:  2
Ha1: 1<
Ha2: < 2.
 This is equivalent to:
– H0:  1 or  2
Ha: 1<<2
 It is called as an interval hypothesis because the equivalence
hypothesis is in the alternative hypothesis and it is expressed as an
interval
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Training workshop: Training of BE assessors, Kiev, October 2009
Interval hypothesis or two one-sided tests
 The new alternative hypothesis is decided with a statistic
that follows a distribution that can be approximated to a tdistribution
 To conclude bioequivalence a P value <0.05 has to be
obtained in both one-sided tests
 The hypothesis tests do not give an idea of magnitude of
equivalence (P<0001 vs. 90% CI: 0.95 – 1.05).
 That is why confidence intervals are preferred
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Training workshop: Training of BE assessors, Kiev, October 2009
Point estimate of the difference
If T=R, d=T-R=0
If T>R, d=T-R>0
If T<R, d=T-R<0
d<0
Negative effect
10 |
d=0
No difference
Training workshop: Training of BE assessors, Kiev, October 2009
d>0
Positive effect
Estimation with confidence intervals
in a superiority trial
It is not statistically significant!
Because the CI includes the d=0 value
Confidence interval 90% - 95%
d<0
Negative effect
11 |
d=0
No difference
Training workshop: Training of BE assessors, Kiev, October 2009
d>0
Positive effect
Estimation with confidence intervals
in a superiority trial
It is statistically significant!
Because the CI does not includes the d=0 value
Confidence interval 90% - 95%
d<0
Negative effect
12 |
d=0
No difference
Training workshop: Training of BE assessors, Kiev, October 2009
d>0
Positive effect
Estimation with confidence intervals
in a superiority trial
It is statistically significant with P=0.05
Because the boundary of the CI touches the d=0 value
Confidence interval 90% - 95%
d<0
Negative effect
13 |
d=0
No difference
Training workshop: Training of BE assessors, Kiev, October 2009
d>0
Positive effect
Equivalence study
Region of
clinical
equivalence
-d
d<0
Negative effect
14 |
+d
d=0
No difference
Training workshop: Training of BE assessors, Kiev, October 2009
d>0
Positive effect
Equivalence vs. difference
Region of clinical equivalence
Equivalent?
Different?
?
Yes
Yes
Yes
?
Yes
Yes
Yes
?
No
?
Yes
Yes
Yes
No
?
-d
d<0
Negative effect
15 |
+d
d=0
No difference
Training workshop: Training of BE assessors, Kiev, October 2009
d>0
Positive effect
Non-inferiority study
Inferiority limit
Inferior?
?
Yes
?
No
No
No
No
No
-d
d<0
Negative effect
16 |
d=0
No difference
Training workshop: Training of BE assessors, Kiev, October 2009
d>0
Positive effect
Superiority study (?)
Superiority limit
?
Superior?
No
No
No
d<0
Negative effect
17 |
No, not clinically and ? statistically
No, not clinically, but yes statistically
?, but yes statistically
Yes, statistical & clinically
Yes, but only the
+d
point estimate
d=0
d>0
No difference
Positive effect
Training workshop: Training of BE assessors, Kiev, October 2009
How to perform the statistical analysis of a
2x2 cross-over bioequivalence study
18 |
Training workshop: Training of BE assessors, Kiev, October 2009
Statistical Analysis of BE studies
 Sponsors have to use validated software
– E.g. SAS, SPSS, Winnonlin, etc.
 In the past, it was possible to find statistical
analyses performed with incorrect software.
– Calculations based on arithmetic means, instead of
Least Square Means, give biased results in
unbalanced studies
• Unbalance: different number of subjects in each sequence
– Calculations for replicate designs are more
complex and prone to mistakes
19 |
Training workshop: Training of BE assessors, Kiev, October 2009
The statistical analysis is not so complex
2x2 BE trial
Period 1
Period 2
Y11
Y12
Y21
Y22
N=12
Sequence 1 (BA)
BA is RT
Sequence 2 (AB)
AB is TR
20 |
Training workshop: Training of BE assessors, Kiev, October 2009
We don’t need to calculate an ANOVA table
Sources of variation
Inter-subject
Carry-over
Residual / subjects
d. f.
SS
MS
F
P
23
1
22
16487,49
276,00
16211,49
716,85
276,00
736,89
4,286
0,375
4,406
0,5468
0,0005
Intra-subject
Formulation
Period
Residual
1
1
22
3778,19
62,79
35,97
3679,43
62,79
35,97
167,25
0,375
0,215
0,5463
0,6474
Total
47
20265,68
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Training workshop: Training of BE assessors, Kiev, October 2009
With complex formulae
2
2
nk


