UIUVERSITY OF NORTH CAROLINA
Depart:.ient of Statistics
C~lal?el
Hill
N. C.
STATISTICS OF BIOASSAY
Notes on lectures ileld during t!le
spring semester (1964) at tIle Statis··
tics Departnent, University of
Ncrt~1 Car:>lina, C~lapel :Hill, N. C.
by
Artdur Linder
Visiting ProfeSsor froil tde University of Geneva (S"litzerland)
October 1964
T~lis researCd was supported by ti1e Mathematics Division of t!le
Air Force Office of Scientific ResearC-'l Grant No. AF-AFOSR-84-63.
to
Institute of Statistics
Mime:) Series No. 404
TABLE OF CONTENTS
PAGE
1.
INTRODUCTION
............................................
1
2.
DIRECT ASSAYS •••••••••••••••• ,..........................
5
3.
INDIRECT ASSAYS •••••••••.••••••.••••••••.••••••••••••.••
9
3.0 Preliminary remarl;:s
9
3.1
3.2
.e
Parallel line assays
13
3.1.1 General case of parallel line assay
13
3.1.2 Symmetrical four-point assay
32
3.1.3 Symmetrical six-point assay
38
3.1.4 Assays in randomized blocks or in latin squares
41
3.1.5 Sensitivity of parallel line assays
47
Slope Ratio Assays
50
3.2.1. General case of slope ratio assays
51
3.2.2
Symmetrical five-point assay
68
3.2.3
Sensitivity of slope ratio assays
80
3.3 Multiple Assays
81
3.3.1. Multiple parallel line assays
82
3.3.2
Multiple slope ratio assays
84
3.3.3
Number of replications in the standarcl
preparation
91
..................................................
92
REFERENCES
•
11
.e
1•
1. Introduction
According to Finney,,"Bio1ogica1 assays are methods for the estimation f'f the
nature" constitution, or potency of a material (or oS a process) by means of the
reaction that follows its application to living matter".
This definition includes qualitative and quantitative a.ssays.
In a qualitative
assay we may wish just to know if a certain substance is present in a preparation"
and a characteristic effect in a liVing organism is used to detect the substance.
In a quantitative assay we want to estimate the degree of some property of a
material through its effect on an an1ma1" a plant" a micro-organism or a part of
living indiVidual, for instance a piece of animal tissue.
We will call the property a "factor" and the reaction in a liVing organism.
which served as an indicator of the amount or concentration of the factor the
·e
"response".
The factor is sometimes also called the" stimulus".
of a biological assay is to compare
",wo
The main purpose
substances with respect to the amount of
the factor they contain" and to deter,aune the relative potency.
The response may be the body wight of an animal" or the weight of a specific
organ (e. g. the. heart" or the kidney)" or the blood-sugar content bone ash percentage, the growth of a bacterial colony, the death of an insect.
The first attempts in using biological assays were not very successful.
The
responses were measured in animal unit .,the unit being the amount or dose of the
factor which produced a specified effect.
A cat unit of digitalis was determined
as the amount of digitalis leading to the stopping of heartbeat of a cat.
This
was very unsatisfactory because this dc.se varies from one cat to another.
Only when it was realized" that it was necessary to adopt standards in this
type of measurement as in the measurement of length" weights and other physical
properties" was progress made.
As the meter is, or rather was until recently" the
distance between two lines on the meter bar preserved in ~ores near Paris (France),
2•
.4It
so also the potency of a specific, carefully prepared and preserved quantity of
digita1is was declared to be the unit of digital1s.
To find out the unlmown potency of a test preparation, the effect of this
preparation is compared to the effect of the standard, by means of a biological
assay.
The Permanent Commission on Biological Standarization, established by the
Health Organization of the League of Nations has set up the first international
biological standard in 1922 for diphtheria antitoxin.
Many such international
standards have since then be established for a great variety of therapeutic substances.
The different assay procedures can be classified in various ways.
The most
important classification is to distinguish
,
-e
(a) direct assays, and
(b) indirect assays.
In a direct assay :dose
of the standard and the test preparation which pro-
duce a specified response are recorded.
The ratio of these doses gives an estimate
of the potency of the test preparation relative to that of the standard.
This can
only be done if the response is such, that the exact dose producing the specified
response can be determined without ambiguity.
This is possible in the cat assay of
digitalis, where the standard or the test preparation is infused at a fixed rate
into the blood-stream of a cat until the heart stops beating.
infUSion is multipl1ed
by
The length of time of
the rate to give the dose.
In many dose-response relationships this method is not available because in
applying a given dose we are only able to observe whether a specified effect has
been produced or not.
If the effect is present, it might also have been produced by
a smaller dosej if the effect is absent, it might also be absent at a higher dese.
It is not possible, in such instances, to vary the dose continuously as in the cat
assay of digitalis.
In these cases, use is made of the methods of indirect assays.
·e
The main steps in an indirect assay are:
(1) to determine the dose-response relationship for both standard and test
preparation.
Example
1/
Response
(weight of
ovaries)
L.-
.__
Dose (Hormone)
Human menopausal
gonadotrophia
(2) It is sUpPOsed that the active ingredient, the factor, in which we are
interested, acts independantly of the other substances in the preparations.
In that case it can be assumed that the test preparation is simply the
same substance as the standard, but at a different concentration or dilu-
-e
tion.
This means that we have to analyse what are called dilution assalS.
The consequence of having
dilution assays is that the dose-response curves
are parallel for standard and test preparation, if log dose is used.
(3) The problem is then to determine an estimate of the relative potency of the
test preparation compared to the standard assuming that the curves (logdose)-response are parallel.
In order to simplify the analysis linearity
of (log-dose)-response curves is assumed.
If'
this assumption is manifestly
not tenable, a transformation of response is tried in order to linearize
the regression.
(4) The log of relative potency R is obtained as the distance between the two
parallel straight lines, measured parallel to abscissae.
Resp<n se
Test
M =
log dose
log R
4.
·e
(5) It is necessary also to determine a measure of variability of the estimate
R.
This is done by calculating fiducial limts for R.
(6) It will also be indispensable to test the assumptions of linearity,
parallelism and homoscedasticity.
Validity of Nat.
~alysis.
Respons
b
T
---l""'''"---+--+--·--close
The description just given assumes that the regressions are linear for log
dose.
This is not always the case.
response.
Sometimes the regressions are linear for dose-
In this case the ratio of the slopes of the regressions serves as an
estimate of the relative potency R.
We have then what is called a slope-ratio
assay.
-e
In the above description it has been assumed that the response can be measured
on a quantitative scale.
This includes measurements like weight, etc., but may also
include cases in which an artificial scale is adopted to measure qualitative changes,
and scores are attr1buted to the grades in the scale.
In many cases the response is, however, quantal" that is we are only able to
record whether a certain reaction has occurred or DOt.
In the study of insell:ticides"
when giving a dose we are only able to count the number of individuals which surviva after being exposed to the insecticide for a given period of time.
case the dose-response curve is usually a sigmoid-curve response
respo"~"'+
"""7""
_
lOO~
dose
In that
A suitable cboice of transformations of dose and reaponse will give a linear
regression necessary for easy computations.
be discussed later on.
There are different possibilities to
The methods of analysis are generally more complicated for
quantal responses than for quantitative responses.
2.
Direct Usays.
Direct assays seem not to be of great practical importance.
Nevertheless we
will discuss an example in order to introduce a few concepts and the notation we
are using.
Example.
una.bain
•
per
kg
Comparison of two tinctures of stropbanhus and a preparation of
Given are tbe fatal doses or tolerances" expressed in c.c. per
body weight of the cat.
kg
or mg.
It is thus assumed that the tolerance varies in
proportion to body weight.
Tincture of Strophantus
A
B
Unabain
preparation
0.01
0.01 c.c. per
kg.
1.55
1.58
1.71
1.44
1.24
1.8~
2.3
2.42
1.85
2.00
2.27
1.70
1.47
•• •
• ••
•••
•••
5.23
9.91
4.76
6.51
6.68
5.76
4.93
4.58
6.69
Total
11.75
13.91
55.05
Mean
1.68
1.99
6.12
7
7
9
Number of cats
~.20
mg per kg
Let me state again that we assume that the effective substance or principle is
acting 1Ddepetld,ntlYof all other substances" which are supposed to be inert.
thus suppose to have a dilution assay.
We
In this case we have simply to determine the
6.
~e
Ratio of the dilutions of the two substances to be compared.
Comparing two preparations of which we know that they are of the same nature
as far as the active principle is concerned, and that the other substances in the
preparations are inert and Without any effect on the response, then this is called
an analytical dilution assay.
A with B.
In other cases we know that the active prinCiples in the two preparations are
not the same, but that they behave as if they were the same, that is as far as the
response is concerned.
In this case the assay is called an comparative dilution
assay, but the method of analysis will be the same as for the analytical dilution
assay.
A or B with Unabain.
If we assume that the doses giving the same response are in a fixed proportion,
we may write
-e
if we consider B as a standard to which we compare A.
Owing to the inevitable
biological variability, we have to consider the mean of several determinations
instead of individual values.
Thus we find an estimate R of P as
R = 1.99 / 1.68
= 1.18
and we conclude that 1 c.c. of A is eqUivalent to 1.18 c.c. of B.
Comparing Unabain to B at a standard we find that
R = 1.99 / 6.12 • 0.325
which means that 1 mg of Ynabain is equivalent in potency to 0.325 of strophanb48
B.
These two estimates of R should be accompanied by an indication on the
errors of estimation.
this respect.
A simple reasoning will show that there are difficulties in
According to our assumption we have
ZA
and from this it follows that
=
P
z.e
.e
If we use the common tests of significance to test the hypothesis that Jp = 1,
and in the calculation leading to fiducial limits, we suppose that
~
= si '
which obviously is not true, as soon as the two preparations have differing
potencies.
On the other hand, if we take logs, we have
log ZA
log P + log
:I:
~
and moreover the variances for the two groups will be equal, apart from random
fluctuations.
In our example we obtain
Logarithms
Observed values
Z
sz
-x
1.68
1.99
6.12
0.1264
0.1136
2.7142
0.217
0.292
0.774
M~an
.e
Strophan m.e
Strophanhus
Unabain
A
B
Var~ance
Mean
Vari~ce
sx
0.007 79'2
0.005 872
0.010 995
The differences between the variances of Uhabain and the two strophannus tinctures
are markedly reduced through the transformation into logs.
For the potency ratio P we get a new estimate log R = M = ~ - x
A
which
yields
and
R
:I:
M ~ - 0.482 and
R
=0.330
M = . 0.075
1.19
for
A to B
for Unabain to B
The estimates are almost the same as those obtained from the observed values.
12. /2 - In order to evaluate fiducial lim!ts for these two estimates, we proceed
as usual.
that
We obtain an estimate of the common variance
(]"2.
From this we know
8•
.e
is distributed with
!
value
t
n
as the degree of freedom of
8
2
Inserting for t a tabular
•
for u degrees of freedom and a significance level P , we have, if
=
t = ... t
or
M - (1l -1l )
1 2
SJ(1/N1)+(1/N2)
" = ~,Mu =M :
.. ;
=
t.s
Ji
1
+
~2=.l
For our example we have for the logarthms
.e
A
Sxi Total
~.
S
xx
-
X lfJean
= s-\Xi~i)2
S2
x
= S2(~
+
i
i 2)
Ollabain
.190
.199
.233
.158
.093
.276
.369
.384
.a67
.301
.356
.230
.167
.342
.1.518
2.047
.718
.996
.678
.814
.825
.760
.693
.661
.825
6.970
.2}7
.29·~
.• 774
.046 751
.035 234
.087 962'
.169 947
.007 792
.005 872
.010 995
.084 794 = 8 2
6
8
6
DF
V(M)
B
(.0493 )2
(.0464)2
20
9•
.e
n;:;20
t. 05
= 2.086
t .05··!VfM)
+
l\)Mu
-.028
.10:;
! .097
+
M
0.075
R
1.19
RL>I\J
.94
.178
-.579 - .385
-.482
.330
.264
1.51
.412
It will be seen that the relative potency of B compared to A) which is
R=1.19 does not differ signific&1tly from 1; that is the two strophantus tinctures
may have the same potency.
The calculations done so far are valid provided the tolerances are proportional
to body weight.
If we had data on doses and body weights separately we would
prefer to adjust the doses for different body weights by analysis of covariance .
.e
There is a criticism which may be raised against direct assays.
In the
strophanhus assay a preparation of lower potency has to be adnlinistered during a
longer time period until it is acting.
in time
If the action of the drug is cumulative
or if there is a time-lag; then the difference in duration will cause a
bias in the comparison of the two drugs.
:;. Indirect Assays.
:;.0 Preliminary remarks.
If it is impossible to observe directly the dose
l~uch
corresponds exactly
to a specified response) as in direct assays, then we have to consider the doseresponse curve of both standard and test preparation and to compare doses giving
the same response.
