What you describes about Tukey's nonadditivity test sounds good to me. In effect, it allows to test for an item by rater interaction. Some words of caution, though:
- Tukey's nonadditivity test effectively allows to test for a linear-by-linear product of two factor main effects.
- The possibility of deriving a total score is irrelevant here, as this particular Tukey's test can be applied in any randomized block design, as described on Stata FAQ, for example.
- It applies in situation where you have a single observation per cell, that is each rater assess only one item (no replicates).
You might recall that the interaction term is confounded with the error term when there're no replicates in an ANOVA design; in inter-rater studies, it means we have only one rating for each rater x item cell. Tukey's test in this case provide a 1-DF test for assessing any deviation from additivity, which is a common assumption to interpret a main effect in two-factor models. Here is a tutorial describing how it works.
I must admit I never used it when computing ICC, and I spent some times trying to reproduce Dave Garson's results with R. This led me to the following two papers that showed that Tukey's nonadditivity test might not be the "best" test to use as it will fail to recover a true interaction effect (e.g., where some raters exhibit an opposite rating behavior compared to the rest of the raters) when there's no main effect of the target of the ratings (e.g., marks given to items):
- Lahey, M.A., Downey, R.G., and Saal, F.E. (1983). Intraclass Correlations: There's More There Than Meets the Eye. Psychological Bulletin, 93(3), 586-595.
- Johnson, D.E. and Graybill, F.A. (1972). An analysis of a two-way model with interaction and no replication. Journal of the American Statistical Association, 67, 862-868.
- Hegemann, V. and Johnson, D.E. (1976). The power of two tests for nonadditivity. Journal of the American Statistical Association, 71(356), 945-948.
(I'm very sorry but I couldn't find ungated PDF version of those papers. The first one is really a must-read one.)
About your particular design, you considered raters as fixed effects (hence the use of Shrout and Fleiss's type 3 ICC, i.e. mixed model approach). In this case,
Lahey et al. (1) stated that you face a situation of nonorthogonal interaction components (i.e., the interaction is not independent of other effect) and a biased estimate of the rating effect -- but, this for the case where you have a single observation per cell (ICC(3,1)). With multiple ratings per items, estimating ICC(3,k) requires the "assumption of nonsignificance of the interaction. In this case, the ANOVA effects are neither theoretically nor mathematically independent, and without adequate justification, the assumption of no interaction is very tenuous."
In other words, such an interaction test aims at offering you diagnostic information.
My opinion is that you can go on with you ICC, but be sure to check that (a) there's a significant effect for the target of ratings (otherwise, it would mean the reliability of measurements is low), (b) no rater systematically deviates from others' ratings (this can be done graphically, or based on the residuals of your ANOVA model).
More technical details are given below.
The alternative test that is proposed is called the characteristic root test of the interaction (2,3). Consider a multiplicative interaction model of the form (here, as an effect model, that is we use parameters that summarize deviations from the grand mean):
$$\mu_{ij}=\mu + \tau_i + \beta_j + \lambda\alpha_i\gamma_j + \varepsilon_{ij}$$
with $\tau$ ($i=1,\dots,t$) the effect due to targets/items, $\beta$ ($j=1,\dots,b$) the effect of raters, $\alpha\gamma$ the interaction targets x raters, and the usual assumptions for the distribution of errors and parameters constraints. We can compute the largest characteristic root of $Z'Z$ or $ZZ'$, where $Z=z_{ij}=y_{ij}-y_{i\cdot}-y_{\cdot j}+y_{\cdot\cdot}$ is the $t \times b$ matrix of residuals from an additive model.
The test then relies on the idea of using $\lambda_1/\text{RSS}$ as a test statistic ($H_0:\, \lambda=0$) where $\lambda_1$ is the largest nonzero characteristic root of $ZZ'$ (or $Z'Z$), and RSS equals the residual sum of squares from an additive model (2).