2
SSTotal   Yijk  Y···
k 1 j 1 i 1
2
2
nk


2
SSW ithin   Yijk  Yi ·k
k 1 j 1 i 1
2
nk


SS Between  2 Yi ·k  Y···
k 1 i 1
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Training workshop: Training of BE assessors, Kiev, October 2009
2
More complex formulae
SS Between  SSCarry  SSint er
SSCarry
2n1n2
2
Y·12  Y·22   Y·11  Y·21 

n1  n2
2
SS Inter
23 |
nk
2
i ·k
2
2
··k
Y
Y
 

k 1 i 1 2
k 1 2nk
Training workshop: Training of BE assessors, Kiev, October 2009
And really complex formulae
SSW ithin  SS Drug  SS Period  SS Intra
SS Drug
2n1n2

n1  n2
SS Period
1

 Y·21  Y·11   Y·22  Y·12 
2

2n1n2

n1  n2
nk
2
1

 Y·21  Y·11   Y·12  Y·22 
2

nk
Y· 2jk
2
2
Y
Y
SS Intra   Yijk2  
 
  ··k
k 1 j 1 i 1
k 1 i 1 2
k 1 j 1 nk
k 1 2nk
2
24 |
2
2
Training workshop: Training of BE assessors, Kiev, October 2009
2
i ·k
2
2
2
Given the following data, it is simple
2x2 BE trial
Period 1
Period 2
Y11
Y12
N=12
Sequence 1 (BA)
75, 95, 90, 80, 70, 85 70, 90, 95, 70, 60, 70
Sequence 2 (AB)
Y21
Y22
75, 85, 80, 90, 50, 65 40, 50, 70, 80, 70, 95
25 |
Training workshop: Training of BE assessors, Kiev, October 2009
First, log-transform the data
2x2 BE trial
Period 1
Period 2
Sequence 1 (BA)
Y11
4.3175, 4.5539,
4.4998, 4.3820,
4.2485, 4.4427
Y12
4.2485, 4.4998,
4.5539, 4.2485,
4.0943, 4.2485
Sequence 2 (AB)
Y21
4.3175, 4.4427,
4.3820, 4,4998,
3,9120, 4.1744
Y22
3.6889, 3,9120,
4,2485, 4.3820,
4.2485, 4.5539
N=12
26 |
Training workshop: Training of BE assessors, Kiev, October 2009
Second, calculate the arithmetic mean of
each period and sequence
2x2 BE trial
Period 1
Period 2
Sequence 1 (BA)
Y11 = 4.407
Y12 = 4.316
Sequence 2 (AB)
Y21 = 4.288
Y22 = 4,172
N=12
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Training workshop: Training of BE assessors, Kiev, October 2009
Note the difference between Arithmetic
Mean and Least Square Mean
 The arithmetic mean (AM) of T (or R) is the mean of all
observations with T (or R) irrespective of its group or
sequence
– All observations have the same weight
 The LSM of T (or R) is the mean of the two sequence by
period means
– In case of balanced studies AM = LSM
– In case of unbalanced studies observations in sequences with
less subjects have more weight
– In case of a large unbalance between sequences due to dropouts or withdrawals the bias of the AM is notable
28 |
Training workshop: Training of BE assessors, Kiev, October 2009
Third, calculate the LSM of T and R
2x2 BE trial
Period 1
Period 2
Y11 = 4.407
Y12 = 4.316
N=12
Sequence 1 (BA)
B = 4.2898
Sequence 2 (AB)
29 |
Y21 = 4.288
Training workshop: Training of BE assessors, Kiev, October 2009
A = 4.3018
Y22 = 4,172
Fourth, calculate the point estimate