It has been found that the dose-response relationship can be theoretically
considered as a monotonic increasing) single-valued real function.
That is) if
o
e
10.
Z is the dose of a stimulus
and the response in any subject is u) then we consider
the expected value in random sampling
E(u)
=U
and the function
U
= F(Z)
is considered as a single-valued real) monotonic increasing function.
For the moment we are considering only quantitative responses.
The case of
quantal response will be treated separately.
In an analytical dilution assay the test preparati on must behave as if it
concentration~.
were the standard preparation at a higher dilution (or lower
consider doses
.e
~~Z2'Z3'"
/.!
If we
for the test preparat10n then the doses producing the
,..,s
same response for the standard must be
Res onse
U
Zl'
Z2'
Z3
...
and given one point on
the test curve; we are able to construct
the whole standard curve.
He have
Z
Zs
where p
= PZT ;
Zs
P = Zrr
=
l/Zrr
l/ZS
is the potency of the test preparation relative to the standard.
It has been found that most of the dose-response-curves can be put in either
of two forms; sometimes after a suitable transfozmation of the response U, where
these transformations are ~f a very simple kind; e.g.
= U (no transformation)
Y = log U
Y = l/U
Y
J..
e
Y
The two forms are
= U2
11.
(1)
Y
=x
+ ~ log Z
(2)
Y
=x
+ ~Z
the first giving the so-called parallel-line assay,
the second the slope-ratio
assay.
If we consider the condition of similari t~l whi ch follows from the fact that
we consider assays as being of the nature of analytical dilution assays, then we
get in case (2) for standard and test preparations
u=y
The condition of similarity requires tha.t
fOr·
both
standard and test preparations
we have straight lines and that they intersect on the point Z
.e
= O. The equations of
these lines may be written as
+~3Z
Ys =
ex
Y =
T
ex + arZ
and
will be the potency ratio.
In tact, it
Y
Zs and
z.r are
doses producing the same effect, we have as a
definition 01' p
On the other hand
from which
~
e
These considerations can be slightly general1 zed by assuming a linear
~
relationship between U = Y and Z = X.
In this case we have
For the case of parallel-line assays" the standard and the test preparation
have equations
.
again as a consequence of the similarity condition in analytical dilution
assays •
.e
x
It may be shown that a slope-ratio assay can always be transformed into
a parallel-line assay.
If we use the general form with ~
ratio assay:
Us
=Q
UT = Q
~
+ f3 SZ
+ f3TC
·
I::
1, we have for a slope-
-e
Taking log(US..a) = IS and log Z we get
IS = log f3S + A log Z
Y = log f3 + A log Z
T
T
which are the equations tor a parallel line assay.
possibilities depends upon homoscedasticity.
The choice between the two
It the slope-ratio assay is
homoscedas~
tic the parallel-line assay cannot have homogeneous variances, and Vice-versa.
'.1 Parallel line assays.
The five statistical points in dealing with parallel line assays are
.e
(1)
linearity of regressionj
(2)
are the regression lines parallel?
(3)
are the regressions homoscedastic?
(4)
determination of Rj
(5)
fiducial limits of R.
The computations as well as the interpretation of results are considerably
simplified if some symmetry is introduced into the design of the assay.
It is
sometimes not easy to obtain this symmetryj and the symmetry of the original
design may be destroyed through accidents.
It is therefore necessary to deal
with the general unsymmetrical case before discussing symmetrical designs.
'.1.1 General case of parallel line assay.
An example will help to illustrate the methods to be used in the general
case.
e
Example.
British Standard Method for the Biological Assay of Vitamin
D, by the Chick Method.
British Standards -Institutioo 1940.
British Standards
.4It
Method for the Biological Assay of Vitamin D by the Chick Method.
3
Standards Institution 1940. British Standards no. 911.
British
Assay of vitam D in cod-liver oil by means of its autirachitic activity
3
in chickens. The response used was percentage bone ash. All responses lay
between 40 and 45 ; therefore we have not to take special safeguards to ensure
linearity and homoscendasticity as we should generally with percentages.
In the following table a metametric transformation of the observed percentages
u is used, to simplify computations,
y
= 10(u
- 30.0),
so that responses of 33.5, 33.0, 29.5 etc. are given as 35, 30, -
etc.
The
transformation
u
= 30.0
+ 0.1 Y
will be used to get back from y to u.
The doses for the standard as well as for the test preparation are taken
in a ratio of 5/3.
are equidistant.
This is a desirable feature because the logs of the doses
We may therefore assign the values -2, 0 and +2 for x in the
Standard and -3, -1, +1, +3 in the test preparation.
Responses in an assay of cod-liver oil for vitamin D
3
15.
.-
BSI - Standard, S
Doses z
Units per 100 g food
g.
per 100 g food
5.46
9.6
16
32.4
54
90
150
-2
0
+2
-3
-1
+1
+3
35
30
24
37
28
73
31
21
-5
62
67
95
62
54
56
48
70
94
42
·116
105
91
94
130
79
120
124
20
39
16
27
-12
2
31
26
57
89
103
129
139
128
89
86
140
133
142
118
137
..
..
T
274
650
859
123
249
820
855
Nj
9
10
8
7
6
8
Yj
30.4 69.0
107.4
T
2047
-y
1783
66.037
73.107
Sx
-2
+2
-x
-.0741
+0.0714
67.8519
139.8571
x
Response y
.e
Cod-liver oil, T
j
S
xx
(27)
60
48
-8
46
77
..
84
101
7 (28)
17.6 41.5 102.5 122.1
The values for x are obtained by taking
xs
= 2(10g us - log 9.6)/log(5/3) for the standard,
xt
= 2 log ut - ~ ( log 90
and
e
+[ log 54]) /log (5/3) for the best preparation.
16.
The first step might be to plot the responses y on the doses
~;
this will
show approximately whether we may assume linearily, parallelism and homoscede81.
ticity; and may also be used to plot parallel straight lines and so measure the
distance Mas an estimate of IJ.
,0
140
120
100
80
60
40
20
0
..c
I
I
IX
.
0
I
.
c
c
0
I
k
I
1
'/.
0
-
><.
Standard
--. Test Preparation
(.
'-
I
I
X
0
)(
I'
0
-20
-3
0
-£
-2
0
+1
+2
+~
x
I
This plot may sometimes be all that is needed; generally however as full statistical analysis will be necessary.
(1) First thing to do is to calculate the linear regression line of y on
x for standard and test preparation separately.
well known.
I will suppose that this is
To illustrate the notation and procedure let us take the cod-liver
oil data.
x
= -3,
-1, +1, +3; N = 28,
= 2047; Y = "B.ll
= +2; x = +0.0714
T = Sy
i
SXi
sx2 = 14(9) + 14(1) = 140
Nsxi = (28)(140)
= 3920
_(sx )2 = 22
= ~4
i
N Sxx
Sxx
= 3916/28
= 3916
=
139.8571
Sxy
2
Sy.J.
= 73
=
382/28
= 2 620,
=
786
215 549
17.
Sx.y.
J. J.
= 3(855 - 123)
= 3.732
•
+
+
1(820 - 249)
N(Sy~)
= (28)(215 549) =
J.
-(Sy i) 2 =
1.571 = 2 767
S
..J1.
1(2047) 2=-4 190 20;4
= 1 845 16;)
= 1 845 163/28
=
7' 382·
xy
Y
b
=
NS 44 =
= 77 476
NS
-1 -2
6.035 372
= Y+
b(x -
x)
= SxylSxx = 2620.786/139.8571 = 18.739
Y = 73.11 + 18.739(x - 0.0714)
Y = 71.77
Test of
~
+
18.739 • x
= O.
Variance
DF
Regression
1
26
Residual
58
MS
F
49 111.130 49 111.130 76.062
1
16 787.549
645.675
...
65 898.679
•••
...
FO•001
13.74
...
...
Total
I
SS (Regression)
= S~Sxx = (2620.786)2/(139.8751) = 49 111.130.
27
We can reject the hypothesis 13 = 0, i.e., the response increases with
increasing doses.
65 898.67S
18.
(2)
Test of
linearit~
y.
2
_ Y.)
J.
S N. (y.
I
J
J.
j
measures deviations from a straight
line.
ss
-+---+-_-+-_+-'-_-+-_~,X
2
1
j
To be compared with
(Y .. _ y.)2
i
JJ.
J.
xj ••• K
j
taking into account the respective degrees of freedom.
An analysis of variance
gives these SS.
S s(y .. _y)2
j
.
J.
JJ.
total SS
= Sj 'S(Yj._'Y.)2
J.
J.
J.
=
+ S N (y._y.)2 + S N.(Y._y)2
j
J.
J
. J J.
J•
J.
SS within doses + nonlinearity SS + regression SS
bet'ween SS
DF
N-l
=
K-l
Computation according to the
follo~dng
2
total SS = S
N-K-l
+
+
formulae
2
= S S Yj . - T /N
YY
j i
J.
= S (T~/N.)
SS(below doses)
SS(within doses)
SS(regression)
J
j
= total
J
2
- T /N
SS - SS(between doses)
= S~Sxx
SS(nonlinearity)
= SS(between
doses) - SS(regression).
For cod-liver oil data:
SS(between doses
=
12372 + 24
= 51
g + ~ + 8758
326,608
2
0
2
1
19.
I
.
Variance
Analysis of variance, test preparation
!
F
SS
MS
DF
I
!
~egreSSion
Means about
. regression
;Between doses
2
3
151 326.608
!l'lithin doses
24
114
572.071
,
149 111.130
I
I
I
Total
i
I
• ••
;
• ••
i
2
21'1.478
1 107.739
!
i
1.824
27
I
I
,
~
FO• 05
i
·..
,{
i
3.40
I
.
>
,
I
"
l
I
I
j
i
,
I
1
;
• ••
• ••
607.170
I
• ••
I,
·..
I
·..
i
• ••
I
I
i
I
65 898.679
I,
• ••
• ••
I
I
There is no reason to assume non-linearity in the regression of responses on
doses.
Some computation for standard preparation
.-
Pt
SS
regreSSiOn
1
24 986.571
24 986.571
:Means about
: regression
1
96.295
96.295
13etween doses
2
25 082.866
•••
Within doses
24
8 356.097
348.171
26
33 438.963
:Variance
,
I
j
MS
i
,.....I
I
, Total
•••
It may be seen that the regression coefficient is different from zero with
high significance and that there is no indication of non-linearity.
(3)
Test of parallelism
The method consists in first calculating separate regressions for
standard and tea" preparation and the corresponding residual SSe
Then
regressions having the same regression coefficient an calculated and the
residual SS for that case.
The difference between these residual SS measures
20.
the amount of non-pallallelism.
bS =
s·xy'Is'xx
If
and bT
= f1'xy'1ft'xx
are the two regression coefficients for individual regressions, the joint
recression coefficient is taken as a weighted mean, the weights being the
and S"xx·
Thus
The SS (resression) is
(S. )2/S'
xx
xy
and
(ft'xy )2/S"xx
for the two individual regressions and
(S. +8" )2/(s, +StI )
xy
xy
xx
xx
for the joint regression.
The difference
(S. )2/s • + (Stl )2/S" _ (S' +S" )2/(S' +8" )
xy
xx
xy
xx
xy xy
xx xx
is the sum of squares for non-parallelism with 1 degree of freedom.
The analysis of variance for each preparation may be co_ined into a
composite analysis of variance according to the following scheme:
S~
,_
21.
Variance
Standard
Regression
~
1
Means a.round regression
K ..2
S
Between doses
KS-l
Within doses
Ns..KS
Ne-l
Total
1
CombinGd
2
Krr..1
K.r.. l
K..2
NT-Kr.r
lI-K
NT-l
B..2
K-4
The two degrees of freedom for regression in the combined analysis correspond
to the sum of squares
and this can be split up into
.e
b2 (S'xx+~'xx)
= SS
(joint regression) with 1 d.f •
and
The analysis of variance for aU N observations may be completed by considering
1 d.f. for the difference between preparations.
Variance
e
D.F.
. SeSe
rT"NT - r /N
Between prepara.tions
1
(TS)2/NS +
Joint regression
1
Parallelism
1
(S' +f!' )2/(S' +S" )
xx xx
xy BY
(S'xy)2/
(S' )2/S' + (S" )2/S" .. (S' +S" )2/(S' +B" )
xy
xx
xy
xx xy xy
xx xx
stxx+(S"xy.)2/~'~
Difference
Linearity
K-4
Between doses
K-l
Within doses
NooK
Total
N..l
:=I
'3 (~/Nj).1J.'2
T,S
Difference
2
S S Yii - I/N
i i
,.
A
22.
The second and fourth row can be used to test the slope of the regression and
linearity simultaneously tor both preparations.
