GORIC(A) weights benchmarks
The GORIC(A) weights benchmarks come from several percentiles of sets
of GORIC(A) weights assuming that the specified set of population values
is true (for the sample size under consideration). More specifically,
the benchmarks are, currently, based on the 5th, 35th, 50th, 65th, and
95th percentiles of the GORIC(A) weights for the preferred hypothesis
and of the ratios of the GORIC(A) weights for the preferred hypothesis
versus the other hypotheses. Bear in mind that research (e.g., a
simulation study) is needed to obtain more insight into how well these
choices work.
Notably,
you can a-priori decide on what percentiles you believe reflect the
different types of support, and pre-register that
together with your informative hypothesis/-es.
You can compare your GORIC(A) weight and/or ratios of GORIC(A) weights of the preferred hypothesis to the benchmarks to draw a conclusion regarding the strength of support for the hypothesis (given the assumed set of population parameter values and given the sample size). If the benchmarks show a maximum/bounded support (see second example below), then there is support for the overlap of two or more hypotheses (which is also signaled by equal log-likelihood values). Otherwise, the GORIC(A) weights (ratios) benchmarks can be used to qualify the height of support (see first example below).
How to use benchmarks
Labelling
We are not in favor of cut-off points (or ‘surrounding anchors’), but we need them when we want to label the height of support via the benchmarks, we propose the following:
| Benchmark (percentile) | Height support |
|---|---|
| below 5th | no support |
| between 5th and 35th | low support |
| between 35th and 65th | medium support |
| between 65th and 95th | high (compelling) support |
| over 95th | very large (tremendous) support |
We advise on using some kind of null model as the assumed population (possibly, using multiple ones). Then, you can see how extreme your finding is (or not) for this null population. The extremer your finding, the more support for your informative, theory-based, inequality-constrained hypothesis.
You can of course use other percentile levels (for which you think the finding is said to be extreme enough etc). But: Do make sure to define these before seeing the data, and preferably also pre-register them (together with your informative hypothesis/-es).
Use minimum effect
You may want to use a minimum effect. One option – the one we also advise, when it is doable – is to specify your hypothesis such that it inspects, for example, minimum differences between parameters (e.g., \(\mu_1 - \mu_2 >\) 0.2 instead of \(\mu_1 > \mu_2\), that is, \(\mu_1 - \mu_2 >\) 0). Notably, one then possible does not need benchmarks anymore: Finding a ratio of GORIC(A) weights of 1 and higher would probably suffice.
A second option – although not something we advise on doing – is to
investigate benchmarks using a minimum effect size (or looking at
multiple ones). Then, you will compare your samples values with the
distribution of (ratios of) GORIC(A) weights under a population with
this minimum effect size.
In the function, you should specify the population parameter values
(reflecting specific effect sizes). In case of an ANOVA model, you can
also specify the effect size level(s) (for Cohen’s \(f\)). Note that the benchmarks differ when
you assume different effect sizes or different population parameter
estimates.
Sensitivity analysis
If of interest, as a sensitivity analysis, you can calculate the benchmarks for multiple sets of population parameter values (or population effect sizes). Note that this may also complicate drawing conclusions (especially when the assumed sets of population parameter values differ much, like when using multiple effect size heights). We advise on doing this for multiple null populations (setting some to all of the inequality restrictions to equalities).
If your preferred hypothesis does not have the highest fit and you want to inspect benchmarks, we also advise on inspecting multiple ratios of population parameter estimates, where some are in agreement with your hypothesis and others in agreement with the data.
Defaults
Once more, we advise on inspecting populations in which some of the inequalities of your hypothesis/-es of interest are set to equalities. This can also be very helpful if the log-likelihood values seem to be close: This way, you can also check for the support of a boundary hypothesis; as discussed in the third example below.
By default, two populations are used: i) a null population in which
the population effect size is – or population parameter values are – set
to 0 and ii) a population based on the observed effect size – or
observed parameter values. One can overrule this by using the
pop_es or pop_est argument, respectively.
Examples
Next, we will discuss two ANOVA examples. More specifically, we will
inspect the case-specific benchmarks values (using the ratio of means as
in the data). We will look at
- an ANOVA example where we evaluate \(H_1:
\mu_1 > \mu_2 > \mu_3\) versus its complement, and \(H_1\) is true;
- an ANOVA example where we evaluate two overlapping hypotheses, namely
\(H_1: \mu_1 > \mu_2 > \mu_3\)
and \(H_2: \mu_1 > \mu_2, \ \mu_3\),
together with the unconstrained, and \(H_1\) is true (and thus the others are as
well, but they are not the most parsimonious one).
