The omnibus F-test may be ignored if you use multiple comparison procedures

I think  trying to be scientific with a small s involves asking critical questions about  common wisdom or common practice. In this post, I would like to focus on multiple comparisons in the context of ANOVA. What does common practice indicate?

Common wisdom suggests doing multiple comparisons only if the F-test is significant

Let's have a look on some practical advice considering multiple comparisons found on the web (R-bloggers.com) and in Field (2015). 

"One way to begin an ANOVA is to run a general omnibus test. The advantage to starting here is that if the omnibus test comes up insignificant, you can stop your analysis and deem all pairwise comparisons insignificant. If the omnibus test is significant, you should continue with pairwise comparisons" (https://www.r-bloggers.com/r-tutorial-series-one-way-anova-with-pairwise-comparisons/)

"When we have a statistically significant effect in ANOVA and an independent variable of more than two levels, we typically want to make follow-up comparisons. There are numerous methods for making pairwise comparisons and this tutorial will demonstrate how to execute several different techniques in R." (https://www.r-bloggers.com/r-tutorial-series-anova-pairwise-comparison-methods/)
And have a look at how the text book I used to use in my statistics course explains it.

"It might seem a a bit unhelpful that an ANOVA doesn't tell you which groups are different from which, given that having gone to the trouble of running an experiment, you probably need to know more than 'there's some difference somewhere or other'. You might wonder, therefore, why we don't just carry out a lot of t-tests, which would tell us very specifically whether pairs of group menas differ. Actually, the reason has already been explained in Section 2.1.6.7: every time you run multiple tests on the same data you inflate the potential Type I errors that you make. However, we'll return to this point in Section 11.5 when we look at how we follow up an ANOVA to discover where the group difference lie." (Field, 2015, p. 442).
Although, in honesty, on p. 459 Field writes:

"The least significance difference (LSD) pairwise comparison makes no attempt to control Type I error and is equivalent to performing multiple t-tests on the data. The only difference is that LSD requires the overall ANOVA to be significant."

This is meant to inform about the relative merits of one post hoc procedure to another in terms of Type I and Type II error.  Crucially, it is not mentioned that the other post-hoc procedures require that the overall ANOVA be significant. (As common wisdom seems to suggest). However,  his flow-chart of the ANOVA procedure (p. 460) clearly suggests multiple comparison procedures should be used as post-hoc procedures (after the ANOVA is significant).

Thus, common "statistical" wisdom seem to suggest that multiple comparison procedures are to be used as post hoc procedures following up a significant omnibus F-test. And the reason is that this two-stepped procedure minimizes the probability of type I errors.

Now, let's ask ourselves whether this common sense is, well, sensible.

Multiple comparisons only after significant F-test affects power negatively

Wilcox (2017) contains some useful information regarding our question. In his discussion of the much used Tukey-HSD procedure (the Tukey-Kramer Method), he references Bernhardson, (1975) who shows that the probability of at least 1 type I error among pairwise comparisons of estimates of equal population means (i.e. true null-hypotheses) is no longer equal to $\alpha$ if the procedure is only carried out following a significant omnibus test. That is, if we use our beloved two step procedure.

The consequence of the two step procedure for the Tukey-HSD is that $\alpha$ is reduced. Thus, if we want our multiple comparisons procedure to generate one type I error or more at most with a probability of  $\alpha = .05$, using the 2 step procedure leads to a lowered $\alpha$. This is of course, bad news, because in the event that not all of the null-hypotheses are true, lowering $\alpha$ increases $\beta$, the probability of not rejecting when the null-hypothesis is false (keeping the sample size constant, of course). In other words, the two step procedure decreases the power of the multiple comparison procedure.

In the words of Wilcox (2017):

"In practical terms, when it comes to controlling the probability of at least one type I error, there is no need to first reject with the ANOVA F test to justify using the Tukey-Kramer method. If the Tukey-Kramer method is used only after the F test rejects, power can be reduced. Currently, however, common practice is to use the Tukey-Kramer method only if the F-test rejects. That is, the insight reported by Bernhardson is not yet well known."  (p. 385).