 F = LSM Test (A) – LSM Reference (B)
 F = 4.30183 – 4.28985 = 0.01198
 Fifth step! Back-transform to the original scale
 Point estimate = eF = e0.01198 = 1.01205
 Five very simple steps to calculate the point estimate!!!
30 |
Training workshop: Training of BE assessors, Kiev, October 2009
Now we need to calculate the variability!
 Step 1: Calculate the difference between periods for each subject
and divide it by 2: (P2-P1)/2
 Step 2: Calculate the mean of these differences within each
sequence to obtain 2 means: d1 and d2
 Step 3:Calculate the difference between “the difference in each
subject” and “its corresponding sequence mean”. And square it.
 Step 4: Sum these squared differences
 Step 5: Divide it by (n1+n2-2), where n1 and n2 is the number of
subjects in each sequence. In this example 6+6-2 = 10
– This value multiplied by 2 is the MSE
– CV (%) = 100 x √eMSE-1
31 |
Training workshop: Training of BE assessors, Kiev, October 2009
This can be done easily in a spreadsheet!
PERIOD
I
R
4,31748811
4,55387689
4,49980967
4,38202663
4,24849524
4,44265126
II
T
4,24849524
4,49980967
4,55387689
4,24849524
4,09434456
4,24849524
Step 1
P2-P1
-0,06899287
-0,05406722
0,05406722
-0,13353139
-0,15415068
-0,19415601
Step 2
Mean d1 =
n1 =
-0,09180516 -0,04590258
6
T
4,3175
4,4427
4,3820
4,4998
3,9120
4,1744
R
3,6889
3,9120
4,2485
4,3820
4,2485
4,5539
Step 2
Mean d2 =
n2 =
32 |
-0,62860866
-0,53062825
-0,13353139
-0,11778304
0,33647224
0,37948962
Step 1
(P2-P1)/2
-0,03449644
-0,02703361
0,02703361
-0,0667657
-0,07707534
-0,09707801
-0,31430433
-0,26531413
-0,0667657
-0,05889152
0,16823612
0,18974481
Step 3
d - mean d
0,01140614
0,01886897
0,07293619
-0,02086312
-0,03117276
-0,05117543
Step 3
squared
0,0001301
0,00035604
0,00531969
0,00043527
0,00097174
0,00261892
-0,25642187
-0,20743167
-0,00888324
-0,00100906
0,22611858
0,24762727
0,06575218
0,0430279
7,8912E-05
1,0182E-06
0,05112961
0,06131926
-0,11576491 -0,05788246
6
Training workshop: Training of BE assessors, Kiev, October 2009
Sum =
Step 4
0,23114064
Step 5
Sigma2(d) = 0,02311406
MSE=
0,04622813
CV =
21,7516218
Step 1: Calculate the difference between periods
for each subject and divide it by 2: (P2-P1)/2
PERIOD
33 |
I
R
4,31748811
4,55387689
4,49980967
4,38202663
4,24849524
4,44265126
II
T
4,24849524
4,49980967
4,55387689
4,24849524
4,09434456
4,24849524
Step 1
P2-P1
-0,06899287
-0,05406722
0,05406722
-0,13353139
-0,15415068
-0,19415601
Step 2
Mean d1 =
n1 =
-0,09180516 -0,04590258
6
T
4,3175
4,4427
4,3820
4,4998
3,9120
4,1744
R
3,6889
3,9120
4,2485
4,3820
4,2485
4,5539
Step 2
Mean d2 =
n2 =
-0,62860866
-0,53062825
-0,13353139
-0,11778304
0,33647224
0,37948962
Step 1
(P2-P1)/2
-0,03449644