Later on (in another section), we will also discuss the following
example:
- an ANOVA example where we evaluate \(H_1:
\mu_1 > \mu_2 > \mu_3\) versus its complement, and the
border \(\mu_1 = \mu_2 > \mu_3\) is
true. Notably, here, both hypotheses are true, \(H_1\) is the most parsimonious one, and we
want to conclude that the border is true.
For a description of interpreting GORIC(A) output, see ‘Guidelines_output_GORIC’ (https://github.com/rebeccakuiper/Tutorials).
Example 1 (ANOVA): \(H_1\) vs its complement
# H1 vs complement - H1 is true
H1 <- "D1 > D2 > D3" # mu1 > mu2 > mu3
# Apply GORIC #
set.seed(123)
results_1c <- goric(fit, hypotheses = list(H1), comparison = "complement")
results_1crestriktor (0.5-90): generalized order-restricted information criterion:
Results:
model loglik penalty goric loglik.weights penalty.weights goric.weights
1 H1 -155.075 2.833 315.816 0.704 0.697 0.845
2 complement -155.939 3.667 319.212 0.296 0.303 0.155
---
The order-restricted hypothesis 'H1' has 5.46 times more support than its complement.Calculating means benchmark for effect-size = 0 (No-effect)
Calculating means benchmark for effect-size = 0.411 (Observed)# benchmarks_1c # use in R file
print(benchmarks_1c, color = FALSE) # use in Rmd file, since Rmd cannot deal with colored text
Benchmark Results
----------------------------------------------------------------------
Preferred Hypothesis: H1
Error probability Preferred Hypothesis vs. complement: 0.155
Number of Groups: 3
Group Sizes: 40, 40, 40
Ratio of Population Means: 4.322, 2.000, 1.000
Population Effect-Sizes (Cohens f): 0.000, 0.411
Observed Effect-Size (Cohens f): 0.411
==================================================================================
Benchmark: Percentiles of Ratio-of-GORIC-weights for the Preferred Hypothesis 'H1'
----------------------------------------------------------------------------------
Population effect-size = 0
Sample 5% 35% 50% 65% 95%
H1 vs. complement 5.464 0.402 1.279 1.687 1.992 2.383
Population effect-size = 0.411
Sample 5% 35% 50% 65% 95%
H1 vs. complement 5.464 2.297 3.643 4.939 8.121 35.607
From the goric output, you can conclude i) that there is
support for \(H_1: \mu_1 > \mu_2 >
\mu_3\) and ii) that \(H_1\) is
0.85 \(/\) 0.15 \(\approx\) 5.46 times more supported than
its complement. The probability that \(H_1\) is not the best is 15.47% (namely,
the goric.weight for the complement of \(H_1\), which is also given in the benchmark
output by $error_prob_pref_hypo, that is, the
Error probability Preferred Hypothesis vs. complement).
This already gives insight into the (un)certainty and, therefore, helps
in qualifying the results. Additionally, the benchmarks can help:
Based on the benchmarks, you can check how plausible your finding is (given the assumed population parameters and given the sample size). If you want to compare your results with the situation in which one or more equalities hold true, use a null population (here, an effect size of 0, indicating that the three means are equal). Then, you obtain insight into how (un)likely / how extreme your finding (based on inequalities) is.
When assuming that there is no effect in the population (i.e., under the null), that is, all three group means are equal, the ratio of GORIC weights of \(H_1\) versus its complement (i.e., 5.46) is larger than the 95th percentile (i.e., 2.38). Hence, our finding is very extreme if the null would be true. Using the table with cut-off values above, this indicates that there is very large (tremendous) support, when assuming no population effect size (given a group size of 40 for each of the 3 groups).
Additionally, when inspecting the distributions of the ratio of GORIC weights of \(H_1\) versus its complement under the null population and under the population based on the observed effect size (see first plot), then we can see that they differ, that is, do not overlap much. This also indicates that our finding is very unlikely if the null would be true.