In conceptual terms,  the fact that the probability of at least one type I error in the multiple comparison procedure is  smaller than $\alpha$ if the F-test rejects is pretty clear, at least to me it is. Suppose we reject if the p-value of the F-test is smaller or equal to 5%. This will also be the probability that we conduct the multiple comparison test over repeated replications of the same experiment. Of that 5%, not every application of the procedure will result in at least one type I error. Indeed, a puzzling fact for many beginning researchers is that the F-test is significant while none of the pairwise comparisons is. In other words, some of those 5%  of the cases in which we perform the procedure following a significant F-test will probably not reject any of the pairwise null-hypotheses, unless it is guaranteed that at least one type I error per application will be made.

(With no adjustment of $\alpha$ for multiple comparisons, this will happen (with high probability so no guarantee) if a huge number of pairwise comparisons are made. For instance, with 99 unadjusted multiple comparisons the probability of at least one type I error is 99%.; this is why it makes sense to demand that the F-test is significant before testing multiple comparisons with the LSD procedure. Although the latter seems to run into trouble with more than 3 groups (Wilcox, 2017).

A quick simulation study

My hunch is that the two-step procedure is unnecessary for the Tukey-Kramer method as well as for other multiple comparison procedures (the exception Fisher's LSD procedure which was designed as a post hoc procedure to be used as a follow up after a significant F-test, as Field (2015) rightly points out), but I only focused on the Tukey-Kramer method. What I did was a simple simulation study with a four group between subjects design (all $\mu$'s equal) and estimated the probability of at least type I error both with and without using the 2 step procedure.

set.seed(456)
#number of groups
ngr = 4

#number of participants
n = 40

#group is a factor
gr <- factor(rep(1:ngr, each=n))

#vector for storing rejections F-test
Reject <- rep(0, 10000)

#vector for storing #rejections multiple
#comparisons
RejectHSD <- rep(0, 10000)

for (i in 1:10000) {

y = rnorm(ngr*n)
mod = aov(y ~ gr)
Reject[i] = anova(mod)$"Pr(>F)"[1] <= .05
PS <- TukeyHSD(mod)$gr[,4]
RejectHSD[i] = sum(PS <=.05)
}

#probability type I error F-test
sum(Reject)/length(Reject)

## [1] 0.0515

#probability at least one type I error Tukey HSD
sum(RejectHSD > 0) / length(RejectHSD)

## [1] 0.0503

#probability at least one type I error given F-tests Rejects
sum(RejectHSD[Reject==TRUE] > 0) / length(RejectHSD)

## [1] 0.0424

Even though a single (relatively tiny) simulation (which, by the way, takes a long time to run, nonetheless), is not necessarily convincing, it does  illustrate the main points of this post. First, the probability of at least one incorrect rejection using the TukeyHSD function is close to .05. With this particular random seed it even performs a little better than the ANOVA F-test: .0503 versus .0515. This illustrates that even without considering whether the omnibus test is significant the main demand of not rejecting too many true null-hypotheses is completely satisfied. So, in practical terms, you can safely ignore the omnibus test if your concerns are about  $\alpha$.

Second, the probability of incorrectly rejecting at least one true pair-wise null-hypothesis after the ANOVA F-test is significant is estimated to be .0424. This shows, that the two-step procedure leads to a larger decrease in the actual type I error probability than is wanted. Even though this may seem good news from the perspective of avoiding type I errors, the down side is that pair wise null-hypotheses that are false (and potentially important) may not be detected.

Conclusion

Common wisdom and practice suggest that multiple comparisons procedures should be done only after a significant omnibus test. We have seen that this is not at all necessary if we use a multiple comparisons procedure that is designed to control the type I error probability. To my knowledge, most of the procedures conventionally thought of as post hoc tests are designed in this manner, the exception being the LSD procedure which does require a significant F-test. For practical purposes, then, do not bother with the omnibus test (note the exception) if you are planning to pair wise compare all the treatment means. 

This practical advice does not mean, of course, that I am suggesting you spend your time comparing all treatment means. Most of the time, focused comparisons are a more fruitful way of analysing your data. But I'll leave that topic for another time. 

References

Field, A. (2013). Discovering Statistics Using IBM SPSS Statistics. 4th Edition. London: Sage. 