-0,02703361
0,02703361
-0,0667657
-0,07707534
-0,09707801
-0,31430433
-0,26531413
-0,0667657
-0,05889152
0,16823612
0,18974481
-0,11576491 -0,05788246
6
Training workshop: Training of BE assessors, Kiev, October 2009
Step 2: Calculate the mean of these differences
within each sequence to obtain 2 means: d1 & d2
PERIOD
34 |
I
R
4,31748811
4,55387689
4,49980967
4,38202663
4,24849524
4,44265126
II
T
4,24849524
4,49980967
4,55387689
4,24849524
4,09434456
4,24849524
Step 1
P2-P1
-0,06899287
-0,05406722
0,05406722
-0,13353139
-0,15415068
-0,19415601
Step 2
Mean d1 =
n1 =
-0,09180516 -0,04590258
6
T
4,3175
4,4427
4,3820
4,4998
3,9120
4,1744
R
3,6889
3,9120
4,2485
4,3820
4,2485
4,5539
Step 2
Mean d2 =
n2 =
-0,62860866
-0,53062825
-0,13353139
-0,11778304
0,33647224
0,37948962
Step 1
(P2-P1)/2
-0,03449644
-0,02703361
0,02703361
-0,0667657
-0,07707534
-0,09707801
-0,31430433
-0,26531413
-0,0667657
-0,05889152
0,16823612
0,18974481
-0,11576491 -0,05788246
6
Training workshop: Training of BE assessors, Kiev, October 2009
Step 3: Squared differences
PERIOD
35 |
I
R
4,31748811
4,55387689
4,49980967
4,38202663
4,24849524
4,44265126
II
T
4,24849524
4,49980967
4,55387689
4,24849524
4,09434456
4,24849524
Step 1
P2-P1
-0,06899287
-0,05406722
0,05406722
-0,13353139
-0,15415068
-0,19415601
Step 2
Mean d1 =
n1 =
-0,09180516 -0,04590258
6
T
4,3175
4,4427
4,3820
4,4998
3,9120
4,1744
R
3,6889
3,9120
4,2485
4,3820
4,2485
4,5539
Step 2
Mean d2 =
n2 =
-0,62860866
-0,53062825
-0,13353139
-0,11778304
0,33647224
0,37948962
Step 1
(P2-P1)/2
-0,03449644
-0,02703361
0,02703361
-0,0667657
-0,07707534
-0,09707801
-0,31430433
-0,26531413
-0,0667657
-0,05889152
0,16823612
0,18974481
-0,11576491 -0,05788246
6
Training workshop: Training of BE assessors, Kiev, October 2009
Step 3
d - mean d
0,01140614
0,01886897
0,07293619
-0,02086312
-0,03117276
-0,05117543
Step 3
squared
0,0001301
0,00035604
0,00531969
0,00043527
0,00097174
0,00261892
-0,25642187
-0,20743167
-0,00888324
-0,00100906
0,22611858
0,24762727
0,06575218
0,0430279
7,8912E-05
1,0182E-06
0,05112961
0,06131926
Step 4: Sum these squared differences
Step 3
squared
0,0001301
0,00035604
0,00531969
0,00043527
0,00097174
0,00261892
Sum =
Step 5
Sigma2(d) = 0,02311406
MSE=
0,04622813
CV =
21,7516218
0,06575218
0,0430279
7,8912E-05
1,0182E-06
0,05112961
0,06131926
36 |
Step 4
0,23114064
Training workshop: Training of BE assessors, Kiev, October 2009
Step 5: Divide the sum by n1+n2-2
Step 3
squared
0,0001301
0,00035604
0,00531969
0,00043527
0,00097174
0,00261892
Sum =
Step 5
Sigma2(d) = 0,02311406
MSE=
0,04622813
CV =
21,7516218
0,06575218
0,0430279
7,8912E-05
1,0182E-06
0,05112961
0,06131926
37 |
Step 4
0,23114064