For our example we get
(a)
Total SS:
S S
j i
Y~i
standard 151 183
Test
215 542
Total
- r!/N
366 732
= -(3830)2/55 .266 707.27'
100 024.727
(b)
Between preparations:
(1783)2/27
(2047)2/28
=117 744.. 037
= 149 650.321
267 304.358
-(3830)2/55
=~6
701.~273
687.085
(c)
Joint regression:
1 302.07
stxx = ·67.8519
.
8" 1:1 2 620.79
xy ..
S~ .. ll9~e 8571
.3 922.86
207.7090
8~y
1:1
b =
3 922.86/207.7090 .. 18.886
SS(Joint regression)
(d)
Para.lle1ism:
::I
(3922.86)2/207.7090
= 74
088.415
~e
Variance
-S.S.
D.F.
Regression Standard
1
24 986.571
Regression Test
1
49 111.130
2
74 097.701
Regression, Joint
1
14 088.514
Nonpara11elism
1
9.286
Total
(e) Between doses:
Standard:
(274)2/9 + (650)2/10 + (859)2/8
= 142 826.903
Test prep: (123)2/7 + (249)2/6 + (eeO)2/8 + (855)2/7 == 200 976 •.22...2
343 803.832
• -266 707.273
77 096.559
The SS (within doses) and SS (1:!.neari1y) can be obtained either by subtraction
or by totalling the corres,onding SS for standard and test preparation.
Thus
we have the following analysis of variance:
VEq'ianc~
e
D.F.
S.S.
M.S.
F
FO• 05
FO•001
Between preparations
1
687.085
687.085
Joint regression
1
74 088.415
14 088.415
Parallelism
1
9.286
9.286
Combined linearity
3
2 311.773
770.591
Between doses
6
77 096.559
•••
•••
•••
• ••
Within doses
48
22 2,28.168
471.670
• ••
• ••
• ••
• ••
•• •
• ••
Total
54 100 024.727
,
..
1.438 4.043
•••
155.104 4.043 12.293
• ••
•••
1,614 2.799
• ••
• • e
This analysis ot variance shows that
(a)
The difference between preparations is not significant.
This is
not essential, but it is in general preferable to have y - it'
.
s
small. The precision of the estimate of JlOtency decreases with
.
increasing
(b)
(y8
-
Yt)"
as we will see later on.
The joint regression coefficient is very highly significant" compared
to b = O.
This is also a desirable feature; other things being equal
a higher value of b means higher precision in the estimate of M and R.
(c)
There is no indication of non-parallelism.
(d)
There is no indication of non-linearity.
(3) Provided there is no heterosce.d8e"te:1ty, the assumptions for the validity
.e
of the assay are fulfilled,
pronounced differences.
We have already seen that the ranges do not show any
It should, of course, be born in mind that the ranges
are based on different sample sizes--varying from 6 to 10.
It would therefore
be better, to plot the estimate· standard deViations (either computed from the
sample, or from the range) as a function of the doses.
Inspection of such a
grar>h would show whether the assumption of hOU1Oscedenltt",yYis fultil.l8d or not.
(If.)
Having thus established the validity of the assay, the next step is to
find M and log R" which is the estimate of the difference between the log doses
produoing the same response,
It is easy to find M as a function of is - xt ' Ys • Yt and b. All we have
to do is to consider the responses for the same log dose X and x •
t
s
y
From geometrical considerations it
folloWS that M = (YT - Ys) /b
if YT and Y are taken for the same
S
value Xr.r
DIeD ~
(say
We have
Y
s
:I
Ys
+ b(x
s
.1s }
and trom this
tor ~
:II
\t = '"
or
Alternatively we arrive at the same expression, it we consider M = Xs
the same response Ys
•
YT.
- JeT
for
This gives
.e
and trom this we get
•
In our example we obtained
xs • -0.0741
.
Ys ..
it • +0,.0714
Yt
= 73,.11
is -it =- .0.1455
;'fs-Yt
=- ·7.07
66.04
b
=- 18.886
therefore
M
e
= -0.1455
+
l~':§~ = .0.1455 + 0.3743
= +0.2288
This is the distance between the parallel straight lines measured in the x
scales (metameter).
We have now to go back to the u-scaJ.e, or rather (log u)-
26.
scale.
The transformat1on to the metemeter was g1ven by the equat10n at the
bottom of page 14) I that 1s
X
s
~
= 2(log Us - log 9.6)/log (5/3)
III
2{log UT - ~ [log 90 + log 54]}/log(5/3) •
From these we get
lDg
'll = log
log
\\r
9'.6 +
Xa
log (5/3)
2
= ~ [log 90 + lDg 54] +
~
log
~5Q)
hence
.e
log R =
~og ~5/3)
M + lDg 9'.6 -
log
M
R
:;I
1
~(lDg
90 + log 54]
+ 0S-OT' with 1 un1t of x-scaJ.e, Os and 0T or1gins
of
~sca1es, e~ssed
in u.
For th1s expression we get
! log(5/3) == 0.1109; lDg 9.6 - ~[log 90 + log 54]
log R • (0.1109)(0.2288) - 0.8610
log R
:II
T .1644
R
III
and
0.1460
:II
:;I
0.8610
0.0254 - 0.8610 = -0.8356
[0.1644 - 1]
as the estimate of the relative potency of the cod-liver oil compared to the
1 mg of cod-liver oil contains 0.146 units of vitamin D "
3
or J. g of cod-liver oiJ. contains 146 units of vitamin D •
3
Vitamin D -Standard.
3
(5)
As a last step in the analysis" we have to calculate bow precise the
above estimate is.
We will obtain fiduc1.cJ. limits of M (and of R) by a method
first proposed by C. I. BliSS" the method being known as Fieller's theorem.
The
procedure is s1mU.ar to that used for direct assays, but more complicated because
in
y - y,
we have an expression (S
T) I where numerator and denominator both depend on
b
th~
varia.ble y.
We 'Will first determine fiducial limits for
.e
M
o
YT - YS
=-_.
...
b
•
As
The fiducial limits for M 81'e obtained by substitution of
We obtain fiduciaJ. J.1mits for ~ as follows.
and b are
and if
JoL
o
JoLT - JoLS
= ---...,;;;-
If the expected vaJ.ues for
YT, Ys
28.
tben M is an estimate of
o
~
0
•
I f we consider
this is normally distributed for any
~
o
, with
and the estimated variance is
.e
blY and Y
T
s
so V( b ~o -
being independently distributed.
The ratio of ( b.
~o
-Y + Y) is distributed as is t 2 with n == N - K d.f.
s
T
- -YT + -)2
Ys
If we insert
for t the tabUlar value + t , we can soJ,ve the equation
- p
2
s)
(b ~
- Y + Y
T
o
for
"0'
2
== t
2
~
s
2
~o
1
1
( - - - - - + - +-)
s~ + s~ NT NS
and thus obtain fiducial limits for Mo.
b
2 ~0
-
(-
We have
-)
2 b YT - YS ~0 +
2
Dividing b,. band rearr8l161ng gives
==
o.
·e
If' we substitute
and divide by (l-g)" we get" after replacing
(YT-YS)/'b
by M
t 2 s2
~2 _ 2 M ~ /(l-g) + M2/(l-g) - ~
o
0 0
The roots of this equation" Mo "L" Mo"U
M ,L M U
0
0"
= M /(l-g) + [
0
.e
-
= [M
o
Now" setting Mo
a
M.." M.. =
-"L --U
b (l-g)
0
s
t
li
0
(l-g)
T
2 -
l
(--m
+~)
ri
...s>l-g
1
+"
2
2
1
1 1/2
(if." + r ) J
b (l-g)
T
S
t
8
1
Ii
T
S·xx + ~'xx
S
0
M - is + 'X.r
- -)2
( ~~~
t s
[M - X + 'i.. ! ....Po...- {(l-g)(.1. + 1:.) +
o
s '1'
b
NS
NT
st
XX
We may note the following three particular COle
(a)
g«
If' we write
= o.
s
T
~.
!..R.- {(l-g)(-N + --N) +
b
o
1
+ S"
XX
~
J
/(l-g)
30•
.e
in the form
we observe that this is t 2 == F with n == 1.. n == N - K divided by the F ratio
2
p
p
.
1
F -= b2(S~ + S~)/s2 for " ..8SJ'el.1on" compared to "within doses." If b differs
significantly from zero.. as it should" the quantity g is likely to be very small,
.
.
i.e. g« 1.
The fiducial limits are in this case
-
(M
t s {_
1+1
...E_ +
b
NS
NT
- X
st +
xx
s
2
;,
·e
s
If we compute according
=4.043/155.104
{
= 0.0260 •
to the exact formula, we get
and
Mu
==
0.8774
and using again log R == (0.1109 M) - 0.8610, we get
-)2
x.r
)1/2
SIt
In our example we have, for p == 0.05
g
+
xx
} == V(M)
31•
.e
log ~ = 1,.0947
log R
o
III
(0.0947-1)
(.0.9053)
~ =
(.0.7637)
0.1244
~
R
If' we consider g
«
1.2363
,
(0.2363-1)
III
0.1723
:I
+0.8589
III
0.1715
0.1460
III
1 1 we obtain
~
III
-0.4013
~
III
0.1243
Mu
Bu
In our example we might calculate fiducial limits according to the formula (a).
·e
This will happen if the arrangement of doses is symmetrical.
Xs
III
~
III
1\1
O.
~
If'
:I
is ...
~ then
is -
~ = 0, and therefore
ts
11
,;
1/2
[M !..R- {(l-g)(rr + 'N) + ---=-=--}
J/(l-g)
1
b
S
T
stxx+ffxx
Often
32.
_e
This formula obtains for highly significant b and symmetrical design of assay.
3.1.2.
S;ymmetrica.l four-pOint assay.
If the linear relationship between log dose and response metameter has been
established
of
be~nd
efficiency.
high
reasonable doubt, then the symmetrical four point assay is
In it the experimental units are evenly distributed over
the t'WO doses of standard and test preparations.
~m:ple.
The relative potency of a hormone BMG, 20 (human menapausa1 gonado-
troph1a) has to be determined by comparison to a standard HMB,A.
is the weight of the uterus of mice.
Twenty animals were evenly distributed
over the two doses and two pHl)&rations.
1/2.
.e
The response
The doses were chosen in the ratio of
For the standard (HMG,A) the doses were 2.0 and 4.0 units per animal" for
the test preparation (JilMG,20) 1.0 and 2.0
mg.
per animal.
The response y - weight
of the Uterus--is given in mg.
Test preparation
Standard
Dose u
2.0
4.0
1.0
2.0
x
- 1
+ 1
- 1
+ 1
Responses Yi
9
35
25
8
33
54
47
58
66
78
36
110
303
Total T
j
Bj
'044
5
353
512
5
22.0
85
159
5
925
5
20
10
10
HSI NT' N
74
56
27
22
413
TS' TT' T
92
46
60.6
70.6
31,8
0
0
0
0
33•
.e
tIbe dose metameters are:
.
.
Xs
for standard
= (log Us - (log 2.0 + log 4.0)/2 }/lo~ 2
for test preparation ~ = (log
u.r - (log 2.0 + log 1.0)/2}/10~ 2
•
We are first considering the analysis according to the general case dis·
cussed earlier.
From this we will be able to find out the simpler straight-
forward analysis of a symmetrical four-point assay.
The first step is to plot the results on a graph.
y~
_Standard
100
80
• • •• Test preparation
60
.e
40
The graph does not seem to indicate
20
non-parallelism or heteroseedastic1ty•.
0
tIbe value of M 'Day be estimated as
~r-----If------+-X
0.5 approximately.
The analysis of variance proceeds as in the general case.
syy
;0:
10,993.75
= total
SB
then
= 7 918.55
SS(preparations)
= (41,2
2
+ 512)2/10 - CJ25 /10
We get
*~
To obtain the SS(regression), we note that
for standard:
S xi
= 0, i = 0, TS = 413, Ys = 41.3
2
S xi
S~
bS
::I
= 303 - 110
=
10.
193
= 193/10 =19.3
(S~)2/s~
for test preparation:
a~ =
10,
=1932/10 = 3,724.9
stlxx = 10, Stlxy = 353 - 159 = 194
b
T
=-
194/10
= 19.4.
(S"xy)2/s"xx = (194)2/10 = 3,763.6
Joint regression:
b =
.e
(194 + 193)/(20 + 10)
SS (joint regression)
SB(non-parallel1sm)
= 3,724.9
+
::I
= 387/20 =19.35
3aT/20 = 7,488.h5
3,763.6 - 7,488.45
From these we obtain the analysis of variance:
Variance
Preparations
Joint regression
Non-parallelism
Between doses
Within doses
fote1
~
D.F.
SS
MS
F
2.• 600
39.137
• ••
1
1
1
3
16
490.• 05
7,488.• 45
0.05
7,978.•55
3,015,20
490,.05
7,488.. 45
0.05
19
10,993.75
•••
....
188.45
.. ..
• •• •
•••
FO• 05
FO.O()J.