Wilcox, R. (2017). Understanding and Applying Basic Statistical Methods Using R. Hoboken, NJ: Wiley,  

Planning for a precise slope estimate in simple regression

In this post, I will show you a way of determining a sample size for obtaining a precise estimate of the slope $\beta_1$of the simple linear regression equation $\hat{Y_i} = \beta_0 + \beta_1X_i$. The basic ingredients we need for sample size planning are a measure of the precision, a way to determine the quantiles of the sampling distribution of our measure of precision, and a way to calculate sample sizes.

As our measure of precision we choose the Margin of Error (MOE), which is the half-width of the 95% confidence interval of our estimate (see: Cumming, 2012; Cumming & Calin-Jageman, 2017; see also www.thenewstatistics.com).

 

The distribution of the margin of error of the regression slope

In the case of simple linear regression, assuming normality and homogeneity of variance, MOE is $t_{.975}\sigma_{\hat{\beta_1}}$, where $t_{.975}$, is the .975 quantile of the central t-distribution with $N - 2$ degrees of freedom, and $\sigma_{\hat{\beta_1}}$ is the standard error of the estimate of $\beta_1$. 

An expression of the squared standard error of the estimate of $\beta_1$ is $\frac{\sigma^2_{Y|X}}{\sum{(X_i - \bar{X})}^2}$ (Wilcox, 2017): the variance of Y given X divided by the sum of squared errors of X. The variance $\sigma^2_{Y|X}$ equals $\sigma^2_y(1 - \rho^2_{YX})$, the variance of Y multiplied by 1 minus the squared population correlation between Y and X, and it is estimated with the residual variance $\frac{\sum{(Y - \hat{Y})^2}}{df_e}$, where $df_e = N - 2$.

The estimated squared standard error is given in (1)

$$\hat{\sigma}_{\hat{\beta_{1}}}^{2}=\frac{\sum(Y-\hat{Y})^{2}/df_{e}}{\sum(X-\bar{X})^{2}}. \tag{1} $$

With respect to the sampling distribution of MOE, we first note the following. The distribution of estimates of the residual variance in the numerator of (1) is a scaled $\chi^2$-distribution:

$$\frac{\sum(Y-\hat{Y})^{2}}{\sigma_{y}^{2}(1-\rho^{2})}\sim\chi^{2}(df_{e}),$$

thus

$$\frac{\sum(Y-\hat{Y})^{2}}{df_{e}}\sim\frac{\sigma_{y}^{2}(1-\rho^{2})\chi^{2}(df_{e})}{df_{e}}.$$

Second, we note that

$$\frac{\sum(X-\bar{X})^{2}}{\sigma_{X}^{2}}\sim\chi^{2}(df),$$

where $df = N - 1$, therefore

$$\sum(X-\bar{X})^{2}\sim\sigma_{X}^{2}\chi^{2}(df).$$

Alternatively, since $\sum{(X - \bar{X})^2} = df\sigma^2_X$, and multiplying by 1 ($\frac{df}{df}$). 

$$df\sigma_{X}^{2}\sim df\sigma_{X}^{2}\chi^{2}(df)/df.$$

In terms of the sampling distribution of (1), then, we have the ratio of two (scaled) $\chi^2$ distributions, one with $df_e = N - 2$ degrees of freedom, and one with $df = N - 1$ degrees of freedom. Or something like:

$$ \hat{\sigma}_{\hat{\beta_{1}}}^{2}\sim\frac{\sigma_{y}^{2}(1-\rho^{2})\chi^{2}(df_{e})/df_{e}}{df\sigma_{X}^{2}\chi^{2}(df)/df}=\frac{\sigma_{y}^{2}(1-\rho^{2})}{df\sigma_{X}^{2}}\frac{\chi^{2}(df_{e})/df_{e}}{\chi^{2}(df)/df}=\frac{\sigma_{y}^{2}(1-\rho^{2})F(df_{e,}df)}{df\sigma_{X}^{2}},$$

which means that the sampling distribution of MOE is:

$$ \hat{MOE}\sim t_{.975}(N-2)\sqrt{\frac{\sigma_{y}^{2}(1-\rho^{2})F(N-2,N-1)}{(N-1)\sigma_{X}^{2}}}. \tag{2} $$

This last equation, that is (2), can be used to obtain quantiles of the sampling distribution of MOE, which enables us to determine assurance MOE, that is the value of MOE that under repeated sampling will not exceed a target value with a given probability. For instance, if we want to know the .80 quantile of estimates of MOE, that is, assurance is .80, we determine the .80 quantile of the (central) F-distribution with N - 2 and N - 1 degrees of freedom and fill in (2) to obtain a value of MOE that will not be exceeded in 80\% of replication experiments.