I assume that A through D are different symptoms, say, and 1 and 2 are the two raters. As you tagged this in Stata, I will build a Stata example. Let us first simulate some data: we have a bunch of subjects with two uncorrelated traits, and a battery of questions, tapping upon these traits. The two raters have different sensitivities to each of the traits: the first rater is a tad more likely than the second rater to give a positive answer on question A, but slightly less likely to give a positive answer on question B, etc.
clear
set seed 10101
set obs 200
* generate orthogonal individual traits
generate trait1 = rnormal()
generate trait2 = rnormal()
* raters' interecepts for individual questions
local q1list 0.3 0.7 -0.2 -0.4
local q2list 0.5 0.5 0 -0.5
* prefixes
local letters a b c d
forvalues k = 1/4 {
local thisletter : word `k' of `letters'
local rater1 : word `k' of `q1list'
local rater2 : word `k' of `q2list'
generate byte `thisletter'1 = ( `k'/3*trait1 + (3-`k')/5*trait2 + 0.3*rnormal() > `rater1' )
generate byte `thisletter'2 = ( `k'/3*trait1 + (3-`k')/5*trait2 + 0.3*rnormal() > `rater2' )
}
This should produce something like
. list a1-d2 in 1/5, noobs
+---------------------------------------+
| a1 a2 b1 b2 c1 c2 d1 d2 |
|---------------------------------------|
| 1 1 0 0 1 0 1 1 |
| 0 0 0 0 0 1 1 1 |
| 0 0 0 0 0 0 0 0 |
| 1 0 0 1 1 1 1 1 |
| 0 0 0 1 1 1 1 1 |
+---------------------------------------+
which I hope resembles your data, at least in terms of the existing variables.
A fully non-parametric summary of the inter-rater agreement can be constructed by converting the binary representation into a decimal representation. The outcome a1=0, b1=0, c1=0, c4=0 is 0000b=0; the outcome in the first observation is 1011b = 11, etc. Let us produce this encoding:
generate int pattern1 = 0
generate int pattern2 = 0
forvalues k = 1/4 {
local thisletter : word `k' of `letters'
replace pattern1 = pattern1 + `thisletter'1 * 2^(4-`k')
replace pattern2 = pattern2 + `thisletter'2 * 2^(4-`k')
tab pattern*
}
This should produce something like
. list a1- d2 pat* in 1/5, noobs
+-------------------------------------------------------------+
| a1 a2 b1 b2 c1 c2 d1 d2 pattern1 pattern2 |
|-------------------------------------------------------------|
| 1 1 0 0 1 0 1 1 11 9 |
| 0 0 0 0 0 1 1 1 1 3 |
| 0 0 0 0 0 0 0 0 0 0 |
| 1 0 0 1 1 1 1 1 11 7 |
| 0 0 0 1 1 1 1 1 3 7 |
+-------------------------------------------------------------+
Now, these patterns are perfectly comparable using kap:
. kap pattern1 pattern2
Agreement Exp.Agrmt Kappa Std. Err. Z Prob>Z
-----------------------------------------------------------------
54.00% 17.91% 0.4396 0.0308 14.25 0.0000
You can play with the sample size or with the differences between raters to produce a non-significant answer :). This kappa suffers from a serious drawback: it does not reflect the fact of having some common items: the patterns 0001 and 0000, even though they match by 75%, would be considered non-matches within this approach. So it is an extremely conservative measure of the inter-rater agreement.
To get fair estimates of all the ICCs, you would need to run a cross-classified mixed model. Let us first reshape the data to make it possible:
generate long id = _n
* reshape the raters
reshape long a b c d , i(id) j(rater 1 2)
* reshape the items
forvalues k = 1/4 {
local thisletter : word `k' of `letters'
rename `thisletter' q`k'
}
reshape long q , i(id rater) j(item 1 2 3 4)
Now, we can run xtmelogit (or gllamm if you like it better) on this data:
. xtmelogit q || _all : R.rater || _all: R.item || _all: R.id, nolog
Note: factor variables specified; option laplace assumed
Mixed-effects logistic regression Number of obs = 1600
Group variable: _all Number of groups = 1
Obs per group:
min = 1600
avg = 1600.0
max = 1600
Integration points = 1 Wald chi2(0) = .