4,.494
4,.494
16.119
16.U9
•••
•••
,. ,.
..
~
.,
., .
....
•••
•••
.e
From the analysis of variance we see that there is no indication of nonparallelism and that the resression coefficient differs from zero.
The three SS for preparations" joint regression and non-parallelism can be
obtained very simply in the case of a symmetrical four-point assay.
Instead of oa1cule.t1n8
we could use the equivalent formula
2
SS(preParations) = (413 .. 512)2/20 == 99 /20
III
9 801/20
·e
and if W
l
= N2 this
g1ves
the required formula
This formula may also be used to calculate the SS(Regression) I
e
SS(regression) ==
t~03 + 353 .. 110 - 159) 2/20
==
(656 - 269)2/20 = (387)2/20
= 149
769/20
= 7 488.45
36.
and for non-parallelism
SS(non-parallelism) = (194 - 193)2/20
= 1/20
It is seen that we have in foot three independent or orthogonal comparisons
between the 4 totals, which may be summarized as follows
Comparison
Sl
S2
(110)
(303)
T
1
(159)
T
2
(353)
Value
Coetficien1;s
.
..
_
Preparations
SS
+
+
Regression
+
Non-parallelism
+
+
+
-99
490.05
+387
7 488.45
+ 1
0.05
Between doses
Estimation of R.
7 978.55
According to the formula for the general case
M = (51.2 - 41.3)/19.35
•
•
I'
= +9.9/19.'5
M = +0.5116
For log R we had the general expression
37.
For our example we get
i = ~ log 2
~(log 2
Os
:=I
0T
=~(log
:I
~(0.3010)
+ log 4)
1 + log 2)
= 0.1505
.
=~(0.3010
1;1
.
+ 0.6(21)
~(0.3010)
= +0.4516
= +0.1505
and therefore
log
R •
(0.1505) (+0.5116) + 0.4516 - 0,1505
=0.0770
log R
lIZ
+ 0.3011
0.3781
Thus, 1 mg of the test preparation MHG,20 contains 2.4 units of the
.e
standard MHG,A.
For the fiducial limits, we may use formula
ts
M-, M._ = [M.:!:..L..
-"1..
--U
b
(b), that is
rf
11
1/2
({l-g)(-N + -N) + --=.:._-]
J/(l-g).
S
T
Sf +S"
xx xx
l-g = 0.886 907
(l/Nl ) + (1/N2) =?2
(l-g)(~ +~) = 0.177 381
•
12.
~/(s~:s~) :a (0.5116)2/20
{ J•
0.190 468, { Jl/2
= 0.013
= 0.436
427
087
38.
·e
l\ = [M - s
Mu = [M + s
(ii. :r28)(~.1.20}(0.436
tO~05{ }l/'~"/b ~
M s ~O~05{ }1/~/b
= -0!1448
M t •••
= +1.1680
s
t o '.05{ }1/2/b J/(1_g)
1 2
to'. 05 ( J / /b)/(1_g)
IL =
log Ru
log
::I
IL
::I
427)/(19.35)
-0:.1448/0.886 907
= +1.1680/0.886
::I
= 0'.6564
-0:.1633
907 = +1.3169
(-O,.1633){0.. 1505) + 0.3011
::I
+0,.2765
(+1.3169)(0.1505) + 0.3011
I:
+0.4993
= 1.890
In this ccee the fiducial 11mits seem to be rather w:J 4e •
.e
As we have seen, in a four-poj.nt assay it is not possible to test linearity.
The next step in symmetrical assays would then be to have three doses in each
preparation, preferably equally spaced.
Example.
Assay of testosterone propionate (L.IQ Pugsley. Endocrinology,
vol. 39, 1946, p. 161-176), increase in (length + height) of combs of capons
.
as response.
Standtad P1'SPeratton
20
40
80
20
40
80
x
-1
0
+1
-1
0
+1
Responses in mm
6
6
12
11
19
14
14
15
14
6
6
6
7
4
12
11
16
18
19
16
15
Doses in mg
5
e
Test preparation
6
7
12
10
7
12
12
10
39•
.e
30
Tj
5
N
j
~9
5
6.0
-YS" -YT
10.4
5
5
-xS" -~
84
378
5
5
30
15
15.2
5,8
n.4
16.8
11.333
10.533
S xi
57
170
15
..NS" NT' N
j
76
158
TS" TT" T
-Y
52
0
0
0
0
It would be possible to measure the increase of the co
in different
ways, for example a.s the product (instead of sum) of length and height.
We
are not concerned with this problem here.
.e
If we use the general formulae given in
3.1.1" we get the following
comparisons between the Six doses:
(a)
Preparations
(b)
Regression
It may be shown that for b = Sx!Sx;x we get ~xy = T - T1, Sx;x = 10,
3
b = (T - T1 )/10 and S~/Sx;x = (T - T )2 /~O. This is the so-called
1
3
3
linear component of regression, with 1 d.!. The remaining degree
of freedom is obtained as
- 2 T + (T + T ) as comparison, curvature
2
1
3
(-2 T + T + T )2/(6)(5) ,.. SS(quadratic component of re2
1
3
gression).
•
(c)
NOn-parallelism
(d)
Interaction between curvature and preparations.
40.
The five comparisons between the six doses can be set out as follows:
-e
T2
+1
T,
-1
T
l
+1
+1
'0
0
+1
-1
0
+1
20
+1
0
-1
..1
0
+1
20
9
4.05
Curvature
+1
..2
+1
+1
..2
+1
60
1
0.02
Curvature preparations
-1
~
-1
+1
..2
+1
60
Comparison
Sl
S2
S,
Preparations
-1
-1
Slope of linear
regression
Non-parallelism
-1
Divisor Comp
SS
4.80
12
101 510.05
-,
0.15
519.07
These are orthogonal comparisons.
18
.e
16
14
,/
,t,
J2
/
,/
>if
There is no indication of non-linearitYI
,..8
but the assumption of parallelism seems
'I'
doubtful.
/
10
,/
8
co
6
4
o
1:./
,/
It is useful to have a formal
U
test.
0
o
1
..1
11
1
o
+1
Analysis of variance.
Total SS = 575.87 j
SS (doses) ~ 5l9.07j
SS (within doses)
= 56.80.
The SS for the five individual d.f. between doses are given abovej
their total checks the SS (between doses).
41.
-e
D.F•.
Variance
SS
Preparations
1
4.80
Slope of linear regression
1
510.05
Non-parallelism
1
4.05
Curvature, joint
1
0.02
Interaction Curve. Prep.
1
0.15
Between doses
5
519.07
Within doses
24
56.80
29
575.87
Total
MS
•••
2.367
:I
s
2
•••
calculation of F is not necElssary; the regression slope is highly significant, there is no indication of lack of linearity or parallelism•
.e
Calculation of M and R.
b ,. 101/20
=5.050;
M 1:1 {Ys-yT)/'b
= (11.333
- 10.533)/5.050
:I
0.800/5.050
= 1.1584
log R • i M + Os - 0T
i ,. log 2 ,. 0.3010,
log R
1:1
OS" 0T = log 40,
(0.3Ol0){0.1584)
Calculation of
fiduc1~1
omitted.Results:
I\,
:I
Os - 0T ,. 0
0.0477; R • 1.12.
limits.
= 0.95;
~
= 1.31.
3.1.3 Assays in randomized blocks or in latin squares.
In the preceding paragraphs the experimental units were allotted at random
to the four or siX dosage groups.
The experiments were all completely randomized,
without any restriction in the design.
This is not always the best design which
can be chosen.
Often it is preferable to take advantage of simularities which
can be
with confidence in the experimental material.
e~cted
The comparisons
between the doses and the preparations can then be made between units Which are
presumably more alike in their responses, so that the results are of bigher
precision.
In biological assays the devices most otten used to increase the precision
of comparisons are (a) to make comparisons between litter--mates (of the same
sex) and (b), if feasible, to use the same animal repeatedly and to make comparisons within animals.
It is to be expected that an animal would react more
uniformly in repeated trials and that Jitter-mates react more similarly than
unrelated animals. (c) To use microbiological reactions on the same agar "PJ,.a.te
as blocks.
The analysis of the assay must take into account the s'truGt!'1J'8 of the
assay.
The error ve,riance s2 must be chosen so as to reflect the variability
between units e.g., within litters, if there were no real differences between
doses and between preparations.
We consider first an example of a symmetrical four-point assay arranged
in randomized blocks.
!..xample.
B.O. SChild.
Estimation of the histamine content of a preparation.
Journ. Physiol., lQ, 1942, p. 115-130.
The hista.m1ne content of a preparation is measured by using a dilution
of the preparation.
A piece ofen1mal tissue is suspended into the solution.
The contraction of the tissue is the response; it is linearly related to the
log of the dose.
twenty t:1mes.
The same piece of tissue may be rinsed and used again up to
The response will however diminish as time progresses.
.e
The four treatments 811 821 T " T2 could be repeated five times each, if
1
we could use the same tissue twenty times. We could arrange these completely
at random.
If we did this I we would IIOt take advantage of the fact
secutive tr
18 are likely to give similar responses.
that con-
We could 1ncrease the
precision of the comparisons by forming five groups of four consecutive trials
and allotting the four treatments at random to the four trials within each
11113 actual desigb used in this experiment was as follows:
I
II
III
rv
V
1
T2
T
1
Te
8
T
1
2
8
8
6
3
T
1
T
l
82
T
l
T
6
2
T
4
8
1
8
2
T2
group
Trial:
1
.e
2
6i
2
1
2
2
6
1
The doses and the responses are given below:
Test preparatiob
Standard
Doses in IJ.g
x
Responses (1/2 mm)
I
II
III
rv
V
tj
T6, T" T
N ,
6
e
NTI N
YS' YT
6xi ,
i 81
0.200
-1
+1
-1
+1
Total
89
84
66
50
122
122
118
110
98
103
104
87
74
73
131
132
136
106
445
450
425
362
327
381
570
441
617
2009
~
'X.r
0.125
112
2009
1058
951
10
76.2
Yi
0.250
0.100
20
10
114.0
88.2
123.4
95.1
105.8
0
0
.
-
44.
Orthogonal comparisons •
Value
T
1
381
8
2
T
l
570
441
2
617
Preparations
.1
-1
+1
+1
Regression
-1
+1
-1
+1
+365
BODparallelism
+1
-1
-1
+1
-13
Comparison
8
SS
1072/20
+107
1:1
572.45
3652/20 • 6,661.25
2
8.45
13 /20 =
7,242.15
b = +365/20 = +18.25
Anawis of variance.
Between doses:
2
SS (doses = (3812 + 5702 + 44 2 + 61r)/5 - 2009 /20 '" 7,242.15
Between groups (blocks):
2 + 4502 +
2
6S (total) = (8e! + 92 +
88(blocks) = (445
·e
+ 3272 )/4 .. 20092/20 = 2,976.70
2
2
+ 106 ) - 2009 /20 = 10,548.95.
The analysis of variance can be set up as tollows:
Variance
Blocks
Preparations
Regression
Nonparallelism
Doses
Error
Total
DF
4
1
1
88
3
2,976,70
572.. 45
6,661.• 25
8.. 45
7,242.. 15
12
"9,10
19
10,548.95
1
MS
744.• 18
572.45
6,661.25
8.45
2
B
F
27..05
20.• 81
24~ .• ;L4
.....
,
....
FO•05 Fo•OOl
3.. 259 9.• 633
4.. 747 18.. 641
4.• 7!4-7 18.641
.. ,- ,.
....
,•••
= 21.51 .t'
.. e,
• ••
•••
•••
~
!Ihe regression coefficient cliffers highly significantly from zero.
no ind:1cation of nonp81"8.ll.elism.
highly significant.
There is
The differences between the five blocks are
-e
of R.
-Estimate
........'--...-.-..-The means
is'
~ being equal, we have for M
and tor log R we get, because of
i :::
~ (log 2) = 0.1505
.
.
Os ::: ~ (log 0.100 + log 0.200)
=~
(-1 + 0.30103 - 1) = 0.15051 - 1
0T = ~ (log 0.125 + log 0,,250)
=~
(0.09691 - 1 + 0.39794 - 1) ::: 0.24742
::: (0.15051)(+0.5863) + 0.15051 - 0.24742
= +0.08824
.e
log R
:2
- 0.09691
.0.00867
R = 0.9803
Thus the test preparation is equivalent in potency to 0.98 units of the standalt"d
preparation tor histamine.
Fiducial 11m!ts.
These are calculated by the usual formulae, care being taken to insert
s
2
= 27.51 with 12 degrees of freedom.
It. = 0.8886
The limits are
Ru = 1.0902
•
It will be seen that the histam5.ne potency of the test preparation is not significantly different from that of the standard preparation.
I
1
46 •
.e
Gain in precision.
The gain in precision tbrough comparing within blocks can be evaluated.
