On doing science, with a small S
I just finished a preprint of an introduction and tutorial to sample size planning for precision of contrast estimates. The tutorial focuses on single factor between and within subjects designs, and mixed factorial designs with one within and one between factor. The tutorial contains R-code for sample size planning in these designs.
The preprint is availabe on researchgate: Click (but I am just as happy to send it to you if you like; just let me know).
Planning for precision requires that we choose a target Margin of Error (MoE; see this post for an introduction to the basic concepts) and a value for assurance, the probability that MoE will not exceed our target MoE. What your exact target MoE will be depends on your research goals, of course.
Cumming and Calin-Jageman (2017, p. 277) propose a strategy for determining target MoE. You can use this strategy if your research goal is to provide strong evidence that the effect size is non-zero. The strategy is to divide the expected value of the difference by two, and to use that result as your target MoE.
Let’s restrict our attention to the comparison of two means. If the expected difference between the two means is Cohens’s d = .80, the proposed strategy is to set your target MoE at f = .40, which means that your target MoE is set at .40 standard deviations. If you plan for this value of target MoE with 80% assurance, the recommended sample size is n = 55 participants per group. These results are guaranteed to be true, if it is known for a fact that Cohen’s d is .80 and all statistical assumptions apply.
But it is generally not known for a fact that Cohen’s d has a particular value and so we need to answer a non-trivial question: what effect size can we reasonably expect? And, how can we have assurance that the MoE will not exceed half the unknown true effect size? One of the many options we have for answering this question is to conduct a pilot study, estimate the plausible values of the effect size and use these values for sample size planning. I will describe a strategy that basically mirrors the sample size planning for power approach described by Anderson, Kelley, and Maxwell (2017).
The procedure is as follows. In order to plan with approximately 80% assurance, estimate on the basis of your pilot the 80% confidence interval for the population effect size and use half the value of the lower limit for sample size planning with 90% assurance. This will give you 81% assurance that assurance MoE is no larger than half the unknown true effect size.
There are two “problems” we need to consider when estimating the true effect size. The first problem is that there is at least 50% probability of obtaining an overestimate of the true effect size. If that happens, and we take the point estimate of the effect size as input for sample size planning, what we “believe” to be a sample size sufficient for 80% assurance will be a sample size that has less than 80% assurance at least 50% of the times. So, using the point estimate gives assurance MoE for the unknown effect size with less than 50% assurance.
To make it more concrete: suppose the true effect equals .80, and we use n = 25 participants in both groups of the pilot study, the probability is approximately 50% that the point estimate is above .80. This implies, of course, that we will plan for a value of f > .40, approximately 50% of the times, and so the sample we get will only give us 80% assurance 50% of the times.
The second problem is that the small sample sizes we normally use for pilot studies may give highly imprecise estimates. For instance, with n = 25 participants per group, the expected MoE is f = 0.5687. So, even if we accept 50% assurance, it is highly likely that the point estimate is rather imprecise.
Since we are considering a pilot study, one of the obvious solutions, increasing the sample size so that expected MoE is likely to be small, is not really an option. But what we can do is to use an estimate that is unlikely to be an overestimate of the true effect size. In particular, we can use as our estimate the lower limit of a confidence interval for the effect size.
Let me explain, by considering the 80% CI of the effect size estimate. From basic theory it follows that the “true” value of the effect size will be smaller than the lower limit of the 80% confidence interval with probability equal to 10%. That is, if we calculate a huge number of 80% confidence intervals, each time on the basis of new random samples from the population, the true value of the effect size will be below the lower limit in 10% of the cases. This also means that the lower limit of the interval has 90% probability to not overestimate the true effect size.
This means that if we take the lower limit of the 80% CI of the pilot estimate as input for our sample size calculations, and if we plan with assurance of .90, we will have 90%*90% = 81% assurance that using the sample size we get from our calculations will have MoE no larger than half the true effect size. (Note that for 80% CI’s with negative limits you should choose the upper limit).
Student of mine recently did a pilot study. This was a pilot for an experiment investigating the size of the effect of fluency of delivery of a spoken message in a video on Comprehensibility, Persuasiveness and viewers’ Appreciation of the video. The pilot study used two groups of size n = 10, one group watched the fluent video (without ‘eh’) and the other group watched the disfluent video where the speaker used ‘eh’ a lot. The dependent variables were measured on 7-point scales.
Let’s look at the results for the Appreciation variable. The (biased) estimate of Cohen’s d (based on the pooled standard deviation) equals 1.09, 80% CI [0.46, 1.69] (I’ve calculated this using the ci.smd function from the MBESS-package. According to the rules-of-thumb for interpreting Cohen’s d, this can be considered a large effect. (For communication effect studies it can be considered an insanely large effect). However, the CI shows the large imprecision of the result, which is of course what we can expect with sample sizes of n = 10. (Average MoE equals f = 0.95, and according to my rules-of-thumb that is well below what I consider to be borderline precise).
If we use the lower limit of the interval (d = 0.46), sample size planning with 90% assurance for half that effect (f = 0.23) gives us a sample size equal to n = 162. (Technical note: I planned for the half-width of the standardized CI of the unstandardized effect size, not for the CI of the standardized effect size; I used my Shiny App for planning assuming an independent groups design with two groups). As explained, since we used the lower limit of the 80% CI of the pilot and used 90% assurance in planning the sample size, the assurance that MoE will not exceed half the unknown true effect size equals 81%.
Here I would like to explain the procedure for sample size planning for one-way and two-way (factorial) between subjects designs. We will consider examples based on and described in Haans (2018).
The first example considers the effect of seating location of students on their educational performance. Seating location is defined as distance from the teacher and operationalized in terms of the row the student is seated in, with first row being the closest to the teacher and the fourth row being the furthest away. 20 Students are randomly assigned to one of the four possible rows, so N = 20, n = 5. The dependent variable is the course grade of the student. (Note: the data and study are hypothetical).
As Haans (2018) explains, one psychological theory explaining the effect of seating position on educational performance is based on social influence. This theory posits that due to the social influence of the teacher, the students that are seated closest to the teacher find themselves in a state of undivided attention. This undivided attention causes their educational performance to be better than the students who are seated further away.
In operational terms, then, we may expect that first row students will have a better average grade than students seated on the other rows. So, the quantitative research question we are interested in is:
“How much do the average grades differ between students seated first row and the students seated on other rows?”
We can estimate this quantity with a Helmert Contrast, where we assign a contrast weight of 1 to mean of the first row grades and weights -1/3 to the means of the grades in the other rows.
Haans (2018) gives us the following results. The contrast estimate equals 2.00 , 95% CI [0.27, 3.73]. In order to interpret this more easily, we divide this estimate by the square root Mean Square Error, to obtain the standardized estimate and standardized confidence interval (not to be confused with the confidence interval of the standardized estimate, but that’s a different story. The result is: 1.26, 95% CI [0.17, 2.36].