For instance, suppose $\sigma^2_Y = 1$, $\sigma^2_X = 1$, $\rho = .50$, $N = 100$, and assurance is .80, then according to (2), 80\% of estimated MOEs will not exceed the value given by:

vary = 1
varx = 1
rho = .5
N = 100 
dfe = N - 2
dfx - N - 1

assu = .80
t = qt(.975, dfe)
MOE.80 = t*sqrt(vary*(1 - rho^2)*qf(.80, dfe, dfx)/(dfx*varx))
MOE.80

## [1] 0.1880535

 

What does a quick simulation study tell us? 

A quick simulation study may be used to check whether this is at all accurate. And, yes, the estimated quantile from the simulation study is pretty close to what we would expect based on (2). If you run the code below, the estimate equals 0.1878628.

library(MASS)
set.seed(355)
m = c(0, 0)

#note: s below is the variance-covariance matrix. In this case,
#rho and the cov(y, x) have the same values
#otherwise: rho = cov(x, y)/sqrt(varY*VarX) (to be used in the 

#functions that calculate MOE)
#equivalently, cov(x, y) = rho*sqrt(varY*varX) (to be used
#in the specification of the variance-covariance matrix for 

#generating bivariate normal variates)

s = matrix(c(1, .5, .5, 1), 2, 2)
se <- rep(10000, 0)
for (i in 1:10000) {
theData <- mvrnorm(100, m, s)
mod <- lm(theData[,1] ~ theData[,2])
se[i] <- summary(mod)$coefficients[4]
}
MOE = qt(.975, 98)*se
quantile(MOE, .80)

##       80% 
## 0.1878628

 

Planning for precision

If we want to plan for precision we can do the following. We start by making a function that calculates the assurance quantile of the sampling distribution of MOE described in (2). Then we formulate a  squared cost function, which we will optimize for the sample sizeusing the optimize function in R.

Suppose we want to plan for a target MOE of .10 with 80% assurance.We may do the following.

vary = 1
varx = 1
rho = .5
assu = .80
tMOE = .10

MOE.assu = function(n, vary, varx, rho, assu) {
        varY.X = vary*(1 - rho^2)
        dfe = n - 2
        dfx = n - 1
        t = qt(.975, dfe)
        q.assu = qf(assu, dfe, dfx)
        MOE = t*sqrt(varY.X*q.assu/(dfx * varx))
        return(MOE)
}

cost = function(x, tMOE) { 

cost = (MOE.assu(x, vary=vary, varx=varx, rho=rho, assu=assu) 
- tMOE)^2
}

#note samplesize is at least 40, at most 5000. 
#note that since we already know that N = 100 is not enough
#in stead of 40 we might just as well set N = 100 at the lower
#limit of the interval
(samplesize = ceiling(optimize(cost, interval=c(40, 5000), 
tMOE = tMOE)$minimum))

## [1] 321

#check the result: 
MOE.assu(samplesize, vary, varx, rho, assu)

## [1] 0.09984381

Let's simulate with the proposed sample size

Let's check it with a simulation study. The value of estimated .80 of estimates of MOE is 0.1007269 (if you run the below code with random seed 335), which is pretty close to what we would expect based on (2).

set.seed(355)
m = c(0, 0)

#note: s below is the variance-covariance matrix. In this case,
#rho and the cov(y, x) have the same values
#otherwise: rho = cov(x, y)/sqrt(varY*VarX) (to be used in the 