A completely randomized layout would result in the following analysis of
Variance
DF
Between doses
3
Within doses
16
19
Total
•
AssUllling absence of all effects, the within doses mean square may be estimated
as
MB (Within doses)
·e
= 15
a
27.5~~ + 2,276 .1°
= 178.385
..
Taking into account the different degrees of freedom, the relative efficiency
of the randomized blocks layout compared to the completely randomized is
.
nl =12, ~ = 16, si =27.508,
.
s~ = 178.385
The gain in precision is very substantial indeed.
The latin square design could be used for example in a symmetrical s1xpoint assay in the following way.
six animals in each litter.
The animals might be chosen from siX litters,
It' the experiments were time-consuming so that only
six could be carried out each day, then a convenient design would be as folloWS
47.
-e
Litter
Day
1
2
3
4
5
6
1
A
E
E
D
C
B
2
B
A
F
E
D
C
3
C
B
A
F
E
D
4
D
C
B
A
F
E
5
E
D
C
B
B
F
6
F
E
D
C
A
This should be properly randomized, interchanging first the rows, then the
columns at random, and finally assigning T , T2 , Ty 51' 52' 53 at random to
l
the letters A, B, C, D, E, F.
Missing Valu::.! in a design ·.;planned as symmetrical can be replaced by the
usual formulae for randomized blocks or Ie-tin squares.
3.1.5 Sensitivity of a parallel line assay.
BOw
should the assay be designed, in order to have a high sensitivity; that
,
is in order to have a small fiducial interval f
The fidUcial interval depends
essent1allyon the expression
+x.r
2 2
( - -)2
1
1
M-x5
P s
I = 2
2{(l-g)(if" + m) +
}
b (l-g)
5
T
5'xx+E!':oc
t
which is the quarter-square of the fiducial interval, that follows from
and the aim will be to keep I as small as possible, with the restriction that
N5 + NT = N.
This is achieved by choosing
48.
-e
(1)
t
(2)
s
small
C~)
b
large
(4)
g small
( 5)
N + If' small
(7)
8'xx + 5'xx large
p
small
1
1
S
T
(6) M .. 'is + ~ small
We assume P to be chosen; then t
freedom of t.
p
depends essentially on n l the degree of
This should be at least equal to 101 preferably more.
In a
symmetrical 2k-point assay t has N..2k degrees of freedom; if there i8 the
restriction that we randomize within
With k
.e
I = .3 1
=2
1
= N.. 2k..l+l.
= 12
1
this would give
= 16" t = 4:
N = 20"
bloCks l then n
we would have a four..point assay, and if we had N
N .. 2k-l+ 1
N
I
K = 5:
n
= 12.. 4...3+1 = 6
=16..4..4+1 = 9
n = 20-4..5+1
= 12
•
With less than 20 units it seems difficult to have a satisfactory assay.
(2)
s will be small if care is taken by the experimenter in the selection of
experimental units" in carrying out the experiment in a proper way.
(3)
b
=8xylSxx is
high if the dose and response metameter are chosen such as
to give a steep line for the dose-response regression.
(4)
g = t~s2/b2(S~+S~) will also be small if t l sand b are suitable.
-e
The fiducial interval is proportional to t
p
,II
needed is that it is sufficiently small, say 0.05.
if the assay is properly designed (i.e., if'
I is pro'POrtional to l/(l-g)
(M-iS+~)2/(S~+S~) is small)
or if g is smallJl proportional to
(11).
For fixed N the smallest value of'
(t+ j:) is attained when NS
it is then
(6)
s and lib j for g all what is
To make
NS + NT
4/N • EN
1M -
value of P.
X·
N
T
N
S
= N:'fr: =s 4/N]
= NT
= N/2,
T
8 T
~ + ~I small, we have to know something about the approximate
If we have from preliminary experience an approximate estimate
of P, say R , then we choose the doses of the test preParation as
o
and if the design is symmetrical, we get for
.e
which will then not be great in absolute value.
+ Sit large we choose the extreme doses of' the standard
xx
xx
as far apart as possible in the linear portion of the dose - reSJlonse curve.
In order ·to have S'
Then the doses for the test preparation may be chosen as indicated in (6).
The symmetrical four-point assay gives the highest value of S' + ft'
xx
xx
for given N.
This must be weighed against the lack of a validity test for
linear!ty in any particular assay.
Comparison of assay techniques.
So far we have seen how the structure of the design affects the sensitivity,
measured by the length of' the fidUcial interval.
Sometimes we wish to compare the effect on sensitiVity of the assay technique,
50•
.e
that is cf the choice of the animal (or subject) of the resl'0nse and possibly
also of the metameter.
For such comparisons we may discard in I those expressions which deJlE!nd
solely on the structure of the assay.
In particular, ~, ~, S~ and S~ are
only deJlE!nding on the design of the assay.
We may therefore use the standard
error of M
to compare different assay techniques.
qua.ilty tor such comparisons.
Thus we see that sib is the essential
The number of experimental units necessary to
obtain a certain precision del'ends essentially on s 2jb2 •
One has to bear in mind that a certain response may also improve the
.e
precision by extending the range of x in which the dose-response curve is linear,
which will increase S~ and 6~.
,.2 Slop: ratio assay.
We have described slo-pe ratio assays in the introductory chapter.
The
condition of similarity requires that if the dose-response curve is linear for
the standard preparation, it baa also to be so for the test preparation.
ul
I
•
Equivalent doses
Zs,
~
51•
.e
The estimate R of
is bribSI where bT and b S are obtained by adjusting linear
regression lines to both standard and test preparation in such a way that they
intersect at the U-axis.
The slope-ratio assays are mostly used in microbiological assays; furtherllW)re mostly U = Y and z =- x.
3.2.t. General case of slope-ratio assays.
SUppose the experimental units are not evenly distributed over the doeea of
each preparation, and that they are also unevenly distributed over the different
doses.
y
:~.
:
The two regression lines have to be
/~
..----
....
~/
/ 1""""'-;
.....
.e
: /.:.-----V.~I
/"......
S.
•
or
I
I
the constants in both lines are equal.
~l~l ~~2
The values observed are
squares is
adjusted, subject to the restriction that
I
X
43 x~x
Y =
s
Si " YSi;
a +
~i'
Os
x
YTi' and the condition of minimum sum of
52•
.e
•
let us define
where N = N + NT (it being understood that x == 0 are included in N and NT).
S
S
We
assume that x
== 0 for all experimental units in the test preparation, and all
Si
~ = 0 for experimental units in standard preparation.
over all N units, and so is
y
y;
==
TIN
&
.e
From (1) we obtain
or
From (2) we obtain
or
e
MUltiplying (4) by S x
Si
'is
we have
j
T • SYSi + SYTi •
and ~ are thus means
53.
and subtracting this from the preceding equation we obtain
If
we define
(because of S xSi~1
.e
then we can write the preceding equation as
(5)
From
or
<:5) we get
be sss + b T SST = SSY •
= 0)
-e
and multiplying (4) by S x.ri and subtracting it from the last equation we
obtain
(6)
b S SST + b T STT = STy •
From (5) and (6) we may calculate b
s
and bT•
Equation (4)" (5) and (6) may be obtained in an entirely different way j
they may in tact be interpreted as the solutions to determine the constants in
the equation of multiple regression
where x
s
and
x.r are
defined for all N experimental units as before.
A preliminary a.na.lysis of variance may be set up" in which the total sum
.e
of squares of the N responses y is split up first into variability between and
and within doses.
The between doses variability can be further subdived into
that on the regression lines" and around these lines.
For the SS (regression)
we have according to the usual theory
SS (regression)
Ex;am:ple.
=bs
Say + b
T
Kent-Jones" D. W. and Meiklejohn, M. (1944).
STy.
Some experiences of
microbiological assays of riboflavin, nicotinic acid and other
nutrient factors.
.Anslyst".§2" 330-36.
Nico'tinic acid in meat extract.
contained 1 g per 5,000 ml.
Solution of meat extract used
Duplicate assay tubes were prepared.
Five doses for standard and three for test preparation, and one
tor "blank" or zero dose.
Total of N • 18 tubes.
The tubes were
inoculated trom e. culture ot Lactobacillus aratimosus" and incubated
a
at 37 C for 7'2 hours. The aCidity at each tube was measured by
55 •
.e
titration with N/14 sodium hydroxide, bromothymo1 being used
as an indicator in a colour comparator.
Responses are given in ml N/14 NaCH.
x
s
I.lg/tube
x
0
y
Sum
d2
Difference
a
ml/tube
"sy
x.ry
...
Blanks
0
0
1
1.5 1.4 2.9
0.1
0.01
0
Standard
0.05
0.10
0.15 .
0
0
0
0
0
0
0
0
0
0
3.5
5.0
6.2
8.0
9.4
6.7
9.7
12.3
15.7
18.9
0.3
0.3
0.1
0.3
0.1
0.09
0.09
0.01
0.09
0.01
0.335
0.970
1.845
3.140
4.725
0
0
0
1.0
1.5
2.0
0
0
0
4.9 4.8 9.7
6.3 6.5 12.8
7.7 7.7 15.4
0.1
0.2
0.0
0.01
0.04
0.00
0
0
0
1.50
9.0
2
•• •
••• 104.1
•••
0.35 11.015 59.70
0.20
0.25
Teat prep.
.e
x.r
Total
3.2
4.7
6.1
7.7
9.5
0
0
0
0
0
9.70
19.20
30.80
Geaphica1 analyses
y
10\
101
9\
I
81
7
6
5
4
2
1
0.10
o
,
0.05
0.20
I
_.
0.15
I
I
0\
1.0
0.···
._.
__
..1·0..•.... _ •.•._-_..
!
1.5
X
s
.•..
2.0
Xrr
J..Ig
ml
per tube
per tube
-- 31 per I-Lg
b
b
3.1 per ml.
bT -
.8
1 I-LgI ml • 0.1
R ::; bT = 3.1/ 31 ( ml )-1/I-Lg )-1 ::; 10
..
L....
1-Lt¥U&.L
8
making allowance for the dilution factor 1 g/5, 000 ml this
meeD,S
that 1 g of
the meat contained about 500 ...., nicotinic a.id.
Analysis of variance
Total SS = 96.865
Between doses SS
Within doses SS
~ (2.92
+
6.72
+ ... +
= 0.35/2 = 0.175
(check v).
S YS + S YT = T = 104.1
S xT1 = 9.0
2
SSS = 0.2750 - (1.5) /18
= 0.2750
SrT = 14.50 - (9.0)2/18 ::; 14.50
SST = -(1.50)(9.0)/18 =
.e
SSY
= 11.015
STY
= 59.70
2
15.42)/2 _ 10~.1 = 96.690
- (1.50)(104.1)/18
- (9.0)(104.1)/18
- 0.1250
- 4.50
= 0.1500
• JO.oo
= -0.75
= 11.015
= 59.10
- 8.615
- 52.05
=2.340
= 7.650
0.154 bs - 0.750 bT ::; 2.34
0'0.15 bS + 10.00 bT • 1.65
SS (regreSsion)
= (31.080)(2.340)
+
(3.096)(7.650)
= 96.412
Preliminary analysis of variance:
Variance
SS
MS
2
96.412
Deviation. from .regression 6
0.278
0.0463
Between doses
8
96.690
•••
Within doses (error)
2
0.115
0.0194
11
96.865
•••
Regression
e
DF
Total
48.206
57.
-e
The SS (deViations from regression with 6 d.f. may be spUt up in a way which
helps to test the validity of the assay.
We can isolate one dot. to test whether
the regl'ession equation remains valid also for the zero doses.
Furthermore one
degree of freedom can be assigned for testing the assumption that the two
regression lines 1Dtersect at x ,. O.
The remaining 4 d.t. represent the curvature
of the two regression lines.
The SS (deViations from regression) and the corresponding M. S. may be used
as a composite test for the three assumptions just mentioned.
F
We have
= (0.0463).(0.0194) = 2.387
,. 6, ~ = 9, F •
,. 3.374. This test is of course not sensitive;
l
O 05
it may be that one of the assumptions is invalid, while the two others are not.
which wit.h n
.e
In order to test these different assumptions separately, we proceed as follows
(a)
Test of difference between a and mean for bL""llks
This can be tested in different ways.
yo •
One is to consider the residuals
X
+ bT~' The mean of the residuals for
S s
the blanks should not differ significantly from the mean of the non-blank residuals.
y - Y about the regression Y • a + b
Let us consider a case where we have N values Yi of which No form one group
and N - No a second group.
The totals are, say.. T and To respectively.
The
problem is how to test the difference between the means T IN and (T - T )/(N - N ).
o
0
0
0
We are interested in determining the sum of squares between these two groups..
because testing the difference between the means is equivalent to taJang the ratio
of the sum of squares between the two groups to a suitable error variance.
The
sum of squares between groups may be obtained by calculattng the sum of square of
a certain linear regression, namely that for which x ;. 1 for the N values in the
o
first group and x ,. 0 for the remaining N - N \-alues.
o
58.