To answer the research question, the estimated difference equals 1.26 standard deviations, which according to rule-of-thumbs frequently used in psychology is a large difference. The CI shows the enormous amount of uncertainty of this estimate: population values between 0.17 (small) and 2.36 (very large) are also consistent with the observed data and our statistical assumptions. So, it seems safe to conclude that it looks like there is a positive effect of seating position, but the wide range of the CI makes it clear that the data do not tell us enough about the size of the effect, the precision is simply too low.
The precision is f = 1.09, which according to my rules-of-thumb is very imprecise (I consider f = 0.65, to be barely tolerable).
So, let’s plan for a replication study with a reasonably precise estimate of f = 0.40, with 80% assurance. (Note: for some advice on setting target Moe: Planning with assurance, with assurance. ) I’ve used the app: https://gmulder.shinyapps.io/PlanningFactorialContrasts/ with the default values for a single factor between subjects design with 4 conditions. According to the app, we need n = 36 participants per condition (making a total of N = 144).
(For more detailed information considering sample size planning for contrast analysis see: http://small-s.science/?p=10 and for some guidelines for setting target MoE: http://small-s.science/?p=14)
Our second example is also taken from Haans (2018). It considered the same phenomenon, the effect of students’ seating distance from the teacher and the educational performance of the students.
A second theory explaining the effect is that the effect is mainly caused by the teacher having decreased levels of eye contact with the students sitting farther to the back in the lecture hall.
To test that theory, a experiment was conducted with N = 72 participants attending a lecture. The lecture was given to two independent groups of 36 pariticpant. The first group attended the lecture while the teacher was wearing dark sunglasses, the second group attented the lecture while the teacher was not wearing sunglasses,. Again, all participants were randomly assigned to 1 of 4 possible rows. The dependent variable was the score on a 10-item questionnaire about the contents of the lecture.
Now, if the eyecontact of the teachter is the causal variable, we may expect that in this experimental setup the difference between the average score of the persons seated on the first row and the averages of the other rows will be smaller for the condition where the teacher wears sunglasses than for the condition in which the teacher does not wear these glasses, as wearing sunglasses prevents eye-contact between the teacher and the students. Our quantitative question is therefore:
“How much does the contrast between the first row and the others rows differ between the conditions with and without sunglasses?”
In other words, we are interested in the size of the interaction effect.
I’ve downloaded the dataset from http://pareonline.net/sup/v23n9.zip (between2by4data.sav) and specified the following syntax in SPSS:
UNIANOVA retention BY sunglasses location
/LMATRIX = “Interaction contrast” sunglasses*location 1 -1/3 -1/3 – 1/3 -1 1/3 1/3 1/3 intercept 0
/DESIGN= sunglasses*location.
The result of the analysis is that the contrast estimate equals 1.0, 95% CI [-0.33, 2.33]. If we standardize this with the within condition variance (the condition being the combination of the levels of the two factors), we get 0.82, 95% CI [-0.27, 1.90].
So, it appears that the difference between the means of the first row and that of the other rows is on average 1.0 points larger in the condition without sunglasses than in the condition with sunglasses. This corresponds to a large difference (dwith = .82). However, the CI also contains negative population difference (albeit that they are smallish), so even though the results are promising for the theory (eyecontact), these negative effects will not persuade a critical reviewer of the study. Indeed, these negative effects contradict the substantive hypothesis.
Again, the confidence interval is so wide, that effects ranging from small negative effects to huge positive effects are considered plausible. Since the results are promising for the theory, a replication study with more precision may be needed to persuade the critics. Let’s plan for a precision of f = .25 with 95% assurance.
I’ve used the app: https://gmulder.shinyapps.io/PlanningFactorialContrasts/ specifying that we have a factorial design with a = 2 levels and b = 4 levels. The result is that for the interaction contrast with f = .25 and assurance = .95, we need 175 participants per combination of the two factors. This means, that a total of N = 1400 must be recruited.
I’ve taken this from the following output.
 |
| Figure 1: Output of sample size planning |
I’ve looked at the “Contrast Summary Tab” to check that interaction A1B1 is the correct one (see Figure 2).
 |
| Figure 2. Summary of contrast weights. |
What’s important in the above figure is that the set of weights for A1B1 matches the set of weights used to get the contrast estimate in SPSS (In the LMATRIX-subcommand), so that’s how we know that A1B1 is the contrast we want. (Note: if you switch the number of levels in the app, that is, use 4 levels for A and 2 for B, the interaction weights will match perfectly).
Reference
Haans, Antal (2018). Contrast Analysis: A Tutorial. Practical Assessment, Research, & Education, 23(9). Available online: http://pareonline.net/getvn.asp?v=23&n=9
One of the most difficult aspects of sample size planning for precision is the specification of a target Margin of Error (MoE). Here, I would like to introduce a simple rule of thumb, in the hope that it helps you in determining a reasonable target MoE.
Here, the rule of thumb is applied to obtaining an estimate of the difference between two independent group means, where the two populations are normally distributed with equal variances.
Sample size planning starts with formulating a goal for the research. A very common goal is to try to determine the direction of an effect. For the goal of assessing the direction of an effect, it helps if the confidence interval of the difference contains only positive or negative values. That is, you want a confidence interval that exludes the value 0, for if that value is included, you would probably conclude that the estimate is consistent with both positive and negative effects. Thus, our first goal is to obtain a confidence interval of the mean difference that excludes the value 0.
Now, a confidence interval excludes 0, if obtained MOE is at most equal to the obtained effect size estimate. Suppose that the estimate equals the true effect of say, 0.50, we want MOE to be at most very close to 0.50, otherwise 0 will be included in the interval. But if our estimate underestimates the true effect, say the estimate equals 0.30, we want MOE to be at most very close to 0.30. Likewise, if we overestimate the effect, MOE can be larger than 0.50.
This means that we cannot say, for instance, we expect that the true effect is .50, so let’s plan for a target MOE that with 80% assurance is at most .50, because this target MOE may be too large for underestimates of the true effect, depending on the extent to which the effect is underestimated. So, in specifying target MOE, we should take into account that underestimates of the effect size occur. (Actually, these underestimates occur with a relative frequency of 50% in a huge collection of direct replications). We can say that we do not only want to exclude zero from the interval, but also that we want that to occur in a large proportion of direct replications. This will be our second goal. I will call the probabiity associated with our second goal, the probability of exclusion (PE)
The rule of thumb is that if we want 80% probability that a random confidence interval excludes zero, we should plan for an expected MOE equal to f = d / √2. (the square root sign is unreadable in my browser; so in words: the effect size divided by the square root of 2; with mathjax: $f = d / sqrt{2}$). Since there is 50% probability that obtained MOE will be larger than expected MOE, this is equal to planning for target MOE = f = d / √2, with 50% assurance or simply without assurance. You can do this in the ESCI-software, but also with the R-functions provided below.