Log likelihood = -697.55526 Prob > chi2 = .
------------------------------------------------------------------------------
q | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
_cons | -.7795316 .9384147 -0.83 0.406 -2.618791 1.059727
------------------------------------------------------------------------------
------------------------------------------------------------------------------
Random-effects Parameters | Estimate Std. Err. [95% Conf. Interval]
-----------------------------+------------------------------------------------
_all: Identity |
sd(R.rater) | .1407056 .1627763 .0145745 1.358408
-----------------------------+------------------------------------------------
_all: Identity |
sd(R.item) | 1.797133 .6461083 .8882897 3.635847
-----------------------------+------------------------------------------------
_all: Identity |
sd(R.id) | 3.18933 .2673165 2.706171 3.758751
------------------------------------------------------------------------------
LR test vs. logistic regression: chi2(3) = 793.71 Prob > chi2 = 0.0000
Note: LR test is conservative and provided only for reference.
Note: log-likelihood calculations are based on the Laplacian approximation.
This is a cross-classified model with three random effects: subjects, raters and items, assuming that they are uncorrelated (which is wrong for this data; see below). Let us now estimate the ICCs:
. local Vrater ( exp(2*_b[lns1_1_1:_cons]) )
. local Vitem ( exp(2*_b[lns1_2_1:_cons]) )
. local Vid ( exp(2*_b[lns1_3_1:_cons]) )
. nlcom `Vrater' / (`Vrater' + `Vitem' + `Vid' + _pi*_pi/3 )
-----------------------------------------------------------------------
q | Coef. Std. Err. z P>|z| [95% Conf. Interval]
------+----------------------------------------------------------------
_nl_1 | .0011847 .0027384 0.43 0.665 -.0041824 .0065519
-----------------------------------------------------------------------
. nlcom `Vid' / (`Vrater' + `Vitem' + `Vid' + _pi*_pi/3 )
------------------------------------------------------------------------
q | Coef. Std. Err. z P>|z| [95% Conf. Interval]
------+----------------------------------------------------------------
_nl_1 | .6086839 .0903816 6.73 0.000 .4315393 .7858285
------------------------------------------------------------------------
. nlcom `Vitem' / (`Vrater' + `Vitem' + `Vid' + _pi*_pi/3 )
-----------------------------------------------------------------------
q | Coef. Std. Err. z P>|z| [95% Conf. Interval]
------+----------------------------------------------------------------
_nl_1 | .193265 .1121376 1.72 0.085 -.0265206 .4130506
-----------------------------------------------------------------------
(Hint: I figured out the names of the parameters by matrix list e(b).)
These are ICCs corresponding to raters, subjects and items, respectively. The zero ICC of the raters actually makes sense in the context of how the data were generated: there is no systematic effect in the sense that one rater consistently rates the condition better or worse than the other rater. There is an interaction between rater and item, but the model does not reflect it. True to life would be something like
xtmelogit q ibn.item##ibn.rater, nocons || id:
With this specification, you would have to get ICCs by an even more complicated mix of the variance components and the point estimates from the fixed effects part of the model.
If you have the patience (or a powerful computer), you can specify intp(7) or something like that to get an approximation more accurate than the Laplace approximation (a single point at the mode of the distribution of the random effects).
Best Answer
The intraclass correlation coefficient (ICC) works by partitioning the rating variance into multiple components, e.g., variance associated with items, variance associated with raters, and variance associated with measurement error. The ICC value is set to equal the fraction of "relevant" variance that is associated with items. Using this approach, when the total rating variance is low, it is almost impossible to achieve a high ICC value. In your example, there is very little variance and thus the ICC values are low despite the raters agreeing on their ratings of 0 for most items.
In this case, due to the low variance, an alternative approach is probably needed. Rather than using the ICC, you might consider using a categorical agreement coefficient with a non-nominal weighting scheme. There are many possible options, including a weighted kappa coefficient or S score.
Click here to view more information and access functions for calculating these coefficients.