Training workshop: Training of BE assessors, Kiev, October 2009
Calculate the confidence interval with
point estimate and variability
 Step 11: In log-scale
 90% CI: F ± t(0.1, n1+n2-2)-√((Sigma2(d) x (1/n1+1/n2))
 F has been calculated before
 The t value is obtained in t-Studient tables with 0,1 alpha
and n1+n2-2 degrees of freedom
– Or in MS Excel with the formula =DISTR.T.INV(0.1; n1+n2-2)
 Sigma2(d) has been calculated before.
38 |
Training workshop: Training of BE assessors, Kiev, October 2009
Final calculation: the 90% CI
 Log-scale 90% CI: F±t(0.1, n1+n2-2)-√((Sigma2(d)·(1/n1+1/n2))
 F = 0.01198
 t(0.1, n1+n2-2) = 1.8124611
 Sigma2(d) = 0.02311406
 90% CI: LL = -0.14711 to UL= 0,17107
 Step 12: Back transform the limits with eLL and eUL
 eLL = e-0.14711 = 0.8632 and eUL = e0.17107 = 1.1866
39 |
Training workshop: Training of BE assessors, Kiev, October 2009
How to calculate the sample size of a 2x2
cross-over bioequivalence study
40 |
Training workshop: Training of BE assessors, Kiev, October 2009
Reasons for a correct calculation of the
sample size
 Too many subjects
– It is unethical to disturb more subjects than necessary
– Some subjects at risk and they are not necessary
– It is an unnecessary waste of some resources ($)
 Too few subjects
– A study unable to reach its objective is unethical
– All subjects at risk for nothing
– All resources ($) is wasted when the study is inconclusive
41 |
Training workshop: Training of BE assessors, Kiev, October 2009
Frequent mistakes
 To calculate the sample size required to detect a 20%
difference assuming that treatments are e.g. equal
– Pocock, Clinical Trials, 1983
 To use calculation based on data without logtransformation
– Design and Analysis of Bioavailability and Bioequivalence
Studies, Chow & Liu, 1992 (1st edition) and 2000 (2nd edition)
 Too many extra subjects. Usually no need of more than
10%. Depends on tolerability
– 10% proposed by Patterson et al, Eur J Clin Pharmacol 57:
663-670 (2001)
42 |
Training workshop: Training of BE assessors, Kiev, October 2009
Methods to calculate the sample size
 Exact value has to be obtained with power curves
 Approximate values are obtained based on formulae
– Best approximation: iterative process (t-test)
– Acceptable approximation: based on Normal distribution
 Calculations are different when we assume products are
really equal and when we assume products are slightly
different
 Any minor deviation is masked by extra subjects to be
included to compensate drop-outs and withdrawals (10%)
43 |
Training workshop: Training of BE assessors, Kiev, October 2009
Calculation assuming that
treatments are equal
N
2  s  Z1b
2
w
2
 Z1a 
2
Ln1.25
2