The first example in the code below, is an illustration of planning for assessing the direction of the effect, with true effect size d = .50. If we want 80% assurance to have only positive values in our confidence interval, we should plan for a target MoE = expected MoE = f = d / √2 = 0.3535. Using the SampleSize-function below, this gives a sample size n = 63, or total sample size = N = 2*63 = 126. The probability that the confidence interval excludes 0 equals approximately 80% (p = 0.7951). So, the rule of thumb of planning for d / √2, seems to work pretty good.
If your research goal is to estimate the value of the effect size in stead of its direction, the rule of thumb can be used as follows. Suppose we do not know the true effect size, but want to have 80% assurance that we have a high probability to be able to distinguish between small (d = .20) and large effects (d = .80). That is, if the true effect is .20 we want the value .80 to be excluded from the confidence interval and if the true effect is .80, we want the value .20 to be excluded from the confidence interval.
We can proceed as follows, the difference between the effect sizes is .80 – .20 = .60. We use this value to determine target MOE. Thus, if we now plan for a target MoE = expected Moe = d / √2), we should have approximately 80% PE that obtained MoE will exclude 0.80 if the true effect is 0.20 and vice versa. The functions below give sample size n = 44, and the probablity of exclusion equals .7947. So, our rule of thumb, seems to work pretty good again. See example 2 in the code below.
Alternatively, we could take the region of practical equivalence (ROPE) into account. Suppose, our equivalence range equals .10 sigma. If we want to have enough precision to distinguish large from small effects, we should plan as follows. We take the difference between a large effect and the upper equivalence value of a small effect or, equivalently, the difference between a small effect and the lower equivalence vaue of a large effect, i.e. .50, and plan for f = .50 / √2. If the effect is large we expect a confidence interval that excludes the equivalence range for the small effect (and vice versa), with 80% probability of exclusion.
But we could also take the difference between the lower equivalence value of a large effect and the upper equivalence value of a small effect, i.e. .40, and plan for f = .40/√2. (See the third example in the code below) This will give us 80% PE that any true value within the ROPE of the one effect will exclude values in the ROPE of the other. For example, if the true effect is .70, and expected MOE equals .40/√2 = .2828, there is approximately 80% probability that the 95% CI excludes .30, which is in the ROPE of a small effect. The expected CI will be .70 +/- .2828 = [0.4172, 0.9828]. Note that the lower limit is larger than the upper limit of the ROPE for d = .20, as we want it to be. Note, however, that if the true effect is small (d = .20), the CI will exclude effects equivalent to large effects, which is consistent with our research goal, but it will not exlude the value 0 or effects equivalent to a medium effect. Indeed, the expected CI will be [-0.0828, 0.4828]. (This is not a problem, of course, since this was not the purpose of our research)
As a final example, suppose we want sufficient precision to distinguish small from medium effects (or large from medium effects). If we take the ROPE perspective, with an equivalence range of +/- .10 sigma, the lower equivalence value of the medium effect equals .50 – .10 = .40 and the upper limit of the small effect equals .30. If we want 80% assurance that the CI will be small enough to distinguish small from medium effects, we should plan for expected MOE f = (.40 – .30)/√2 = 0.0707. Using the functions below, this requires a sample size n = 1538. (See the final example in the code below).
In summary, the rule of thumb is to divide the effect size d by √2 and plan for an expected MoE equal to this value. This will give you a sample size that gives approximately 80% assurance that the CI will not contain 0. In the case of distinguishing effect sizes, one option is to divide the difference between the lower equivalence value of the larger effect and the upper equivalence value of the smaller effect by the square root of 2 and plan for an expected MoE equal to this value. This will give you a sample size that gives approximately 80% PE that the CI of the estimated true value of one effect excludes the values in the ROPE of the other effect.
Do you want at least 90% PE? Use the square root of three, in stead of the square root of two, in determining target MoE.
eMoe = function(n) {
eMoe = qt(.975, 2*(n - 1))*sqrt(2/n)
return(eMoe)
}
cost <- function(n, tMoe) {
(eMoe(n) - tMoe)^2
}
sampleSize <- function(tMoe) {
optimize(cost, interval=c(10, 5000), tMoe = tMoe)$minimum
}
# FIRST EXAMPLE
# plan for 80% assurance of excluding 0
# i.e. estimate the direction if true effect
# equals .50
d = .50
#application of rule of thumb:
f = .50 / sqrt(2)
#sampleSize (uses ceiling() to round up):
n = ceiling(sampleSize(f))
n
## [1] 63
# Probabiity of Exclusion (here taken to be equivalent to # power for two-sided t-test (since true direction is unknown)) df = 2*(n - 1) ncp = f / sqrt(1/n) #or ncp = d / sqrt(2/n) pt(qt(.025, df), df, ncp) + 1 - pt(qt(.975, df), df, ncp)
## [1] 0.7951683
# SECOND EXAMPLE: # distinguish between small and large effect sizes: d = .80 - .20 f = d / sqrt(2) n = ceiling(sampleSize(f)) n
## [1] 44
df = 2*(n - 1) ncp = f / sqrt(1/n) #or ncp = d / sqrt(2/n) #PE: pt(qt(.025, df), df, ncp) + 1 - pt(qt(.975, df), df, ncp)
## [1] 0.79467
# EXAMPLE 3: distinguish small and large with ROPE # ROPE small and large: rope.small = c(.10, .30) rope.large = c(.70, .90) d = rope.large[1] - rope.small[2] f = d / sqrt(2) n = ceiling(sampleSize(f)) n
## [1] 98
df = 2*(n - 1) ncp = f / sqrt(1/n) #or ncp = d / sqrt(2/n) #PE: pt(qt(.025, df), df, ncp) + 1 - pt(qt(.975, df), df, ncp)
## [1] 0.7956414
# Example 4: distinguish medium from small # or medium from large with ROPE rope.medium = c(.40, .60) d = rope.medium[1] - rope.small[2] f = d / sqrt(2) n = ceiling(sampleSize(f)) n
## [1] 1538
df = 2*(n - 1) ncp = f / sqrt(1/n) #or ncp = d / sqrt(2/n) #PE: pt(qt(.025, df), df, ncp) + 1 - pt(qt(.975, df), df, ncp)
## [1] 0.7916783
In this post, I will introduce some of the ideas underlying sample size planning for precision. The ideas are illustrated with a shiny-application which can be found here: https://gmulder.shinyapps.io/PlanningApp/. The app illustrates the basic theory considering sample size planning for two independent groups. (If the app is no longer available (my allotted active monthly hours are limited on shinyapps.io), contact me and I’ll send you the code).
In general, then, you should try to show (to others, if not to yourself) that it is reasonable to assume normally distributed populations, with equal variances and random sampling, before you decide that the p-value of your t-test, the width of your confidence interval, and the results of sample size calculations are believable.
The populations in the app are normal distributions. By default, the app shows two such distributions. One of the distributions, the one I like to think about as corresponding to the control condition, has μ = 0, the other one has μ = 0.5. Both distributions have a standard deviation (σ = 1). The standardized difference between the means is therefore equal to δ = 0.50.