Notably, the gray line denotes a ratio of GORIC(A) weights of 1, which means that the support for both hypothesis is equal. In that case, you would be indecisive. The reason to plot this line is to remember that a ratio of GORIC(A) weights ranges from 0 to infinity, where 0 to 1 denotes the same support as 1 to infinity for the opposite comparison (e.g., a support of 0.2 for \(H_1\) versus \(H_c\) is the same as a support of 1/0.2=5 for \(H_c\) versus \(H_1\). One can also plot the log transformation of the ratio:
This then ranges from minus infinity to infinity, where minus infinity
to 0 is comparable to 0 to infinity. Hence, equal support is now denoted
by 0, that is, (again) the grey line in the plot.
Log-likelihood check:
Before inspecting the height of the support, one may want to establish
whether there is support for the overlap or boundary of hypotheses.
Since we evaluate an informative hypothesis \(H_1\) versus its complement, we should
check whether there is support for a boundary hypothesis (in which one
or more inequalities in \(H_1\) is
replaced by an equality). For this, one should inspect the
log-likelihood / fit values of the hypotheses. When these are close
(i.e., the ratio of loglik weights is close to 1 or the difference in
loglik values is close to 0), then there is support for (one of their)
boundaries. In this case, the loglik values are -155.07 and -155.94,
with corresponding loglik.weights of 0.7 and 0.3 (and thus
a difference of approximately 0.86 and a ratio of approximately 2.37).
Since it is hard to judge what is close, one can inspect the benchmarks
for the ratio of log-likelihood (loglik) weights and/or for the
differences in log-likelihood values. This should then be done for a
null population in which such a boundary is true. We discuss this in the
next section.
For now, we assume that the loglik values are not close.
Population information:
In the data generation, we used a ratio of population means of 3:2:1;
implying that \(H_1\) is correct. More
specifically, we used population mean values of approximately 0.92,
0.61, and 0.31. This implies that Cohen’s \(f\) is .25; thus, there is a medium
population effect size (which are in the same order as hypothesized). We
then sampled 40 observations for each of the three groups, ran an ANOVA
(with three groups), and applied the GORIC. Note that Cohen (1992)
suggest that a minimum group size of 52 is needed to find a medium
effect when doing null hypothesis testing.
Notably, the sample/observed effect size is .411 (with sample means of
1.12, 0.51, and 0.25), which can be seen as a high effect size. This
also explains why we, despite the medium population effect size, find
tremendous support for our hypothesis.
When we would sample more observations, the GORIC(A) weight for \(H_1\) converges to 1 (denoting full support
for \(H_1\)). Note that the benchmarks
for the GORIC(A) weight for \(H_1\)
will remain the same for a null population and will go to 1 for a
non-null population. Note that the error probability then goes to 0, and
that the ratio of GORIC(A) weights of \(H_1\) versus its complement then goes to
infinity.
Example 2 (ANOVA): Overlapping hypotheses
# H1, H2, and unconstrained (default) - subset/overlap, that is, H1 is true
H1 <- "D1 > D2 > D3" # mu1 > mu2 > mu3
H2 <- "D1 > D2" # H2: D1 > D2, D3 # mu1 > mu2, mu3
# Apply GORIC #
set.seed(123)
results_12u <- goric(fit, hypotheses = list(H1, H2))
results_12urestriktor (0.5-90): generalized order-restricted information criterion:
Results:
model loglik penalty goric loglik.weights penalty.weights goric.weights goric.weights_without_unc
1 H1 -155.075 2.833 315.816 0.333 0.548 0.548 0.661
2 H2 -155.075 3.500 317.149 0.333 0.281 0.281 0.339
3 unconstrained -155.075 4.000 318.149 0.333 0.171 0.171
---
The order-restricted hypothesis 'H1' is not weak, with 1.948 times more support than H2. vs. H1 vs. H2 vs. unconstrained
H1 1.000 1.948 3.211
H2 0.513 1.000 1.649
unconstrained 0.311 0.607 1.000Calculating means benchmark for effect-size = 0 (No-effect)
Calculating means benchmark for effect-size = 0.411 (Observed)
Benchmark Results
----------------------------------------------------------------------
Preferred Hypothesis: H1
Error probability Preferred Hypothesis vs. complement: 0.155
Number of Groups: 3
Group Sizes: 40, 40, 40
Ratio of Population Means: 4.322, 2.000, 1.000
Population Effect-Sizes (Cohens f): 0.000, 0.411
Observed Effect-Size (Cohens f): 0.411
==================================================================================
Benchmark: Percentiles of Ratio-of-GORIC-weights for the Preferred Hypothesis 'H1'
----------------------------------------------------------------------------------
Population effect-size = 0
Sample 5% 35% 50% 65% 95%
H1 vs. H2 1.948 0.451 1.576 1.831 1.948 1.948
H1 vs. unconstrained 3.211 0.689 2.097 2.464 2.796 3.211
Population effect-size = 0.411
Sample 5% 35% 50% 65% 95%
H1 vs. H2 1.948 1.938 1.948 1.948 1.948 1.948
H1 vs. unconstrained 3.211 3.195 3.211 3.211 3.211 3.211
From the GORIC output, we can see that both hypotheses are not weak and that \(H_1\) is the best. Additionally, we can see that the log-likelihood values are exactly the same, and thus there is support for the overlap of the hypotheses which is \(H_1\) here (since \(H_1\) is a subset of \(H_2\)).