-e
x
S Yi
1
T
NOoN values
0
T..T
Sum tor N values
No
N0 values
0
Means
Y.o ..
0
We obtain
T0 /N0
SJet == N0 •
Sxy
b ..
Yi(T..To )/(N..N0 )
y = TIN
0
T
llII
rIB
=N0 (NOoN0 )/N
0
T0 .. N0 T/N = (NT0 -N0 T)/N
sx/sxx .. (NT -N T)/N (NOoN )
000
0
and SS (l'egression) == S;;Sxx is
S2/S
x
Bad we chosen x
xx
- N T)2 /rm (NOoN ) •
.. (NT
0
0
0
0
0 tor the N values of the first group and x
o
of the other values I we would get
==
se
IS xx
~
·e
=1
for those
= [N(T - T0 ) - (N - N0
)TJ/NN
(N .. N )
00
::I
[NT - NT0
- NT0
+ N TJ/NN
(N - N0 )
0
which is the same as that obtained above.
On the other hand, if we calculate the sum of square between the two lroups I
'\ole
get
rI/N
o 0
rI/N
+ (TOoT )2/(N_N ) 0
0
N(N-N )T2 + NN (TOoT )2 _ N (NOoN )T2
o 0
0
0
0
0
= ---------=NR==-r(..
N_...N~)----------o
2
rT - NNr
000
0
+NNT2
0
2NNTT
0
NN (NOoN)
o
= (NTo
.. N0 T) 2 /NN0 (NOoN0 )
again the same as the SS (regression).
0
0
+NNT2
.. NNr +rf-r!
000
0 0
-e
To test the discrepancy of the blanks, we have to work on the residuals
from the regression equation Y = a + bSx + b
S
O ' Accord1ng to standard
multiple regression theory this means that we have to consider the multiple
Xir
regression with three independent variables xs '
x = 1 for the blank and x
o
0
a
and x ' where we choose
o
0 for all other observations. The difference
between the SS for all three variables xs '
xs '
x.r
~I
xo ' and that for the two variables
is the SS measuring the discrepancy between a and YOI the mean of the
blanks.
In our example we calculate
·e
and finally
Soy = 2.9
= 208.2/18
• -8.6667 [= S(x0 i - x0 )(Yi - y)].
Thus the equations tor bol b ' bT (where the two last are different from those
S
calculated previously) are
0.1500 b S - 0.7500 bT - 0.1667 bo
= 2.,400
= 7.6500
-0.7500 bS + 10.0000 bT - 1.0000 bo
-0.1667 bS - 1.0000 bT + 1.7778 b0
and their solutions turn out to be
::I
-8.6667
60.
·e
The sum of squares for regression is
b
S
= (30.1253){2.3400)
SSy + bT ~ + bo Soy
+ (2.9874)(7.6500) + (0.3704)
(8.6667)
with 3 d.f., while with two variables xs'
~
above we "had, with 2 d.f.,
Consequently
SS (discrepancy of blanks)
(b)
III
96.556 • 96.412 = 0.144 (1 d.f.)
Test for intersection and nonlinearity of regressions.
In (a) we have taken Care of that part of the non-linearity which occurs
·e
between the blanks and all other observations.
It remains therefore to be seen
whether there is any non-linearity in that part of the regressioDs which lies
between the oon- zero doses.
At the same time we may also investigate whether
the two regression lines intersect on the y axis.
We first calculate the sums of squares on and around the two indiVidual
regressions, discarding the blanks.
Standard preparation:
SS (between doses) .. (6.7
2 + 9.72 + 12.32 + 15.72 + 18.9'j/2
1oa2 .. 46.296 (4 d.f.)
,. - 6
For the regression S6, we consider the x-values
61.
·e
8y
-2
+2
6.7
9.7
12.3
15.7
18.9
-13.4
- 9.7 -23.1
0
+15.7
+37.8 +53.5
0
63.3
+30.4
-1
0
+1
S
(x)(Sy)
x
= (30.4)2/20 = 924.16/20 ~ 46.208 (1
=46.296 - 46.208 = 0.088 (3 d.f.)
SS (regression)
S8 (non linearity)
d.f.)
Test preparation:
2
.
2
2
2
I
37.9
S8 (between doses) = (9.7 + 12.8 + 15.4 ) 2 - 6
·e
Regression:
Sum
SS (regression)
x
Sy
x(Sy)
-1
0
+1
9.7
12,.8
15.4
- 9.7
0.0
+15.4
0
37.9
+ 5.7
= (5.7)2/4 = 8.122
= 8.143
(2 d.f.)
(1 d.f.)
SS (non linearity) = 8.143 - 8.122 • 0.021 (1 d.f.)
Standard and test preparation
SS (non linearity)
= 0.088
+ 0,021
= 0.109
(4 d.f.)
This completes the analysis of variance because it is now possible to obtain
by subtraction the SS (intersection) with 1 d.f.
62.
·e
Var18nce
SS
DF
Blanks
Nonlinearity
Regression
8
1
4
2
96.690
0,.144
0,.109
96.412
Difference (1ntersection)
1
0.025
Between doses
'!he comprehensive analysis of variance can therefore be set up as follows:
Variance
Regression
Blanks
Intersection
.
-
Nonlinearity
Between doses
Within doses
Total
D.F.
S.S.
M.S.
F
FO• 05
2
1
1
4
8
9
17
96.412
0.144
0.025
0.109
96.690
0.175
96.865
48.206
0.144
0.025
0.027
• ••
• ••
7.423
• ••
5.117
• ••
•• •
• ••
• ••
0.0194
• ••
·..
• ••
• ••
• ••
• ••
• ••
From this analysis of variance it may be inferred that the regressions are linear,
that they intersect at a point on the y-axis I but that the assumption of no
difference between
large, because
yo
yo ...
and a is not correct.
,
1.450 and a = 1.645.
'!he difference is in fact not very
In view of the fact that the blanks lie below the regression lines, it seems
preferable to estimate the potency ratio without using the data from dose zero.
Estimation of potency ratio
l.
We have first to determine SSS' SST'
blanks.
B.rr and
Say' STy' leaving out the
'!his gives
2
Sss • 0.275 • (1.50) /16 ... 0.2750 .. 0.1406 ... 0.1344
·e
STT
=
2
14.500 - (9.0) /16
= 14.5000
- 5.0602 :: 9.4398
= -0.8438
- 0.8438
SSy
= 1."1..• 0150
- (1.50)(101•.2)/16
=
11.0150 -9.4875 = 1.5275
= 59.7000
STy = 59.7000 - (9.0)(101.2)/16
- 56.9250
= 2.7750
and therefore
0.1344 bS - 0.8438 bT = 1.5275
-0.8438 bS + 9.4398 bT
= 2.7750
•
The solutions are
identical with those obtained With x
This should be so
becal~se
s"
~
and x •
o
the simple regression between x
o
and y takes out
aU the differences between the two groups of blanks and non-blanks.
consider the regression of the residuals with X and
s
same as the regression of the non-blanks With X and
s
XT this
XT.
If we
can only give the
The estimate R of p is
or
496
b'1'
R = b = (2.9874)/(30.1253) = 0.0992 I-1g/ml
S
I-1g nicotinic acid per g for the potency of the meat extract.
To find the fiducial limits of R it is necessary to inve t the matriX
!
e
I
Sss
SST
Sao
SST
STT
STo
SSo
STo
Soo
'\
!
I
64.
and this gives in our case
(
16.9663 1.51685 2.44382 \\
(C) =, 1.5185 0.•241573 0.278090 ,)
\ 2.44382 0.278090 0.948034 ,:
I
Usually, (C) is computed from (S) and the b ' b , b (if necessary) obtained
S T o
as
bS
= cSS
SSy + cST STy + csp SOy
= (16.9663)(2.3400)
+
(1.51685)(7.6500) + (2.44382)(-8.6667)
=30.1253
• (1.51685)(2.3400) + (0.241573)(7.6500) + (0.278090)(-8.6667)
=2.9874
= (2.44382)(2.3400) + (0.278090)(7.6500) + (0.948034) (-8.667)
= -0.3704
Furthermore, the variances ot b
V(b S)
where we may use s
2
= cSSs,
2
•
V(bT)
S
and b
= cTTs2
1
T
and their covariance are
COV(b S' bT)
=cSTS 2
0.0194 with 9 d.t. although this includes 1 d.f. trom
the blanks.
Fiducial limits ot R.
To find the fiducial limits R
o
that
e
and consequently
and
~
we apply Fieller' s theorem.
We note
·e
For any value of I' we have
E(bSP - bT )
and
= 13SP -
13T
= o.
V(bSf) - bT ) = JV(b ) - 21' COv(b SI bT ) + V(bT ).
S
b S and bT being normally distributed l it follows that this holds also for
bSP - bT and therefore we have
2
with n equal to the degree of freedom of s.
Inserting t • t pl we find
·e
or
dividing by b~ this gives
=0
If we introduce
and remembering that
we get
•
66.
2
cST
2
C
TT
P (l-g) - 2p (R - --c SS g) + R - --c SS g • 0
or
2
cST
2
c TT
p - 2p(R---g)/(l-g) + (R - --g)/(l-g)
c
c
SS
SS
=0
and therefore
(FL,Fl_)
-L -U
= (R
CST
cST
- --g)/(l-g) -+ f(R - --g)
C
c
SS
SS
2
- (R
2
c TT
- --g)(l-g)}
c
SS
1/2
/(l-g)
The expression for g may also be written as
t
g =
which shows that t
2
2
S
bs/s
2
2
is divided by the variance ratio which is used to test
p
the significance of bS.
If b differs from zero in such a way that the
S
difference is highly significant, then g should be small. In that event we
may use the much simpler formula
67.
This is the usual approXimate formula for a ratio (bi'b )' namely
S
For our example the fiducial limits will be calculated both by the general
8I1cl by the simpler formula.
For g we obtain
p =
t
g
0.05, n = 9
o•05 = 2.262
= 0.0019
and the general formula yields
Ru
(R = 0.09915)
= 0.1026
•
Multiplying by 5,000 gives
~ = 478 ~g,
Ru = 513
J.Lg
(R =
496
J.Lg) •
The simpler formula would give the same results.
If we had taken the whole of the data., including the blanks, for
combining R and its fiducial limits, the results would have been
R
1:11
498 and
I\, = 482, Iu
= 514,
a result only slightly different from that obtained Without blanks.
68•
.e
3.2.2
Symmetrical five-point assay.
The simplest slope-ratio assay would be a three-point assay with blanks
and one dose for standard preparation and one for the test preparation.
This
design is unlikely to be of use since the assumptions for validity cannot be
tested.
In a symmetrical five-point assay the experiment units are evenly dis-
tributed over blanks, two doses for standard and two doses for the test preparation.
J'urthermore the doses should be equally spaced for the test
preparation and the standard j but the doses need not be the same for both
preparations.
Some of the calculations are simplified if a s,rm:netrical design is
·e
adopted.
An example will make the procedures clear.
Example.
Assay of the riboflavin content of malt.
Wood, E. C. The theory of
certain analytical procedures, with particular reference to microbiological assays.
-
Analyst, 71, 1-14.
milliliters 0.1 N NaOH.
Z
x
xt
Zt
s
s
(g)
(~g)
x
y
0
Responses (y) Utter in
Total
4.50
4.90
7.80
4.10
8.35
20.00
32.30
17.00
=B
=8
= 821
= Tl
6.10
6.10
24.45
I:
Blanks
0
0
0
Standard
0.10 0
1
(RiboflaVin) 0.20 0
2
Test prepara- 0 0.025 0
tion
(Malt)
0 0.050 0
0
0
0
1
1
0
0
0
1.• 90
4.85
8.35
4.00
2.25
5.00
8.20
4.40
2.00
2.20
5.25
2
0
6.05
6.20
Total
12
0
1.20 0.300 12
7..95
102.10
The first step in the analysis is to make a graph showing the results
of the individual observations and to adjust two straight lines.
T
2
·e
Resources
y
8
6
4
o
.---_.-+--_.--..,..--
o
, I
2
1
b
s
-.3
X(doses)
bT -
This has to be referred back to the units of z.
We have
a.nd therefore
from which we get
R -
4 (~) - 2.1 ~g/g •
Analysis of variance
The total sum of squares and that wi thin and between
a.s usual.
doS8S
are obtained
70 •
.e
SS(total)
= 79.275
19 d.f.
SS(between doses)
= 78.733
4 d.f.
SS(within doses)
=
15 d.f.