The default populations are presented in Figure 1 below.
 |
| Figure 1: Two normal distributions. The distribution to the left has μ = 0, the one to the right has μ = 0.5 The standard deviation in both distributions equals σ = 1. The standardized difference δ and the unstandardized difference between the means both equal 0.50. |
The other default setting in the app is a sample size (per group) of n = 20. From the sample size and the specification of the populations, we can deduce the probability density of the different values of the estimates of the difference between the population means. The estimate is simply the difference between the sample means.
This so-called sampling distribution of the mean difference is depicted on the tab next to the population. Figure 2 shows what the sampling distribution looks like if we repeatedly draw random samples of size n = 20 per group from our populations and keep track of the difference between the sample means we get in each repetition.
 |
| Figure 2: Sampling distribution of the difference between two sample means based on samples of n = 20 per group and random sampling from the populations described in Figure 1. |
Note that the mean of the sampling distribution equals 0.5 (as indicated by the middle vertical line). This is of course the (default) difference between the population means in the app. So, on average, estimates of the population difference equal the population difference.
The lines to the left and the right of the mean indicate the mean plus or minus the Margin of Error (MOE). The values corresponding to the lines are 0.5 ± MOE. 95% of estimates of the population mean difference have a value between these lines.
Conceptually, the purpose of planning for precision is to decrease the (horizontal) distance between these lines and the population mean difference. In other words, we would like the left and right lines as close to the mean of the distribution as is practically acceptable and possible.
The tab next to the sampling distribution tab contains a figure representing the sampling distribution of the t-statistic. The sampling distribution of t can be deduced on the basis of the population values and the sample size. In the app, it is assumed that t is calculated under the assumption that the null-hypothesis of zero difference between the means is true. The sampling distribution of t is what you get if you repeatedly sample from the populations as specified, calculate the t-statistic and keep a record of the values of the t-statistic.
The sampling distribution of the t-statistic presented in Figure 3 contains two vertical lines. These lines are located (horizontally) on the value of t that would lead to rejection of the null-hypothesis of equal population means. In other words, the lines are located at the critical value of t (for a two-tailed test).
 |
| Figure 3: Distribution of the t-statistic testing the null-hypothesis of equal population means. The distribution is based on sampling from the populations described in Figure 3. The sample size is n = 20 per group. The lines represent the critical value of t for a two sided t-test. The area between the vertical lines is the probability of a type II error. The combined areas to the left of the left line and to the right of the right line is the power of the test. |
The area between the lines is the probability that the null-hypothesis will not be rejected. In the case of a true population mean difference (which is the default assumption in the app), that probability is the probability of an error of the second kind: a type II error.
The complement of that probability is called the power of the test. This is, of course, the area to the left of the left vertical line added to the area to the right of the right vertical line. Conceptually, the power of the test is the probability of rejecting the null-hypothesis when in fact it is false.
Figure 3 clearly demonstrates that if the true mean difference equals 0.50 and the sample size (per group) equals n = 20, that there is a large probability that the null-hypothesis will not be rejected. Actually, the probability of a type II error equals .66. (So, the power of the test is .34).
With respect to sample size planning for precision, the app by default takes half of a standard deviation (f = .50) as the target MOE. Besides, planning is with 80% assurance. This means that the default settings search for a sample size (per group), so that with 80% probability MOE will not exceed 0.50 (Note that the default value of the standard deviation is 1, so an f of .50 corresponds to a target MOE of 0.50 on the scale of the data; Likewise, were the standard deviation equal to 2, an f of .50 would correspond to a target MOE of 1.0).
As described above, planning with the default values gives us a sample size of n = 37 per group, with an expected MOE of 0.46. In the tab next to the planning results, a figure displays what you can expect to find on average, given the planned sample size and the specification of the population. That figure is repeated here as Figure 4.
 |
| Figure 4: Expected results in terms of point and interval estimates (95% confidence intervals). This is what you will find on average given the population specification in Figure 1 and using the default values for sample size planning. |
Figure 4 displays point and interval estimates of the group means and the difference between the means. The interval estimates are 95% confidence intervals. The figure clearly shows that on average, our estimate of the difference is very imprecise. That is, the expected 95% confidence interval ranges from almost 0 (0.50 – 0.46 = 0.04) to almost 1 (0.50 + 0.46 = 0.96). Of course, using n = 20, would be worse still.
A nice thing about the app (well, I for one think it’s pretty cool) is that as soon as you ask for the sample sizes, the sample size in the set population values form is automatically updated. Most importantly, this will also update the sampling distribution graphs of the difference between the means and the t-statistic. So, it provides an excellent way of showing what the updated sample size means in terms of MOE and the power of the t-test.
Let’s have a look at the sampling distribution of the mean difference, see Figure 5.
 |
| Figure 5: Sampling distribution of the mean difference with n = 37 per group. Compare with Figure 2 to see the (small) difference in the Margin of Error compared to n = 20. |
If you compare Figures 5 and 2, you see that the vertical lines corresponding to the mean plus and minus MOE have shifted somewhat towards the mean. So here you can see, that almost doubling the sample size (from 20 to 37) had the desired effect of making MOE smaller.
I would like to point out the similarity between the sampling distribution of the difference and the expected results plot in Figure 4. If you look at the expected results for our estimate of the population difference, you see that the point estimate corresponds to the mean of the sampling distribution, which is of course equal to the populations mean difference and that the limits of the expected confidence interval correspond to the left and right vertical lines in Figure 5. Thus, on average the limits of the confidence interval correspond to the values that mark the middle 95% of the sampling distribution of the samples mean difference.
Since we specified an assurance of 80%, there is an 80% probability that in repeated sampling from the populations (see Figure 1) with n = 37 per group, our (estimated) MOE will not exceed half a standard deviation. Thus, whatever the true value of the populations mean difference is, there is a high probability that our estimate will not be more than half a standard deviation away from the mean. This is, I think, one of the major advantages of sample size planning for precision: we do not have to specify the unknown population mean difference. This is in contrast to sample size planning for power, where we do have to specify a specific population mean difference.
Speaking of power, the results of the sample size planning suggest that for our specification of the populations mean difference (Cohen’s delta = 0.50) the power of the test equals 0.56. Thus, there is a probability of 56% that with n = 37 per group the t-test will reject. The probability of a type II error is therefore 44%.
Figure 6 shows the distribution of the t statistic with n = 37 per group and a standardized effect size of 0.50.
 |
| Figure 6. The distribution of the t-statistic testing the null-hypothesis of equal population means. The distribution is based on the population specification in Figure 1 and sample sizes of n = 37 per group, with true effect size equal to 0.50. The probability of a type II error is the area of under the curve between the two vertical lines. The power is the area under the curve beyond the two lines. Compare with Figure 3 to see the differences in these probabilities compared to n = 20. |
Now suppose that the unstandardized mean difference between the population means equals 2 and that the standard deviation equals 2.5. I just filled in the set population values form, setting the mean of population 2 to 2.0 and the standard deviation to 2.5. And I clicked set values.
