Power for analyzing Likert items

First for some other updates of interest to folks on the blog. On CRIME De-Coder a blog post, You should be geocoding crime data locally. I give python code to create a local geocoding engine using the arcpy library.

This is a bit more techy, so would typically post this on this blog instead of the CRIME De-Coder one. But, currently the web is sorely lacking in good advice for local geocoding solutions. Vast majority of sites discuss online geocoding APIs (e.g. google or the census geocoder), which I guess are common for web-apps, but they do not make sense for crime analysis. For the few webpages that are actually relevant to describe local solutions (that do not involve calling an online web API), all the exmples use PostGIS that I am aware of. PostGIS is both very difficult to setup and has worse results compared to ESRI. So I know ESRI is a paid for service, but they have reasonable academic and small business pricing (and most police departments already have access), so to me this is a reasonable use case. If you need to geocode 100k cases, the license fee for ESRI is worth it at that point relative to using the web engines.

Definitely do not spend thousands of dollars if you need to batch geocode a few million records. That is something that is worth getting in touch with me about. And so hopefully that gets picked up by search engines and drives a bit more traffic to my consulting website.

A second example I posted some python code to help construct network experiments. So the idea here is you want to spread out the treated nodes so you have a specific allocation of treated, connected to treated (what I call spillover here), and those not connected to treated (the leftover control group). This python code constructs linear programs to accomplish certain treated/not-touched proportions. So this graph shows if you choose to treat 1 person, but have constraints on 1,2,3 leftover.

And then you can apply this to bigger networks, here the network is 311 nodes, and 90 are treated and I want a total of 150 not treated.

Idea derivative from some work Bruce Desmarais discussed on Twitter, but also have thought about this in some discussion with Barak Ariel in focused deterrence style interventions. So hopefully that comes in handy.

My linear programming formulation is not as svelte as the optimal treatment assignment with spillovers formulation, it is 3*N + 2*E decision variables and 5*N + 2*E constraints (where N is the number of nodes and E is the number of un-directed edges). I have a feeling my formulation is redundant, so if I write my constraints smarter can be more like 2N decision variables and 2N + E constraints.

But for my examples I show it solves quite fast as is (and maybe solvers get rid of the cruft in pre-solve), so not worried about that at the moment. Don’t know the typical size networks people use, but I suspect it will work just fine and dandy on typical machines for networks even as large as 10k. (Generally if I keep the model to under 100k decision variables it is only a few minutes to solve the types of problems I show on this blog.)

Power with Likert items

The other day for a grant application we needed to conduct power analysis. Our design was t-test of mean differences for a simple treated/control group randomized experiment with the outcome being a Likert score survey response. (I know, most of the time people create latent scores with Likert items, analyzing the individual items is fine IMO and simpler to specify for a pre-registration analysis.) I am sure others have needed to do similar things, but I could not find code online to help out with this. So I scripted up a simulation in R to do this, and figured sharing would be useful.

So the rub with Likert data, and why you can’t use typical power calculations, is that they have ceiling effects. If most people answer on average 4.5 out of the your scale up to 5, it is difficult to go much higher. Here I simulate data that has that skew (so ceiling effects come into play), and then go through the motions of doing the t-test. So first for some setup, I have a function that rounds and clips data to limited sets of integers, plus some plotting functions.

# power analysis of Likert data
library(ggplot2)

# custom theme
theme_cdc <- function(){
  theme_bw() %+replace% theme(
     text = element_text(size = 16),
     panel.grid.major= element_line(linetype = "longdash"),
     panel.grid.minor= element_blank()
) }

set.seed(10) # setting the random seed

# rounding/clipping data to integers 
clipint <- function(x,min=1,max=5){
    rx <- round(x)
    rx <- ifelse(rx < min,min,rx)
    rx <- ifelse(rx > max,max,rx)
    return(rx)
}

This following function generates the p-values and standard errors, what I will use later in my simulation. Here I use a t-test of mean differences, but it would be fairly easy to say swap out with an ordinal logistic regression if you prefer that. Probably the bigger deal is the simulation generates data using a normal distribution, and then post rounds/clip the data. There is probably a smarter way to generate the data according to the logistic model and ordinal intercepts (Frank Harrell discusses such things a bit on his blog). But this at least takes into account that the data will be skewed, even in the control group, to have more positive outcomes and thus take the ceiling into account.