0.542
For the regression coefficients we have to calculate
With
N = 20,
we obtain
2
SSS = 20 - (12) /20 = 20 • 7.2 = 12.8
2
STT=20- (12) /20
S
ST
=
2
0- (12) /20
= 20
=
- 7.2 = 12.8
0 - 7.2
= -7.2
SSy = 84.60 -
(12)(~~2.l0)
STy = 65'.90 -
(12)(~~.!91 = 65'.90
= 84'.60 - 61.26 = +23.34
- 61.26 = + 4.64
12.8 bS - 7.~ b T =23.34
-7.2 b S + 12.8 bT = 4.64
From this we obtain
SS(regression) • (23.34)(2.9657) + (4.64)(2.0307)
= 78.642
•
The regression coefficients may also be obtained directly from the totals
l and T2 • Assuming that there are r experimental units per dose 1
so that the total number of values y is N = 5r, we get
B, Sl' S2" T
SSS = l6r/5,
and
STT
= l6r/5,
SST = -9r/5
71•
.e
Ssy = (-3B + 281 + 782 - 3T1 - 3T2)/5
STy
= (-3B
- 3S1 - 3S2 + 2T1 + 7T2 )/5 •
The solutions of the equations
may be written as
b S • (ST'!' SSy - SST STy)/(SSS ST'!' -
1ST )
b T .. (Sas STy - SST SSy)/(SSS ST'!' -
a2ST )
-e
b S = (-15 B + Sl + 17 S2 - 6T1 + 3T2 )/35r
b
For our example
T
'We
=
(-15
1 + 3 S2 + T1 + 17
B - 6S
T2 )/35r •
have to substitute
which gives
as obtained previously.
Sum of Squares for "blanks'
e
The SS for blanks may also be expressed as a function of B, Sl' S2' Tl and
T •
2
Accord1ut! . to the general formulas in 3.2.1 we have to consider the multi-
72.
·e
ple regression of y a.gainst the three indePendent variables xs'
~
and x.
The
sum of squares for "blanks" is then
In order to obta.in the explicit formula" we have first to calculate the
matriX (S) and its inverse (C).
We have already found
Soo = 4r/5, Sso
.e
= -3r/;"
6Ta • -3r/5 •
The matriX (6) is thus
( 16r/5
(S) =
-9r/5
. -3r/5
\
-Cft/5
l6r/5
-3r/5
-}r/5 ~
-3r/5
4r/5.
and its inverse (C):
·ll/lOr
(C)
=(
I
9/l0r
9/10r ll/lOr
15/10r 15/l0r
We have obtained Say and
l5/10r !
35/l0r /
I
s.r.y previously as
=:I
(-3B + 281 + 782 - 3Tl - 3T2)/5
STy =
(-3B - 361 - 382 + 2T1 + 7T2 )/5
8Sy
and we find for 8
oy
l5/lor \
.e
According to standard multiple regression theory we obtain b
o
b o ==
and as c
00
:II
Ceo
as
Ssy + cT0 S-Ty + c00 Soy
35/10r, we have
or
The espre8sion in brackets may be written as
from Which it is seen that this SS tests in fact the nonlinearity of the
regression lines for the standard and the test preparation jointly.
Sum of squares for intersection
The test for intersection can be obtained directly 1n the general assay
as well as in a symmetrical five·point assay, where its expression is very
simple.
The sum of squares for intersection is given by1be difference in the
sum of squares tor the seParate regressions tor standard and test preparation
and that tor the joint regression, disregarding blanks.
For the standard preparation we have, if there are r replicates tor each
dose:
X
SYi
SXSiY1
1
8
2
S
2.
~
2S
2
Sl+S2
Sl-teS2
s
N ... 2r
Sum
3r
I
SSS ... 5r • (3r)2/2r == r/2
SSy == Sl+eS2 • (3 r )(Sl+S2)/2r
Say a ( ... St+82)/2
74•
•~
Hence:
For the test preparation, we get
Consider now the joint regression (multiple regression) for
X
s
and XTr'
disregarding blanks.
s
x.r
Sy!
SX y
s
SXTry
1
0
Sl
Sl
0
2
0
S2
2S2
0
0
1
0
T
0
Tl
T2
2T
2
3r Sl+S2+Tl iT2 Sl+2S2 Tl +2T2
X
•
4~
Sum
2
3r
SST =
5r - (3r)2/4r
2
-(3r) /4r
SSY
= llr/4
= -9r/4
•
SST b S + STT b T = STy
we get
2
= (Sl+5S2-3Tl-3T2)/4
STy = (-381 - 3S2+Tl f:5 T2 )/4
The equations for b S and bT are
and if
N = 4r
0
SSS = 5r - (3r)24r = llr/4
STT =
1
SSSSTT - 8ST = 10 r
2
/4
75 •
.e
and
The sum of squares for the mUltiple regression is thus:
2
SS(multiple regresa1cm) ... (S1+52+Tl+T2) /4r
+ (51+552-3Tl-3T2)(-451+752-6Tl+3T2)/40r
+ (-351-352 -+lrl +5T2 ) (-6S1+352..4T1+7T2 )/40r
with 3 d.t.
The sum of squares tor intersection is therefore given by
222
2
SS ( intersection) ... [ 20(5 +6 ) +eO(-Sl+S2) +20(T +T ) +eO(-T1+T2 )
1 2
1 2
-10(5 +52 +T -+1r ) 2
1
1 2
-(51+5S2-3T1-3T2) (-4S1+7S2-6T1+3T2)
-(-3S1-3S2-T1+5T2) (..661+3S2..4T1+7T2»)40r
... [16Si +
4S~ + 16~ + 4~ - 166162 .. 16T1T2
..32S1T1 + 1661T2 + 1662T1 - aS2T2 J/40r
=
[4S~+S~+4~~.. 4S1S2..4T1T2
..a61T +46 T +4S T -26 T )10r
1 1 2 2 l
2 2
16•
.e
SS(1ntersect1on)
= [(2S1-S2)
- (+2T1-T2)]2/10r
•
= [2(Sl-T1)
- (S2- T2)]2/10r
From the last expression it is seen that the sum of squares for intersection
measures the degree to which the differences in the response deviate from the
proportion of 1:2.
The coefficients and diVisors for the calculation of b S and bT (from the
complete data) and for the sum of squares for blanks and for 1nt9irsection are
given in thefo1lowing table
Contrast
bS
·e
bT
Blanks
Intersection
+11
T
1
-6
T2
+3
-6
+3
+1
+11
35r
-2
+1
-1
..2
+1
+1
14r
lOr
B
Sl
S2
-15
+1
-15
2
0
~
-2
Divisor
35r
The contrasts for bS is not orthogonal to that for b • The contrast for
T
blanks is orthogonal to that for intersection and both are orthogonal to those
for bS and b •
T
In our example we obtain fur the SS ror blanks and for intersection:
SS(blanks)
= [(12)(8.35)
- (2)(20.00) + (32.30) - (2)(11.00) + (24.45)/(14)(4)
= (0.55)2/56 = 0.005
SS(intersection)
= [(2)(20.00)
2
= (1.85) /40
- (32.30) - (2)(11.00) + (24.45)]/(10)(4)
= 0.086
On the other hand the sum of squares for regression is, as already stated,
77.
-e
SS(regression)
= (2.9657)(23.34)
+
(2.0307)(4.64)
=
78.642
The con:rplete analysis of variance is thus:
--variance
S.S.
Regressj.on
2
78.6h2
Blanks
Interi3ection
1
Between doses
Wi. thin doses
4
0.• 005
0.,)86
78.733
--.....;,0..:5:±?
79.2:5
1
19
Total
39.• 321
0.005
0.086
...
--
,'" .. _._-
~
!YLS.
... .
.;...;..-
2
s
::I
Q.&361
•••
F
108.922
0.1
2.3f3
·...
• ••
·..--
E6t~te......2.:t£~d fidu~U~_~'!
.e
The estimate of p is (bT/b ) provided there is no change in units from
S
Zs
s and Xrr'
and zT to x
If there is a change) that is if
Then
In our example we have
and therefore
and
R = (4)(0.6847)
=
2.739 •
18.
-e
The test preparation of malt contains 2.139 'tJ.g/g units of the riboflavin standard.
The estima.te 2.1 obtained by the graphical analysis compares favorably
with the exact va.lue.
The fiducial limits of (b~bS) are reasily obtained from the general
expression in 3.2.1, assuming again that the number of replicates is r and that
Xs and
Xrr
are 1 at the lower, equal to 2 at the higher level.
It is necessary
to consider the matrix
(S)
l6r/5
-9r/5
-9r.5
l6r/5
=(
)
and its inverse
l6/35r
9/35r
9/35r
l6:35r
(C) = (
·e
)
and inserting in
where css is the element in the matriX
CSS
(C)
=(
CST
)
CST
CTT
obtained fromthe regression of y on Xs and
Xrr'
(Note that in the example of
XT
3.2.1 the estimate b~bS was obtained from the regression on xs '
and xo '
because of the significance of the MS for blanks). With css = l6/35r we get
and as fiducial limits for the ratio (bT/b s )
79.
If g is mnall, as is very often the case in microbiological assays, it
will be sufficient to use the variance V(bT/b S) of the ratio b~bS
and the fiducial limits will then be given CS
We may therefore use the formula for the variance of (b~bS)' for which we
find
·e
V(bT/b )
S
= (2)(0.0361)2{8
V(bT/b S)
and
.(9)(0.6847) +(8)(0.6847)2J/(35)(4)(2.9657)2
= 0.00032766
!
.
[V(b~S)]2 • 0.018 106
which is the standard error for (b~bS). The standard error for R is
1
[V(R»)~
1
= ~ [V(bT/b S)]~
T
r-
so that the standard error of R is, in our example,
(4)(0.018 106)
and the fiducial limits of R
degrees of freedom)
= 2.739
= 0.072424
are found as (to•05
=2.131 with 15
so.
A symmetrical five-point assay could also be arranged in a randomized
Each block would ecotain 5 experimental units assigned at
blocks design.
random to the five doses.
The interaction mean square between blocks and doses
will be used as s2 in computing fiducial limits.
3.2.3 Sensitivity of slope ratio assays.
From the general formula for the fiducial limits in
3.~.1,
we see that the
quarter-square of the fiducial interval for R is
2
g(c
TT
cClfl
- ~a)]
cSS
where
g
= t .p2
2
s 2 c S6/b S·
R depends on the relative potency of the two preparations; it cannot be chosen
arbitrarily.
If we want to shorten the fiducial interval,
w.
have to see to it
that
(1) t p is small;
(2)
s
C.~)
b
is small;
S
is large;
(4) g is small;
(5)
(1)
[~TT -
2RCST +c ssR2
-
g(c TT -
c~T/cSS)] is small.
As in the case of a parallel line assay we find that at least twenty experi-
mental units are necessary to ensure that the degree of freedam for error is not
too small.
(2)
Care in the selection of assay conditions and in the execution of the assay
will help to keep s
2
small.
In some cases a randomized blocks design may help to
81.
_e
achieve the same goal.
b as measured on the original units zs' will be greatest, if the highest
S
dose of the standard is chosen to have the largest value consistent with linearity
(3)
of the regression line.
(4)
If the conditions under (1), (2) and (3) are satisfied, g will also be small,
proVided CSS is small.
(5) This eJ/iPl:ession would be small, if the doses were all chosen near zero and
near the highest value of
Zs
and zT for which the regressions remain linear.
Such
a choice would, however, prevent us from making tests on the assumptions for
validity.
As a reasonable compromise" the five-point assay has a relatively high
sensitiVity, but permits also to test the assumptions for validity of the assay.
As in the case of parallel line assays, we may also be interested to investi-
.e
gate the effect of the choice of experimental subjects, assay techniques and measurements on the fiducial interval.
If g is small, as it should be in a good assay,
the standard error of R is proportional to sibS.
in terms of
zs'
There again b must be expressed
S
2 2
the original units for the standard. The comparison of S /b S for
two different procedures, enables the experimenter to compare the relative efficiencies of the two procedures in terms of the number of experimental units needed
to attain the same precision.
3.3 Multiple assays.
Sometimes the experimenter wishes to assay several test preparation against
the same standard.
chapters.
This could be done in the manner indicated in the previous
But it is more efficient to incorporate these comparisons in one assay.
For this case we will speak of a multiple assay.
-_
3.3.1 Multiple parallel line assays.
The procedures discussed in 3.1 may easily be extended for mUltiple assays.
As an example we consider a part of data given by Bliss (1952) p. 518, where two
preparations of vitamin D oil are assayed against the U.S.P. Reference Cod Liver
Oil No.2.
in rats.
The responses yare healing scores for treatment by the line tests
The test preparations were assumed to have potencies of one unit in
1.1546 x 10-6 g of A and one unit in 0.8060 x 10-6 g of B.
The rats were
selected from litters, and 6 animals fram the same litter were assigned at ran40m
.-
S2
Doses in Units Z
81
4
X
..1
+1
Responses, y
Total
Difference
Al
4
8
A2
8
Bl
4
B2
8
+1
..1
+1
..1
Litter
e
1
4
2
6
3
4
4
4
5
0
6
10
8
10
10
10
8
2
4
26
56
TotaJ.