Let us plan for a target MOE of f = 0.5 standard deviations with 80% assurance. Click get sample sizes in the sample size planning form. In this case, target MOE equals 1.25.
The results are not very surprising. Since the f did not change compared to the previous time, the results as regards the sample size are exactly the same. We need n = 37. Again, this is what I like about sample size planning, no matter what the unknown situation in the population is, I just want my margin of error to be no more than half a standard deviation (for example).
But the power did change (of course). Since the standardized population mean difference is now 0.80 (= 2.0 / 2.5) in stead of 0.50, and all the other specifications remained the same, the power increases from 56% to 92%. That’s great.
However, the high probability of rejecting the null-hypothesis does not mean that we get precise estimates. On average, the point estimate of the difference equals 2 and the 95% confidence limits are 0.85 and 3.15 (the point estimate plus or minus 0.46 times the standard deviation of 2.5). See Figure 7.
 |
| Figure 7: Expected results using n = 37 when sampling from two normal populations with equal standard deviations (σ = 2.5) and mean difference of 2.0. The standardized effect size equals 0.80. Note the imprecision of the estimates even though the power of the t-test equals .92. |
In short, even though there is a high probability of (correctly) rejecting the null-hypothesis of equal population means, we are still not in the position to confidently conclude what the size of the difference is: the expected confidence interval is very wide.
In a previous post (here), we saw how we can determine sample size for obtaining, with assurance, a precise interaction contrast estimate. In that post we considered a 2 x 2 factorial design. In this post, I will extend the discussion to the mixed model case. That is, we will consider sample size planning for a precise interaction estimate in case of a design with 2 fixed factors and two random factors: participant and stimulus (item). (A pdf version of this post can be found here: view pdf. )
In order to keep things relatively simple, we will focus on a design where both participants and items are nested under condition. So, each treatment condition has a unique sample of participants and items. We will call this design the both-within-condition design (see, for instance, Westfall et al. 2014, for detailed descriptions of this design). We will analyse the 2 x 2 factorial design as a single factor design (the factor has a = 4 levels) and formulate an interaction contrast.
Let’s start with p participants and q stimuli. We randomly assign n= p/a participants and m = q/a stimuli to each of the a treatment levels. The ANOVA table for the design is presented in Table 1.
UNIANOVA score BY cond pp ss /RANDOM=pp ss /METHOD=SSTYPE(3) /INTERCEPT=INCLUDE /CRITERIA=ALPHA(0.05) /DESIGN=cond pp WITHIN cond ss WITHIN cond /CONTRAST(cond) = SPECIAL(1, -1, -1, 1).
 |
| Figure 1: SPSS GLM ANOVA output |
Take a look at Table 1 and focus on the Participant row. The expected Mean Square (EMS) associated with Participant is 
. Now, suppose that due to some freak accident of nature there are no differences in the mean scores (averaged of stimuli) of each participant. In that case, 
. This means that under these circumstances the expected mean square associated with participants is simply an estimate of the error variance with 
degrees of freedom, because 
, if . Of course, the other estimate of the Error variance is MSError and this estimate is based on 
degrees of freedom. The logic of the F-test is that under the null-hypothesis, in our case that , the ratio of these two estimates of the error variance follows an F-distribution with and degrees of freedom.
Now focus on the Treatment row in Table 1. The expected mean square associated with Treatment equals 
. If we now suppose that there is no difference between the treatment means, that is 
, MSTreatment does not estimate 
, but 
. Note that no other source of variance has an expected mean square that is equal to the latter figure. That is, in contrast to our test of the Participant factor, where under the null-hypotheses two Mean Squares estimate the error variance, i.e. MSParticipant and MSError, no mean square is available to form an F-ratio to test the Treatment effect.
But a linear combination of MSParticipant, MSStimulus and MSError, does provide an estimate with expected value . Namely, the sum of the participant and stimulus mean squares minus mean square error: ![[m\sigma^2_p + \sigma^2_e] + [m\sigma^2_s + \sigma^2_e] - [\sigma^2_e] = m\sigma^2_p + n\sigma^2_s + \sigma^2_e](https://small-s.science/wp-content/ql-cache/quicklatex.com-9cd6bb15506816413e74641fee23cf13_l3.png "Rendered by QuickLaTeX.com")
. It is exactly this linear combination of mean squares that is used in the F-ratio to obtain an error term against which to test the Treatment effect in Figure 1: 
. We will also use this figure to obtain the variance (and standard error) of our contrast estimate.
If you take a closer look at Figure 1, in particular the degrees of freedom column, you will notice that the degrees of freedom associated with the error term that is used to test the Treatment effect is a fractional number and not a nice round number that you would expect to get if you only consider the degrees of freedom in Table 1. The cause of this fractional number is that we cannot simply use the degrees of freedom of the mean square used to test the treatment effect, because that mean square does not exist. Indeed, we had to combine three mean squares in order to obtain an estimate of the error term for the Treatment effect. The consequence of this is that we will have to use an approximation of the degrees of freedom associated with that error term.
SPSS (and my precision app) use the Satterthwaite procedure to approximate the degrees of freedom of the error term. That approximation is as follows (notice that the numerator is equal to the linear combination of mean squares used to obtain the error term).
![\[df=\frac{(MSp+MSs-MSe)^{2}}{\frac{MSp^{2}}{df_{p}}+\frac{MSs^{2}}{df_{s}}+\frac{MSe^{2}}{df_{e}}}.\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-13f753be88a6e13890953c2465e22a05_l3.png "Rendered by QuickLaTeX.com")
Thus, using the results in Figure 1.
MSp = 6.403 MSs = 10.137 MSe = 1.470 dfp = 44 dfs = 20 dfe = 220 df = (MSp + MSs - MSe)^2 / (MSp^2/dfp + MSs^2/dfs + MSe^2/dfe) df
## [1] 37.35559
Now that we have obtained the error variance of a treatment effect by using a linear combination of mean squares and a Satterthwaite approximation of the degrees of freedom we are able to figure out the margin of error (MOE) of our contrast estimate. Just as in the simple between subjects design we discussed previously we obtain MOE by multiplying the standard error of the estimate with a critical value of t. The critical value of t is the .975 quantile of the central t-distribution with the Satterthwaite approximated degrees of freedom (if you are looking for something other than a 95% confidence interval, you will have to use another critical value, of course). The following code gives the critical value of t for a 95% confidence interval (change the value of C if you want something other than a 95% interval).
C = .95 alpha = 1 - C critT = qt(1 - alpha/2, df) critT
## [1] 2.025542
The standard error of the contrast estimate 
can be obtained as follows.