s  Ln 1  CV
2
w
2

CV expressed as 0.3 for 30%
 Z(1-(b/2)) = DISTR.NORM.ESTAND.INV(0.05) for 90% 1-b
 Z(1-(b/2)) = DISTR.NORM.ESTAND.INV(0.1) for 80% 1-b
 Z(1-a) = DISTR.NORM.ESTAND.INV(0.05) for 5% a
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Training workshop: Training of BE assessors, Kiev, October 2009
Example of calculation assuming that
treatments are equal
 If we desire a 80% power, Z(1-(b/2)) = -1.281551566
 Consumer risk always 5%, Z(1-a) = -1.644853627
 The equation becomes: N = 343.977655 x S2
 Given a CV of 30%, S2 = 0,086177696
 Then N = 29,64
 We have to round up to the next pair number: 30
 Plus e.g. 4 extra subject in case of drop-outs
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Training workshop: Training of BE assessors, Kiev, October 2009
Example of calculation assuming that
treatments are equal
 If we desire a 90% power, Z(1-(b/2)) = -1.644853627
 Consumer risk always 5%, Z(1-a) = -1.644853627
 The equation becomes: N = 434.686167 x S2
 Given a CV of 25%, S2 = 0,06062462
 Then N = 26,35
 We have to round up to the next pair number: 28
 Plus e.g. 4 extra subject in case of drop-outs
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Training workshop: Training of BE assessors, Kiev, October 2009
Calculation assuming that
treatments are not equal
N
2  s  Z1 b  Z1a 
2
w
LnT
2
 R   Ln1.25
2
T R  1
 Z(1-b) = DISTR.NORM.ESTAND.INV(0.1) for 90% 1-b
 Z(1-b) = DISTR.NORM.ESTAND.INV(0.2) for 80% 1-b
 Z(1-a) = DISTR.NORM.ESTAND.INV(0.05) for 5% a
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Training workshop: Training of BE assessors, Kiev, October 2009
Example of calculation assuming that
treatments are 5% different
 If we desire a 90% power, Z(1-b) = -1.28155157
 Consumer risk always 5%, Z(1-a) = -1.644853627
 If we assume that T/R=1.05
 The equation becomes: N = 563.427623 x S2
 Given a CV of 40 %, S2 = 0,14842001
 Then N = 83.62
 We have to round up to the next pair number: 84
 Plus e.g. 8 extra subject in case of drop-outs
48 |
Training workshop: Training of BE assessors, Kiev, October 2009
Example of calculation assuming that
treatments are 5% different
 If we desire a 80% power, Z(1-b) = -0.84162123
 Consumer risk always 5%, Z(1-a) = -1.644853627
 If we assume that T/R=1.05
 The equation becomes: N = 406.75918 x S2
 Given a CV of 20 %, S2 = 0,03922071
 Then N = 15.95
 We have to round up to the next pair number: 16
 Plus e.g. 2 extra subject in case of drop-outs
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Training workshop: Training of BE assessors, Kiev, October 2009
Example of calculation assuming that
treatments are 10% different
 If we desire a 80% power, Z(1-b) = -0.84162123
 Consumer risk always 5%, Z(1-a) = -1.644853627
 If we assume that T/R=1.11
 The equation becomes: N = 876.366247 x S2
 Given a CV of 20 %, S2 = 0,03922071
 Then N = 34.37
 We have to round up to the next pair number: 36
 Plus e.g. 4 extra subject in case of drop-outs
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Training workshop: Training of BE assessors, Kiev, October 2009
How to calculate the CV
based on the 90% CI of a BE study
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Training workshop: Training of BE assessors, Kiev, October 2009
Example of calculation of
the CV based on the 90% CI
 Given a 90% CI: 82.46 to 111.99 in BE study with N=24
 Log-transform the 90% CI: 4.4123 to 4.7184
 The mean of these extremes is the point estimate: 4.5654
 Back-transform to the original scale e4.5654 = 96.08
 The width in log-scale is 4.7184 – 4.5654 = 0,1530
 With the sample size calculate the t-value. How?
– Based on the Student-t test tables or a computer (MS Excel)
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Training workshop: Training of BE assessors, Kiev, October 2009
Example of calculation of
the CV based on the 90% CI
 Given a N = 24, the degrees of freedom are 22
 t = DISTR.T.INV(0.1;n-2) = 1.7171
 Standard error of the difference (SE(d)) = Width / t-value =
0.1530 / 1.7171 = 0,0891
 Square it: 0.08912 = 0,0079 and divide it by 2 = 0,0040
 Multiply it by the sample size: 0.0040x24 = 0,0953 = MSE
 CV (%) = 100 x √(eMSE-1) = 100 x √(e0.0953-1) = 31,63 %
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Training workshop: Training of BE assessors, Kiev, October 2009