#functions that calculate MOE)
#equivalently, cov(x, y) = rho*sqrt(varY*varX) (to be used
#in the specification of the variance-covariance matrix for 
#generating bivariate normal variates)

s = matrix(c(1, .5, .5, 1), 2, 2)
se <- rep(10000, 0)
samplesize = 321
for (i in 1:10000) {
theData <- mvrnorm(samplesize, m, s)
mod <- lm(theData[,1] ~ theData[,2])
se[i] <- summary(mod)$coefficients[4]
}
MOE = qt(.975, 98)*se
quantile(MOE, .80)

##       80% 
## 0.1007269

 

References

Cumming, G. (2012). Understanding the New Statistics. Effect Sizes, Confidence Intervals, and Meta-Analysis. New York: Routledge
Cumming, G., & Calin-Jageman, R. (2017). Introduction to the New Statistics: Estimation, Open Science, and Beyond. New York: Routledge.
Wilcox, R. (2017). Understanding and Applying Basic Statistical Methods using R. Hoboken, New Jersey: John Wiley and Sons.

Planning for a precise interaction contrast estimate

In my previous post (here),  I wrote about obtaining a confidence interval for the estimate of an interaction contrast. I demonstrated, for a simple two-way independent factorial design, how to obtain a confidence interval by making use of the information in an ANOVA source table and estimates of the marginal means and how a custom contrast estimate can be obtained with SPSS.

One of the results of the analysis in the previous post was that the 95% confidence interval for the interaction was very wide. The estimate was .77, 95% CI [0.04, 1.49]. Suppose that it is theoretically or practically important to know the value of the contrast to a more precise degree.  (I.e. some researchers will be content that the CI allows for a directional qualitative interpretation: there seems to exist a positive interaction effect, but others, more interested in the quantitative questions may not be so easily satisfied).  Let's see how we can plan the research to obtain a more precise estimate. In other words, let's plan for precision.

Of course, there are several ways in which the precision of the estimate can be increased. For instance, by using measurement procedures that are designed to obtain reliable data, we could change the experimental design, for example switching to a repeated measures (crossed) design, and/or increase the number of observations. An example of the latter would be to increase the number of participants and/or the number of observations per participant.  We will only consider the option of increasing the number of participants, and keep the independent factorial design, although in reality we would of course also strive for a measurement instrument that generally gives us highly reliable data. (By the way, it is possible to use my Precision application to investigate the effects of changing the experimental design on the expected precision of contrast estimates in studies with 1 fixed factor and 2 random factors).

The plan for the rest of this post is as follows. We will focus on getting a short confidence interval for our interaction estimate, and we will do that by considering the half-width of the interval, the Margin of Error (MOE). First we will try to find a sample size that gives us an expected MOE (in repeated replication of the experiment with new random samples) no more than a target MOE. Second, we will try to find a sample size that gives a MOE smaller than or equal to our target MOE in a specifiable percentage (say, 80% or 90%) of replication experiments. The latter approach is called planning with assurance.

Planning for Precision: A confidence interval for the contrast estimate

In a previous post, which can be found here, I described how the relative error variance of a treatment mean can be obtained by combining variance components.  I concluded that post by mentioning how this relative error variance for the treatment mean can be used to obtain the variance of a contrast estimate. In this post, I will discuss a little more how this latter variance can be used to obtain a confidence interval for the contrast estimate, but we take a few steps back and consider a relatively simple study.

The plan of this post is as follows. We will have a look at the analysis of a factorial design and focus on estimating an interaction effect. We will consider both the NHST approach and an estimation approach. We will use both 'hand calculations' and SPSS.

An important didactic aspect of this post is to show the connection between the ANOVA source table and estimates of the standard error of a contrast estimate. Understanding that connections helps in understanding one of my planned posts on how obtaining these estimates work in the case of mixed model ANOVA.  See the final section of this post.

The new statistics: a five-day course

Last week, I taught a 5-day-course for the LOT (Landelijke Onderzoeksschool Taalwetenschap; Netherlands National Graduate School of Linguistics; www.lotschool.nl) introducing the new statistics to PhD-students working in linguistics and related fields of research. Links to the course materials can be found in this post (apologies for the many typos).

The day-to-day program was as  follows.