![\[\hat{\sigma}_{\hat{\psi}}=\sqrt{\sum c_{i}^{2}\hat{\sigma}_{\bar{X},Rel}^{2}},\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-489e315828278a9811848f3ce71bf10f_l3.png "Rendered by QuickLaTeX.com")
where I have used the symbol 
to refer to the relative error variance of the treatment mean (which in this design is equal to the absolute error variance, but that’s another story), and 
refers to the contrast weight of treatment mean i. The relative error variance of the treatment mean is obtained by dividing the error variance that is used to test the treatment effect by the total number of observations in each treament, 
. Thus, using the results in Figure 1.
![\[\hat{\sigma}_{\bar{X},Rel}^{2}=\frac{MS_{p}+MS_{s}-MS_{e}}{nm}=\frac{15.07}{72}=0.2093.\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-0d81fe2ec65ea2c26c84682713e499bf_l3.png "Rendered by QuickLaTeX.com")
If we want to estimate an interaction contrast for the 2 x 2 design, we may, for example, specify contrasts weights 
. Let’s use the results in Figure 1 to calculate what MOE is for this particular contrast.
#sample sizes per treatment n = 12 m = 6 #obtained mean squares (see Figure 1): MSp = 6.403 MSs = 10.137 MSe = 1.470 #Relative error variance: VarT = (MSp + MSs - MSe) / (n*m) #contrast weights: weights = c(1, -1, -1, 1) #standard error of contrast estimate SEcontrast = sqrt(sum(weights^2)*VarT) #Satterthwaite degrees of freedom: dfp = 44 dfs = 20 dfe = 220 df = (MSp + MSs - MSe)^2 / (MSp^2/dfp + MSs^2/dfs + MSe^2/dfe) #critical T critT = qt(.975, df) #Margin of Error MOE = critT * SEcontrast SEcontrast; MOE
## [1] 0.9149985
## [1] 1.853368
Even though SPSS GLM Univariate allows you to specify a mixed model design and tests the treatment effect with a linear combination of mean squares, the procedure does not use the correct error variance if you want to estimate the value of a contrast (using the CONTRAST subcommand), it uses MSError instead. In our example, then, SPSS uses an error variance that is an order of magnitude smaller than the correct error variance: 
, with 220 degrees of freedom and not 
, with 37.357 (see Figure 1). The consequence of this is, of course, that the 95% CI is much narrower than it should be.
Running the syntax above Figure 1 gives the output in Figure 2. The results can be reproduced as follows. The standard error of the contrast is the result of 
, the critical value of t is the .975 quantile of the central t distribution with 
, which equals 
. The value of MOE is therefore 
. With a contrast estimate of 
, the 95% CI equals ![-0.587 \pm 0.5633 = [-1.1503, -0.0237]](https://small-s.science/wp-content/ql-cache/quicklatex.com-ea5612615f1baf096374644a223d775f_l3.png "Rendered by QuickLaTeX.com")
. In comparison, using the correct value of MOE gives us ![[−2.4404, 1.2664]](https://small-s.science/wp-content/ql-cache/quicklatex.com-ef97651e8790dc304790e8a8724ebed7_l3.png "Rendered by QuickLaTeX.com")
.
 |
| Figure 2: SPSS GLM Univariate Contrasts Output |
MIXED score BY cond /CRITERIA=CIN(95) MXITER(100) MXSTEP(10) SCORING(1) SINGULAR(0.000000000001) HCONVERGE(0, ABSOLUTE) LCONVERGE(0, ABSOLUTE) PCONVERGE(0.000001, ABSOLUTE) /FIXED= cond | SSTYPE(3) /METHOD=REML /TEST= 'interaction' cond 1 -1 -1 1 /RANDOM=INTERCEPT | SUBJECT(pp) COVTYPE(VC) /RANDOM=INTERCEPT | SUBJECT(ss) COVTYPE(VC).
 |
| Figure 3: Contrast Estimate SPSS Mixed |
Even though the result in Figure 3 is hard to interpret without substantive detail (the data are made up) it is clear that the precision of the estimate is, well, suboptimal. As an indication: the estimated within treatment standard deviaion is about 1.74, so the estimated difference between differences (interaction) is close to a value of Cohen’s d of 
, approximate 95% CI ![[-1.40, 0.73]](https://small-s.science/wp-content/ql-cache/quicklatex.com-867264624a612f401b2c1cea536372a9_l3.png "Rendered by QuickLaTeX.com")
, which according to the rules of thumb is a medium negative effect, but consistent with anytihing from a huge negative effect to a large positive effect in the population, as the approximate CI shows. (I have divided the point estimate and the confidence interval in Figure 2 by 1.74, to obtain Cohen’s d and an approximate confidence interval). Clearly, then, our precision can be optimized.
Suppose that you are very fond of the both-within-condition design (BwC-design) and you plan to use it again in a replication study, You could of courseopt for a design with better expected precision, but based on the data and estimates at hand, that involves a lot of assumptions, but I will show you how you can do it in one of the next posts. If you plan for precision using the BwC-design, you need the following ingrediënts.
1. A figure for your target MOE. Let’s set target MOE to .40.
2. A specification of the percentage of assurance. Let’s say we want 80% assurance that target MOE will not exceed .40.
3. Estimates (or guesstimates) of the person variance 
, the stimulus variance 
, and the error variance . We will have a look in the next section.
4. Functions for calculating the relative error variance, degrees of freedom, MOE and determining the required sample sizes for Participants and Stimuli. These are all present in the Precision App, so I will use the application, but I will show how the results of the sample sizes relate to the information above.
We need to specify the values of three variance components. These variance components can be estimated on the basis of the mean squares and sample sizes obtained with SPSS GLM Univariate, we can use SPSS MIXED to obtain direct estimates or any other way to estimate variance components, such as GLM VARCOMPS (which has several estimation procedures). I like to use SPSS MIXED or LME4. and not a dedicated program for variance components, because most of the times the main purpose of the analysis I am doing is obtaining contrast estimate or F-tests, so most of the times variance components estimates are a handy by-product of my main analysis. For demonstrative purposes, I will show how it can be done with the GLM univariate output and I will show how the results match those of SPSS MIXED.
Take a look at Figure 1. The estimate of is simply MS(Error) = 1.47. For obtaining an estimate for the variance component associated with participants, we set the obtained mean square equal to the expected mean square (see Table 1). Thus, 
. Rearranging and using 1.47 as an estimate for leads to 
. Likewise, the estimate for 
. Thus, our estimates are 
, 
, and 
.
In order to obtain direct estimates you can use SPSS Mixed (or GLM Varcomps, or whatever you like). If you run the SPSS syntax in Figure 3, you will find estimates of the variance components under the heading Covariance Parameters in your SPSS output. See Figure 4. Note that the standard errors are pretty large, so the point estimates are not very precise. But since it is the only information we have, we will consider the point estimates to be the best we have.
 |
| Figure 4: Variance components estimates |
(Note: it is not necessary to fill it in in contrast 3).