If we would now check the GORIC(A) weights benchmarks (especially for non-nulls), we would find again that there is support for the overlap (here, \(H_1\)): Namely, the benchmarks for multiple percentiles (here, from the 35th percentile on) have the same value as our finding, that is, the sample value.
Notably, if there is support for the overlap or boundary, it is not
meaningful to use the benchmarks to label the height of support for
\(H_1\), since the ratio of support
reached its maximum. Hence, we do not proceed with labeling the height
of support (by comparing the sample (ratio of) GORIC weights value(s) to
the GORIC weights benchmarks under the null population). Instead, we
conclude that there is support for the overlap.
If of interest, one can next inspect the support for the overlap, here
\(H_1\), versus its complement. When
that does not result in support for a boundary, then the height of the
support for \(H_1\) versus its
complement can be labelled using the benchmarks. This is then helpful
information for future research: Generating a new theory and having
another value to compare future results to.
Log-likelihood check:
Here, one clearly finds that the loglik / fit values are exactly the
same. Hence, the ratio of loglik weights is exactly 1 and the difference
in loglik values is exactly 0. Consequently, there is support for the
overlap. In this case, we do not need to check the log-likelihood
benchmarks.
Note that the log-likelihood benchmarks (under a null population) give
insight into the distribution of loglik weights ratios and of the loglik
differences in case some or all of the group means would be the
same.
Population information:
In the data generation, we used a ratio of population means of 3:2:1;
implying that \(H_1\) is correct. More
specifically, we used population mean values of approximately 0.92,
0.61, and 0.31. This implies that Cohen’s \(f\) is .25; thus, there is a medium
population effect size (which are in the same order as hypothesized). We
then sampled 40 observations for each of the three groups, ran an ANOVA
(with three groups), and applied the GORIC. Note that Cohen (1992)
suggest that a minimum group size of 52 is needed to find a medium
effect when doing null hypothesis testing. Notably, the sample/observed
effect size is .411 (with sample means of 1.12, 0.51, and 0.25), which
can be seen as a high effect size.
When we would sample more observations, it does not (really) affect the
GORIC(A) weights for \(H_1\), \(H_2\), and the unconstrained: It will
converge to the bounds (i.e., the maximum support) it can take on. The
benchmarks for the GORIC(A) weights will also attain the maximum value
as will the ratio of weights; and it will for each positive population
effect size. For a null population, the GORIC(A) weight benchmarks will
remain the same.
In this example – as can be seen from both the tables and the plots –
the difference in log-likelihood values and the ratio of log-likelihood
weights lies between the 5th and the 95th percentile of the
corresponding benchmarks (for all null populations). Based on this, we
conclude that the log-likelihood values of 
Also in this example, with a much higher sample size, the sample
difference in log-likelihood values and the sample ratio of
log-likelihood weights lies between the 5th and the 95th percentiles of
the corresponding benchmarks (for all null populations). Based on this,
we conclude that the log-likelihood values of 
In this example, the sample difference in log-likelihood values and the
sample ratio of log-likelihood weights are larger than the 95th
percentile of the benchmarks for the null distributions ‘PE_123eq’ and
‘PE_13eq’ (see the benchmarks output table). For the other two null
distributions – ‘PE_12eq’ and ‘PE_23eq’, where the means of groups 1 and
2 and that of 2 and 3, respectively, are set equal – the sample value is
in between the 65th and 95th percentile.
In this example with a higher sample size, the sample difference in
log-likelihood values and the sample ratio of log-likelihood weights are
higher than the 95th percentile of the corresponding benchmarks (for all
null populations; see the benchmarks output table). Based on this, we
conclude that the log-likelihood values of