# this uses OLS to do t-test of mean differences
# generates normal data, but then rounds/clips
sim_ols <- function(n,eff=0.5,int=4,sd=1){
    df <- data.frame(1:n)
    df$treat <- sample(c(0,1),n,replace=TRUE)
    df$latent <- int + eff*df$treat + rnorm(n,sd=sd)
    df$likert <- clipint(df$latent)
    m1 <- lm(likert ~ treat,data=df)
    cd <- coef(summary(m1))
    pval <- cd[2,4]
    se <- cd[2,2]
    return(c(pval,se))
}

Now we can move onto the simulations, this evaluates sample sizes from 100 to 2000 (in increments of 50), effect sizes of 0.1, 0.3, and 0.5, and repeats the simulations 10k times. I then see the power (how often the two-tailed p-value is less than 0.05), as well as the standard error (precision) of the estimates. Effect sizes are in terms of the original Likert scale values, what I take to be much easier to reason about. (I have seen power analyses here use Cohen’s D, which you really can’t get a very large Cohen’s D value due to ceiling effects with this data.)

# running power estimates over different
# sample sizes and effect sizes
samp_sizes <- seq(100,2000,50)
eff_sizes <- c(0.1,0.3,0.5)
rep_size <- 10000
df <- expand.grid(samps_sizes=samp_sizes,eff_sizes=eff_sizes,pow=c(NA),se=c(NA))
for (i in 1:nrow(df)){
    ss <- df[i,1]
    es <- df[i,2]
    stats <- replicate(rep_size,sim_ols(n=ss,eff=es))
    smean <- rowMeans(stats)
    df[i,3] <- mean(stats[1,] < 0.05) # alpha 0.05
    df[i,4] <- smean[2]
}

df$eff_sizes <- as.factor(df$eff_sizes)

The graph of the power shows what you would expect, so with a few hundred samples you can determine an effect size of 0.5, but with a smaller effect size (on the order of 0.1) you will need more than 2k samples.

# Graph of power
powg <- ggplot(data=df,aes(x=samps_sizes,y=pow)) + 
        geom_line(aes(color=eff_sizes)) + 
        geom_point(pch=21,color='white',size=2,aes(fill=eff_sizes)) +
        labs(x='Sample Sizes',y=NULL,title='Power Estimates') +
        scale_y_continuous(breaks=seq(0,1,0.1)) + 
        scale_x_continuous(breaks=seq(100,2000,250)) + 
        scale_color_discrete(name="Effect Sizes") +
        scale_fill_discrete(name="Effect Sizes") + 
        theme_cdc()

Unfortunately, in reality with most survey measures of police data, e.g. rate your officer 1 to 5, a 0.5 effect is a really large increase. In my mapping attitudes paper, some demographics shift global attitudes at max by 0.2, and I doubt most interventions will be that dramatic. So I like plotting the precision of the estimator, which the effect size doesn’t really make a dent here (it could with more severe ceiling effects).

# Graph of Standard Errors (for Precision)
precg <- ggplot(data=df,aes(x=samps_sizes,y=se,color=eff_sizes)) + 
         geom_line(aes(color=eff_sizes)) + 
         geom_point(pch=21,color='white',size=2,aes(fill=eff_sizes)) +
         labs(x='Sample Sizes',y=NULL,title='Precision Estimates') +
         scale_x_continuous(breaks=seq(100,2000,250)) + 
         scale_color_discrete(name="Effect Sizes") +
         scale_fill_discrete(name="Effect Sizes") + 
         theme_cdc()

With field experiments when considering post police contacts (general attitude surveys you have more wiggle room, but still they cost money to survey) you really can’t control the sample size. You are at the whims of whatever events happen in the police departments daily duties. So the best you can do is make approximate plans for “how long am I going to collect measures before I analyze the data”, and “how reasonably precise will my estimates be”.

This particular grant I make arguments we care more about a non-inferiority type test (e.g. can be sure attitudes are not worse, within a particular error level, given our treatment is more cost-effective than business as usual). But if we did an intervention specifically intended to improve attitudes, you may be talking more like 5,000+ contacts to detect an effect of 0.1 (given likely sample non-response), which is still a big effect.

You would gain power by doing a scale (e.g. summing together multiple items or conducting a factor analysis), assuming the intervention effects the underlying latent trait, and piecemeal individual items. But that will have to be for another day simulating data to do that end-to-end analysis.