, for assurance the value .80 and the values , 
, and 
for, respectively, Residual variance, Participant intercept variance and Stimulus intercept variance. Fill in the value 0 for all the other variances. See Figure 5. |
| Figure 5: Setting values in the Precision App. |
 |
| Figure 6a: Output for planning target MOE = 0.40 |
 |
| Figure 6b: Outpur for planning target MOE = 0.50 |
, to refer to the residual variance in the BwC-design, we can say 
. Normally, the precision app sums these two components to get a value for the residual variance in the BwC-design, and you will obviously get the same result if you specify the residual variance as the sum and the participant-by-stimulus variance as 0. Likewise, 
, and 
, where 
and 
are the variances associated with the interaction of treatment and participant and treatment and stimulus, respectively.
, and there is 80% assurance that MOE will not exceed 
. Note, again, that the assurance MOE is somewhat larger than target MOE, because a sample of 804 participants requires a sample of more than 500 stimuli to get the target MOE with 80% assurance and 500 stimuli is the maximum number of stimuli the app considers when minimizing the number of participants. |
| Figure 7: Expected and Assurance MOE for the interaction contrast (contrast 3) using 804 participants and 500 stimuli |
, 
, 
, 
, 
, and 
. The Satterthwaite degrees of freedom are 
. The standard error of the contrast equals 
. The critical value for t equals 
. Expected MOE is, therefore, 
(the tiny difference with the results from the app is due to rounding errors).
-distribution. That is, we assume with assurance 
, that the quantile of the sampling distribution of the relative error error variance is 
. Now, the degrees of freedom are 
, the assurance 
, and the .80 quantile of with 1092.66 degrees of freedom equals 
. Since the relative error variance equals 
, the .80 quantile of the error variance equals 
. And this means that assurance MOE equals 
. Again, the difference with the results from the Precision App are due to rounding error. In this post, I will show you a way of determining a sample size for obtaining a precise estimate of the slope 
of the simple linear regression equation 
. 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.
, where 
, is the .975 quantile of the central t-distribution with 
degrees of freedom, and 
is the standard error of the estimate of . is 
(Wilcox, 2017): the variance of Y given X divided by the sum of squared errors of X. The variance 
equals 
, the variance of Y multiplied by 1 minus the squared population correlation between Y and X, and it is estimated with the residual variance 
, where 
. (1) ![\[\hat{\sigma}_{\hat{\beta_{1}}}^{2}=\frac{\sum(Y-\hat{Y})^{2}/df_{e}}{\sum(X-\bar{X})^{2}}. \]](https://small-s.science/wp-content/ql-cache/quicklatex.com-712f2857df79a9cc19ad5198431f2877_l3.png "Rendered by QuickLaTeX.com")
-distribution: ![\[\frac{\sum(Y-\hat{Y})^{2}}{\sigma_{y}^{2}(1-\rho^{2})}\sim\chi^{2}(df_{e}),\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-5fb82f0623b2088e7771fbf7e7d4803c_l3.png "Rendered by QuickLaTeX.com")
![\[\frac{\sum(Y-\hat{Y})^{2}}{df_{e}}\sim\frac{\sigma_{y}^{2}(1-\rho^{2})\chi^{2}(df_{e})}{df_{e}}.\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-6f1e262a579b65243e1854e59a4243d6_l3.png "Rendered by QuickLaTeX.com")
![\[\frac{\sum(X-\bar{X})^{2}}{\sigma_{X}^{2}}\sim\chi^{2}(df),\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-fe2f47a4c99a2963559b96659fb59616_l3.png "Rendered by QuickLaTeX.com")
, therefore ![\[\sum(X-\bar{X})^{2}\sim\sigma_{X}^{2}\chi^{2}(df).\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-e793ae0023912bb5ee2c4d42856eb6a8_l3.png "Rendered by QuickLaTeX.com")
, and multiplying by 1 (
). ![\[df\sigma_{X}^{2}\sim df\sigma_{X}^{2}\chi^{2}(df)/df.\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-0f86831fd6d9113dcff83251b70eb992_l3.png "Rendered by QuickLaTeX.com")
distributions, one with degrees of freedom, and one with degrees of freedom. Or something like: ![\[\hat{\sigma}_{\hat{\beta_{1}}}^{2}\sim\frac{\sigma_{y}^{2}(1-\rho^{2})\chi^{2}(df_{e})/df_{e}}{df\sigma_{X}^{2}\chi^{2}(df)/df}=\frac{\sigma_{y}^{2}(1-\rho^{2})}{df\sigma_{X}^{2}}\frac{\chi^{2}(df_{e})/df_{e}}{\chi^{2}(df)/df}=\frac{\sigma_{y}^{2}(1-\rho^{2})F(df_{e,}df)}{df\sigma_{X}^{2}},\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-cdc57f8bf40a827845ec2f30dcfb8751_l3.png "Rendered by QuickLaTeX.com")
(2) ![\[\hat{MOE}\sim t_{.975}(N-2)\sqrt{\frac{\sigma_{y}^{2}(1-\rho^{2})F(N-2,N-1)}{(N-1)\sigma_{X}^{2}}}. \]](https://small-s.science/wp-content/ql-cache/quicklatex.com-f0b396b6b94f61a97099bec31546363b_l3.png "Rendered by QuickLaTeX.com")
, 
, 
, 
, 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
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)$coefficients[4]
}
MOE = qt(.975, 98)*se
quantile(MOE, .80)
## 80% ## 0.1878628
vary = 1
varx = 1
rho = .5
assu = .80
tMOE = .10
MOE.assu = function(n, vary, varx, rho, assu) {
varY.X = vary*(1 - rho^2)
dfe = n - 2
dfx = n - 1
t = qt(.975, dfe)
q.assu = qf(assu, dfe, dfx)
MOE = t*sqrt(varY.X*q.assu/(dfx * varx))
return(MOE)
}
cost = function(x, tMOE) {
cost = (MOE.assu(x, vary=vary, varx=varx, rho=rho, assu=assu)
- tMOE)^2
}
#note samplesize is at least 40, at most 5000.
#note that since we already know that N = 100 is not enough
#in stead of 40 we might just as well set N = 100 at the lower
#limit of the interval
(samplesize = ceiling(optimize(cost, interval=c(40, 5000),
tMOE = tMOE)$minimum))
## [1] 321
#check the result: MOE.assu(samplesize, vary, varx, rho, assu)
## [1] 0.09984381
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
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.
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.
Let us get back to some of the SPSS output we considered in the previous post to get the ingredients we need for sample size planning. First, the ANOVA table.
 |
| Table 1. ANOVA source table |
We are interested in estimating and optimizing the precision of an interaction contrast estimate. The first things we need are an expression of the error variance needed to calculate the standard error of the estimate and the degrees of freedom that were used in estimating the error variance. In general, the error variance needed is the same error variance you would use in performing an F-test for the specific effect, in this case the interaction effect.
Thus, we note the error variance used to test the interaction effect, i.e. mean square error, and the degrees of freedom. The value of mean square error is 3.324, and the degrees of freedom are 389. Note that this value is the total sample sizes minus the number of conditions (393 – 4 = 389), or, equivalently, the total sample sizes minus the degrees of freedom of the intercept, the main effects, and the interaction (393 – (1 + 1 + 1 + 1) = 389). I will call these degrees of freedom the error degrees of freedom, dfe.