6
8
6
4
10
8
8
10
10
8
2
4
3
6
3
0
10
8
10
10
8
10
42
36
43
48
41
30
30
54
18
56
82
84
74
240
240
+30
+24
+38
92
The first step would as usual be the graphical analysis. We omit this.
For the analysis of variance we proceed as usual, excluding the differences
between ll'tters from the experimenta.l error.
The five degrees of freedan between
doses can be split up into 1 degree of freedom for the combined slope, 2 d.f.
between preparations and 2 d.f. for non-parallelism.
e
This gives a complete
analysis of va.riance for testing some of the validity assumptions.
83
.e
Variance
s.s.
M.S.
235.111
4.667
8.222
235.111
2.333
4.111
D.F.
1
Combined slope
Preparations
Non-parallelism
2
Between doses
Between litters
Error
5
5
25
248.000
35
342.000
Total
2
32.333
61.667
...
,
6.467
2. 467=-82
•••
The combined slope is very highly significant, while there is no sign of signit10ance between preparations and for non-parallism.
Tb estimate the potency-ratio we compute the mean regression coefficient
for all three preparations as
.e
Furthermore, we have
and according to the general formula in 3.1.1,
82 - 84
12
and
Also, we get for
i
=- ~(0.3010) = 0.1505
Os = ~(lOg
4 + log 8)
.
= 0.7526
36
."92
84
0A
= ~(lOg
4 + log 8)
=0.7526
0B
1
=2(log
4 + log 8)
= 0.7526
thus
log RA := (0.1505) (iO.0652)
log ~
= iO.009
13
= (0.1505)(~0.2609) = -0.039 26
•
Finally
R = 1.021
A
~
= 0.9136
The fiducial limits may be calculated according to the formula given
in 3.1.1, where s2
= 2.467.
This 1s left to the student as an exercise.
3.3.2 Multiple slope ratio assays.
There again, as in 3.3.1, we will only consider one particular case.
There will be no difficulty to deal with other multiple slope ratio assays,
once the general procedure is understood.
The case which we present is an
extension of the symmetrical five point assay.
The standard and three test
preparation (A, B, C) are compared, using two doses in the ratio of 1:2.
The
example is taken from C. I. Bliss (1952), p. 569, but using only the data from
two, instead of four doses.
It 1s interesting to note that our three estimates
lie within the fiducial limits given
by
Bliss from the whole data.
The three
test preparations A, B, C, were extracts of tomato and gourd tissues.
standard contained 0.04 and 0.08
~g
The
of calcium pantothenate, and the tubes
of the unknowns contained 0.10 and 0.20 ml of plant extract.
y were measured as titer to pH 6.8 with pH meter
The
res~nses
85
,e
\
Zc
Zs ZA
J.1g m1 m1 m1
Blanks
Standard
... .
X
A
0
1
2
0
0
0
0
0
0
6
0
0
0
• •• •••
• ,e •
0.4 • ••
0.8 • ••
Test pre- A.• ,•.• 0.• 1
parations
,. ,.' 0.2
B...... • ••
X
s
• •• • ••
••• • ••
...
• ••
,
• •• • ••
0.1 • ••
,. ,. ,e. 0.2
0.1
c.•." • ••
•• • • •• • •• 0.2
.,-
...
·..
,
,
Total
\
Xc
X
0
0
1
0
0
0
0
0
0
0
0
2
0
0
0
0
0
1
2
0
0
6
1
2
0
0
0
0
6
0
0
0
0
0
1
2
6
y
o
1.3
3.6
6.1
4.1
6.4
3.2
4.7
3.2
4.6
Sy
1.7 3.0 ::I B
3.6 7.2 = Sl
6.3 12.4 S2
3.7 7.8 ::I A1
6.7 13..1 A2
3.2 6.4 ::I Bl
4.9 9.6 = B2
3.0 6.2 = C1
5.1 9.7 = c2
75.4
::I
::I
The first step in the analysis would again be graphical.
We leave this
to the student.
The analysis of variance will serve to test the assumptions of validity,
and to calculate the assay error.
This analysis may be verformed according
to the general rules, or we may 81so establish the explicit expressions 1 as we
did in the case of the symmetrical five-point assay.
In any case we obtain
SS(tota1)
= 40.898;
SS(doses)
::I
40.508; SS(error)
= 0.390
•
Using the general approach developed in 3.2.1, we have
axs
= SXA
SSS
= SAA = ~ = Scc
SSA = SBB
::I
:=
SXa = SXc
Sse
::I
Ssy = +6.867; SAy
::I
6
= 8
~C
SAB = SAC
::I
= +8.867;
SBy
and the equations for the regression coefficients are
::I
-2
= +0.467;
SCy
= +0.467
86
.e
8 bS - 2bA - 2'bB - 2b e = 6.867
-2 b S + BbA - 2bB - 2b e = 8.867
-2 bS - 2bA + BbB - 2be = 0.467
-2 bS - 2bA - 2bB + Bb e = 0.467 •
From these equations we get
The sum of squares for regression is then
SS(regression)
= (2.3533)(6.867)
+ (2.5533)(9.867) + (1.7133)(0.467)
+ (1.7133)(0.467)
= 40.398
,e
(4 d.t.)
According to the general method we obtain the SS for blanks
between the 5S for regression of
regression of
y
on XS'
A'
X
Xe'
y
on
X ' X ,
A
S
just given.
Xc
Xe'
Xc
and
XO'
al3
the difference
'and the 5S for
We note that
From the equations
-0.667b0 -2
0.667bA - 0.667bB - 0.667bC = -5.378
bS - 2
bA - 2
b - 2
be = +6.867
B
bA - 2
bE - 2
be = +8.867
bS + 8
-0.667b0 -2
bS - 2
-0.667bO -2
bS - 2
1.778bo -0.667b s
-0.667b O +8
-
bA + 8
bA - 2
bB - 2
bB + 8
be
= +0.467
be
= +0.467
we get
and
SS(regression)
= (2.42)(6.867)
+ (2.62)(8.867) + (2)(1.78)(0.467)
- (0.20) (5.378)
= 40.437 (5 d.f.)
87
Comparing the two sums of squares for regression, the SS(blanks) is
SS(blanks)
= 40.437 - 40.398 = 0.039
(1 d.f.)
This value is affected by rounding errors, we will see that a more exact
value is SS(blanks)
= 0.036.
The general method for computing the sum of squares for intersection consists in calculating the multiple regression of y on X X , ~1 Xc and the
S' A
four individual regressions of y on X y on XA, y on ~ and,. on XC' omitting
S'
the blanks in all cases.
Each individual regression gives a SS with 2 degrees
of freedom, so that all four account for 8 degrees of freedom,
The multiple
regression on ihe other hand accounts for 5 degrees of freedom.
The difference
gives the sum of squares for intersection With 3 degrees of freedom.
It is
possible to give a sum of squares for each of the three degrees of freedom,
as will see later.
From the data from which the blanks are omitted, we get
SXS
=SXA ::
SX
SSS :: SAA :: ~B
= SXc
:: 6 , Sy
= 72.4
= SCC = 7.75
SSA = SSB :: Sse = SAB = SAC:: SBC = -2.25
SSy :: +4.85; SAy
= +6.85;
~y
= -1.55;
SCY
= -1.55
•
The equations for the regression coefficients are set up as usual and yield
the solutions
the same values as obtained previously when X was used as a supplementary
o
variable. The sum of squares accounted for by the multiple regression (taking
also into account the constant) is
SS(regression)
= (2.42)(4.85)
= 351.776
+ (2.62)(6.85)
(5 d.f.)
= (2)(1.78)(1.55)
+ (72.4)2/16
88
The four separate simple regressions, give the following SS for regression,
again tak1ng into account the constants:
Regression
Sum of Beres
Standard
[(19.6)2 + (5.2)2]/4
Test preparation A
[(20.9)2 + (5.3)2]/4
= 102.800
= 116.225
B
[(16.0)2 + (3.2)2]/4 = 66.560
c
[(15.9)2 + (3.5)2]/4
=
66.265
---;;;",;,;;~-------
8ella.rate simple regressions, total
= 351.850
8
Multiple regression
= 351. 776
5
Intersection
=
3
0.074
The analysis of variance can be set up as follows:
~e
Variance
D.F.
Regression
Blanks
Intersection
Between doses
Within doses (error)
Total
4
1
3
8
9
17
S.S.
M.S.
40.398
10,.099
0.036
0.025
0.036
0.074
40.508
0.390
40.898
•••
0.0433
•••
The mean squares tor blanks and for intersection are smaller than the
error mean square, while the regression mean square is highly significant.
The assay is thus a valid one.
If
the calculations given just now are carried out algebraically, using
the symbols B, Sl' S2' Al , A2 , Bl , B2 , Cl , C2 as shown in the table of the
results, and if again r is the number of replications, assumed to be the
same tor the blanks and for all doses, the following results are obtained:
89
~e
B
Sl
S2
A
1
A
2
B
1
B
2
C
1
C
2
Divisor
-5
+1
+7
-2
+1
-2
+1
-2
+1
15r
-5
-2
+1
+1
+7
-2
+1
-2
+1
15r
-5
-2
+1
-2
+1
+1
-2
+1
15r
-5
-2
+1
-2
+1
-2
+7
+1
+1
+7
.15r
+4
-2
+1
-2
+1
-2
+1
-2
+1
:36r
Intersection 1
0
-2
+1
-2
+1
+2
-1
+2
-1
20r
2
0
-2
+1
+e
-1
+2
-1
-2
+1
20r
:3
0
-2
+1
+2
-1
-2
+1
+2
-1
20r
Contrast
Regresaion Coetficients
bs
b
A
b
B
b
C
Sum of ppres
Blanks
1.be contnsts for the four regression coefficients are not mutually orthogonal.
The contrasts for blanks and for interseettoo. are mutually orthogonal,
and also orthogonal to the contrasts for the regression coefficients.
The SS for blanks can be put into the form
SS(blanks) = [(B-2S +S ) + (B-2Al+~) + (B-2B +B2 ) + (B-2Cl+C2)]2/l5r,
l 2
l
which shows that it measures the deviations from l1nearily jointly for the four
preparations.
The SS for intersections may be spl1t up into three orthogonal components.
These correspond to three orthogonal contrasts between the four expressions
The three orthogonal contrasts are
Contrast
...
1
2S -6
l 2
2A -A
l 2
2B -B
l 2
2C -C
l 2
2
:3
+
+
+
+
+
+
90
.e
Inserting the values from our example, we have
SS(b1anks)
= [(4)(3.0)-(2)(7.2)+(12.4)-(2)(7.8)+(13.1)-(2)(6.4)+(9.6)-(2)(6.2)
+(9.7)]2/(35 )(2)
= (1.6) 2/72 = 2.56/72 = 0.036
and similarly for the intersection SS:
1 (1.4)2/40
= 0.D49
2 (0.0)2/40 = 0.000
3
(1.0)2/40
= 0.025
Total
0.074
Using the explicit formulae, the regression coefficients and the sums of
It
squares for blanks and for intersection are easily obtained.
To estimate the relative potency for the substances A, B and C, we first
calculate
= (2.5533)/(2.3533) = 1.085
bB/b S' = bC/b S = (1.7133)/(2.3533) = 0.730
bA/b S
and in view of
we have
so that
RA = (.04)(1.085)
= 0.4340; BB = RC = (0.4)(0.730) = 0.2920.
The fiducial limits of bibs can be shown to be
91
where
The standard error of R is obtained as the product of the standard error of
d. 3 •3
Number of replications in the standard preparation.
In
3.3.1 and 3.3.2 we gave examples of assays with two and three test
preparations.
In both cases we had the same number of replicates in the
standard and in each of the test preparations.
For the parallel line assay it has been shown in 3.1.5 that it is best
to choose the same number of replicates in the standard as in the test prepara-
•
tion, if there is only one test preparation.
If, however, these are several test preparation, the best policy is no
more to have the same number of replication in the standard as in the test
preparation.
In that case I if C is the number of test preparation S
is the best choice.
If C
=4 this would me6n that the number of replicates
r s for each dose of the standard should be double the number of replicates r T
in each dose of each of the 4 test preparations.
In the case of a multiple slope ratio assay, the sa.::ne policy should be
adopted as that just mentioned for parallel line assays, though the theoretical
basis for this recommendation seems less obvious.
REFERENCES
[1]
Bliss, C. I. (1952) The Statistics of Bioassay, With special reference
to the Vitamins. New York, Academis Press, Inc.
[2] Burn, J.' H., Finney, D. T. and Goodwin, L. G. (1950).
zation. London University Press, 2nd edt
[3] Emmens, C.
w.
(1948).
Principles of Biological Assay.
Biolo§ical StandardiLondon, Chapman
and Hall.
[4]
Finney, D. T. (1952). Statistical Method in Biological Assa~.
Charles Griffin (New York, 1IB1'ner Publishing Co.).
London,