1. Important concepts underlying statistics, like population paremeters, sampling, sanpling distribution, standard error and the margin of error. The primary means of developing these concepts was working with ESCI (www.tiny.cc/itns). The lab assignments are primarily based on Cumming and Calin-Jageman's (2017) "Introduction to the new statistics".  The lab-assignments can be found here: www.tiny.cc/newstats. A pdf-version of the presentation can be found here: http://tiny.cc/newstats-presentation
2. Continuation of day 1. For students that finished the first assignment and to accommodate differences in backgrounds, new lab assignments focusing on statistical assumptions underlying the crucial concepts. Some of these assignments are based on Cumming and Calin-Jageman (2017) and ESCI, others work with R. The lab-assignments can be found here:  www.tiny.cc/newstatsla2
3. Lecture only. In the lecture we reviewed the basic concepts discovered in the first two days. The concept of a confidence interval was introduced and the p-value. Furthermore, we discussed  NHST by considering (at a procedural level and not so much on a statistical/philosophical level) how the procedure relates to its foundations: Fisher's significance testing and Neyman and Pearson Hypothesis Testing. We basically saw that NHST is inconsistent with both of these foundations. We also discussed misinterpretations of p-values. The presentation can be found here: www.tiny.cc/newstatsday3. I also made available the lecture notes: www.tiny.cc/newstatsday3ln.
4. Lecture only. This day was about effect sizes. We considered the unstandardized difference between means,  Cohen's d, and the case level effect size measures Cohen's U3 and the Common Language Effectsize. The powerpoint presentation is at www.tiny.cc/newstatsday4.
5. On the last day the students worked on new lab assignments focusing on interpretations of significance, the use of p-values and effect sizes in published work and working with effect size measures based on SPSS ANOVA output. These assignments can be found here: www.tiny.cc/newstatsday5.

Planning for Precision: Introduction to variance components

Both theory underlying the Precision application and the use of the app in practice rely for a large part on specifying variance components. In this post, I will give you some more details about what these components are, and how they relate to the analysis of variance model underlying the app.

What is variance?

Let's start with a relatively simple conceptual explanation of variance. The key ideas are expected value and error.  Suppose you randomly select a single score from a population of possible values. Let's suppose furthermore that the population of values can be described with a normal distribution. There is actually no need to suppose a normal distribution, but it makes the explanation relatively easy to follow.

As you probably know, the normal distribution is centered around its mean value, which is (equal to) the parameter μ. We call this parameter the population mean.

Now, we select a single random value from the population. Let's call this value X.  Because we know something about the probability distribution of the population values, we are also in the position to specify an expected value for the score X. Let's use the symbol E(X) for this expected value. The value of E(X) proves (and can be proven) to be equal to the parameter μ. (Conceptually, the expectation of a variable can be considered as its long run average).

Of course, the actual value obtained will in general not be equal to the expected value, at least not if we sample from continuous distributions like the normal distribution. Let's call the difference between the value X and it's expectation E(X) = μ. an error, deviation or residual: e = X - E(X) = X -  μ.

We would like to have some indication of the extent to which X differs from its expectation, especially when E(X) is estimated on the basis of  a statistical model.  Thus, we would like to have something like E(X - E(X)) = E(X -  μ). The variance gives us such an indication, but does so in squared units, because working with the expected error itself always leads to the value 0  E(X -  μ) = E(X) - E( μ) =  μ -  μ = 0. (This simply says that on average the error is zero; the standard explanation is that negative and positive errors cancel out in the long run).

The variance is the expected squared deviation (mean squared error) between X and its expectation: E((X - E(X))²) = E(X -  μ)²), and the symbol for the population value is σ².

Some examples of variances (remember we are talking conceptually here):
- the variance of the mean, is the expected squared deviation between a sample mean and its expectation the population mean.
- the variance of the difference between two means:  the expected squared deviation between the sample difference and the population difference between two means.
- the variance of a contrast: the expected squared deviation between the sample value of the contrast and the population value of the contrast.

It's really not that complicated, I believe.

Planning for Precision: simulation results for four designs with four conditions

This is the third post about the Planning for Precision app (in the future I'll explain the difference between Planning for Precision and Precision for Planning). Some background information about the application can be found here: http://the-small-s-scientist.blogspot.com/2017/04/planning-for-precision.html. 