MOE can be obtained by multiplying a critical t-value with the same degrees of freedom as the error degrees of freedom with the standard error of the estimate.
The standard error of the contrast estimate is
![\[\hat{\sigma_\psi}= \sqrt{\sum{c_i^2MS_e/n_i}},\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-e711016f1b800409ee4d1632159091d7_l3.png "Rendered by QuickLaTeX.com")
where is the contrast weight for the i-th condition mean, and 
the number of observations (in our example participants) in treatment condition i. Note that 
is the variance of treatment mean i, the square root of which gives the familiar standard error of the mean.
The contrast weights we used to estimate the 2 x 2 interaction were {-1, 1, 1, -1}. So, the expression for MOE becomes
![\[MOE = t_{.975}(df_e)\sqrt{\sum{c_i^2MS_e/n_i}}=t_{.975}(df_e)\sqrt{4MS_e/n_i} = 2t_{.975}(df_e)\sqrt{MS_e/n_i}.\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-c566353662177824c69974b4d59785d4_l3.png "Rendered by QuickLaTeX.com")
Thus, suppose we have the independent 2×2 factorial design, 
, and the true value of Mean Square Error is 3.324, then MOE for the contrast estimate equals
![\[MOE = 2*t_{.975}(396)*\sqrt{3.324/100} = 0.7071\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-98e408fbb1899034420a41dc86f58115_l3.png "Rendered by QuickLaTeX.com")
.
Note that this is the value of MOE we obtain on average in repeated replications with new samples, if we use sample sizes of 100 (total number of participants is 400) and if the true value of the error variance is 3.324. The value is close to the value we obtained in the previous post (MOE = 0.72) because the sample sizes were very close to 100 per group.
Now, we found the original confidence interval too wide, and we have just seen how 100 participants per group does not really help. MOE is only slightly smaller than our originally obtained MOE. We need to set a target MOE and then figure out how many participants we need to get that target MOE.
(Here are some updated rules of thumb: https://the-small-s-scientist.blogspot.com/2018/11/contrast-tutorial.html)
In the absence of theoretical or practical considerations about the precision we want, we may want to use rules of thumb. My (very first proposal for) rules of thumb are based on the default interpretations of Cohen’s d. Considering the absolute values of d ≤ .10 to be negligible d = .20 small, d = .50 medium and d = .80 large. (I really do not like rules-of-thumb, because using them is a sign that you are not thinking).
Now, suppose that we interpret the confidence interval as a range of plausible values for the true value of the effect size. It is not at all clear to me what such a supposition entails, but let’s simply take it for granted right now (please don’t). Then, I think it is reasonable to say that being able to distinguish between small and negligible effects sizes is relatively precise. Thus a MOE of .05 (pooled) standard deviations can be considered precise because (on average) the 95% CI for the small effect sizes is [.15, .25], assuming we know the value of the standard deviation, so negligible effects will not be deemed plausible values on average, since effect sizes smaller than .10 are outside the interval.
By essentially the same reasoning. if we cannot distinguish between large and negligible effects, we are not estimating things very precisely. Therefore, a MOE of .80 standard deviations can be considered to be not very precise. On average, the CI for an existing large effect, will be [0, 1.60], so it includes both negligible and very large effects as plausible values.
For medium (does it make sense to speak of medium precision?) precision I would like to suggest .20-.25 standard deviations. On average, with this value for MOE, if there is a medium effect, small effects and large effects are relatively implausible. In the case of small effects, medium precision entails that on average both effects in the opposite direction and medium effects are among the plausible values.
Of course, I am interpreting the d-values as strict boundaries, but the scale is not categorical, but continuous. So instead of small, large effect sizes, it’s better to speak of smallish and largish effect sizes. And as soon as I find a variant for medium effects sizes I will also include that term in the list.
Note: sample size planning may indicate that precision of MOE = .20-.25 standard deviations is unattainable. In that case, we will simply have to accept that our precision does not lead to confident conclusions about the population effect size. (Once I showed one of my colleagues my precision app, during which he said: “that amount of precision requires a very large sample. I do not like your ideas about sample size planning”).
(By the way, I am also considering rules-of-thumb for target MOE that include assurance. Something like: high precision is when repeated experiments have a high probability of distinguishing small and negligible effects; in that case the average MOE will be smaller than .05).
![\[MOE = 2*t_{.975}(df_e)\sqrt{(MS_w / n_i)},\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-1984a751953fc6b0698d72e7bf2c550b_l3.png "Rendered by QuickLaTeX.com")
where 
, since we are considering the 2×2 design.
, since we want the sample size: ![\[0.4558 = 2*t_{.975}(4(n_i - 1)\sqrt{(MS_w / n_i)},\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-9a775e47b691309421ea2f9d94c016c2_l3.png "Rendered by QuickLaTeX.com")
determines both the standard error and the degrees of freedom (and thereby the critical value of t), the equation may be a little hard to solve. So, I will create a function in R that enables me to quite easily get the required sample size. (It is relatively easy to create a more general function (see the Precision App), but here I will give an example tailored to the specific situation at hand).
First we create a function to calculate MOE:
MOE = function(n) {
MOE = 2*qt(.975, 4*(n - 1))*sqrt(3.324/n)
}
Next, we will define a loss function and use R’s built-in optimize function to determine the sample size. Note that the loss-function calculates the squared difference between MOE based on a sample size n and our target MOE. The optimize function minimizes that squared difference in terms of sample size n (starting with n = 100 and stopping at n = 1000).
loss <- function(n) {
(MOE(n) - 0.4558)^2
}
optimize(loss, c(100, 1000))
## $minimum ## [1] 246.4563 ## ## $objective ## [1] 8.591375e-18
Thus, according to the optimize function we need 247 participants (per group; total N = 988), to get an expected MOE equal to our target MOE. The expected MOE equals 0.4553, which you can confirm by using the MOE function we made above.
![\[MOE_{\gamma} = 2*t_{.975}(df)*\sqrt{MS_w/n_i*\chi^2_{\gamma}(df)/df},\]](https://small-s.science/wp-content/ql-cache/quicklatex.com-f5945fe8ef0dad8f2bf28d70a7553fae_l3.png "Rendered by QuickLaTeX.com")
where 
is the assurance expressed in a probability between 0 and 1.
MOE.gamma = function(n) {
df = 4*(n-1)
MOE = 2*qt(.975, df)*sqrt(3.324/n*qchisq(.80, df)/df)
}
loss <- function(n) {
(MOE.gamma(n) - 0.4558)^2
}
optimize(loss, c(100, 1000))
## $minimum ## [1] 255.576 ## ## $objective ## [1] 2.900716e-18
Thus, according to the results, we need 256 persons per group (N = 1024 in total) to have a 80% probability of obtaining a MOE not larger than our target MOE. In that case, our expected MOE will be 0.4472.