In this post, I want to present the simulation results for 4 designs with 4 conditions. The designs are: the counter balanced design (see previous post), the fully-crossed design, the stimulus-within-condition design, and the stimulus-and-participant-within-condition design (the both-within-condition design). I have not included the participants-within-condition design, because this is simply the mirror-image (so to say) of the stimulus-within-condition design.

In one of my next posts, I will describe some more background information about planning for precision, but some of the basics are as follows. We have a design with 4 treatment conditions, and what we want do is to estimate differences between these condition means by using contrasts. For instance, we may be interested in the (amount of) difference between the first mean, maybe because it is a control-condition with the average of the other three conditions: μ₁ - (μ₂ + μ₃ + μ₄)/3 =  1*μ₁ - 1/3*μ₂ -1/3*μ₃ - 1/3*μ₄.  The values {1, -1/3, -1/3, -1/3} are the contrast weights, and for the result we use the term ψ.

The value of ψ is estimated on the basis of estimates of the population means, that is, the sample means or condition means. Due to sampling error, the contrast estimate varies from sample to sample and the amount of sampling error can be expressed by means of a confidence interval. Conceptually, the confidence interval expresses the precision of the estimate: the wider the confidence interval, the less precise the estimate is.

The Margin of Error (MOE) of an estimate is the half-width of the confidence interval, so the confidence interval is the estimate plus or minus MOE. We will take MOE as an expression of the precision of the estimate (the less the value of MOE the more precise the estimate).  Now, if you want to estimate an effect size, more precision (lower value of MOE; less wide confidence interval) is better than less precision (higher value of MOE; wider confidence interval).  The app let's you specify the design and the contrast weights and helps you find the minimum required sample sizes (for participants and stimuli) for a given target MOE. (You can also play with the designs to see which design gives you smallest expected MOE).

Crucially, if you plan for precision, you also want to have some assurance that the MOE you are likely to obtain in you actual experiment will not be larger than you target MOE. Compare this with power: 80% power means that the probability that you will reject the null-hypothesis is 80%. Likewise, assurance MOE of 80% means that there is an 80% probability that your obtained MOE will be no larger than assurance MOE.

The simulations (with N = 10000 replications) estimate Expected MOE as well as Assurance MOE for assurances of .80, .90, .95, and .99, for 4 designs with 4 treatment conditions, with a total number of 48 participants and 24 stimuli (items).  The MOEs are given for three standard constrasts: 1) the difference between the first mean and the mean of the other three, with weights {1, -1/3, -1/3, -1/3}; 2) the difference between the second mean and the mean of conditions three and four, with weights {0, 1, -1/2, -1/2}; 3) the difference between the third and fourth condition means, with weights {0, 0, 1, -1}.

I will present the results in  separate tables for the 4 designs considered and include percentage difference between expected values of assurance MOE and the estimated values estimated values.

The fully crossed design 

The results are in the following table.

The percentage difference between the expected quantiles (= assurance MOEs for given insurance;  i.e. q.80 is expected or estimated  80% Assurance MOE) and the estimated quantiles are: .80: 0.11%; .90: 0.05%; .95: -0.14%; 99: -0.05%.

The counter balanced design 

The results are presented in the following table. 

The percentage difference between the expected quantiles and the estimated quantiles are: .80: 0.03%; .90: 0.13%;  .95: 0.09%, .99: -0.23%.

The stimulus-within-condition design 

The following table contains the details. 

The percentage difference between the expected quantiles and the estimated quantiles are: .80: -0.11%; .90: -0.33%;  .95: -0.55%, .99: -0.70%.  

Both-participant-and-stimulus-within-condition design 

Here is the table. 

And the percentage differences are: .80: -0.34%; .90: -0.59%;  .95: -0.82%;  .99: -1.06%. 

Conclusion

The results show that the simulation results are quite consistent with the expected values based on mixed model ANOVA. We can see that the differences between expected and estimated values increase the less the number of participants and items per condition. For instance, in the both within condition design 12 participants respond to 6 stimuli in one of the four treatment conditions. The fact that even with these small samples sizes the results seem to agree to an acceptable degree is (to my mind) encouraging. Note that with small samples the expected assurance MOES are slightly lower than the estimates, but the largest difference is -1.06% (see the MOE for 99% assurance).