Despite not being an academic anymore, I have in the hopper more grant collaborations than I did when I was a professor. The NIJ season is winding down, so probably too late to collaborate on any more of those. But if you have other ideas and need quant help with your projects, always feel free to reach out. I enjoy doing these side projects with academics when reasonable funding is available.

Using precision to plan experiments (SPSS)

SPSS Version 28 has some new nice power analysis facilities. Here is some example code to test the difference in proportions, with sample sizes equal between the two groups, and the groups have an underlying proportion of 0.1 and 0.2.

POWER PROPORTIONS INDEPENDENT
  /PARAMETERS TEST=NONDIRECTIONAL SIGNIFICANCE=0.05 POWER=0.6 0.7 0.8 0.9 NRATIO=1 
    PROPORTIONS=0.1 0.2 METHOD=CHISQ ESTIMATE=NORMAL CONTINUITY=TRUE POOLED=TRUE
  /PLOT TOTAL_N.

So this tells you the sample size to get a statistically significant difference (at the 0.05 level) between two groups for a proportion difference test (here just a chi square test). And you can specify the solution for different power estimates (here a grid of 0.6 to 0.9 by 0.1), as well as get a nice plot.

So this is nice for academics planning factorial experiments, but I often don’t personally plan experiments this way. I typically get sample size estimates via thinking about precision of my estimates. In particular I think to myself, I want subgroup X to have a confidence interval width of no more than 5%. Even if you want to extend this to testing the differences between groups this is OK (but will be conservative), because even if two confidence intervals overlap the two groups can still be statistically different (see this Geoff Cumming reference).

So for example a friend the other day was asking about how to sample cases for an audit, and they wanted to do subgroup analysis in errors between different racial categories. But it was paper records, so they needed to just get an estimate of the total records to sample upfront (can’t really control the subgroup sizes). I suggested to estimate the precision they wanted in the smallest subgroup of interest for the errors, and base the total sample size of the paper audit on that. This is much easier to plan than worrying about a true power analysis, in which you also need to specify the expected differences in the groups.

So here is a macro I made in SPSS to generate precision estimates for binomial proportions (Clopper Pearson exact intervals). See my prior blog post for different ways to generate these intervals.

* Single proportion, precision.
DEFINE !PrecisionProp (Prop = !TOKENS(1)
                      /MinN = !TOKENS(1)
                      /MaxN = !TOKENS(1)
                      /Steps = !DEFAULT(100) !TOKENS(1) 
                      /DName = !DEFAULT("Prec") !TOKENS(1) )
INPUT PROGRAM.
COMPUTE #Lmi = LN(!MinN).
COMPUTE #Step = (LN(!MaxN) - #Lmi)/!Steps.
LOOP #i = 1 TO (!Steps + 1).
  COMPUTE N = EXP(#Lmi + (#i-1)*#Step).
  COMPUTE #Suc = N*!Prop.
  COMPUTE Prop = !Prop.
  COMPUTE Low90 = IDF.BETA(0.05,#Suc,N-#Suc+1).
  COMPUTE High90 = IDF.BETA(0.95,#Suc+1,N-#Suc).
  COMPUTE Low95 = IDF.BETA(0.025,#Suc,N-#Suc+1).
  COMPUTE High95 = IDF.BETA(0.975,#Suc+1,N-#Suc).
  COMPUTE Low99 = IDF.BETA(0.005,#Suc,N-#Suc+1).
  COMPUTE High99 = IDF.BETA(0.995,#Suc+1,N-#Suc).
  END CASE.
END LOOP.
END FILE.
END INPUT PROGRAM.
DATASET NAME !DName.
FORMATS N (F10.0) Prop (F3.2).
EXECUTE.
!ENDDEFINE.

This generates a new dataset, with a baseline proportion of 50%, and shows how the exact confidence intervals change with increasing the sample size. Here is an example generating confidence intervals (90%,95%, and 99%) for sample sizes from 50 to 3000 with a baseline proportion of 50%.

!PrecisionProp Prop=0.5 MinN=50 MaxN=3000.

And we can go and look at the generated dataset. For sample size of 50 cases, the 90% CI is 38%-62%, the 95% CI is 36%-64%, and the 99% CI is 32%-68%.

For sample sizes of closer to 3000, you can then see how these confidence intervals decrease in width to changes in 4 to 5%.

But I really just generated the data this way to do a nice error bar chart to visualize:

*Precision chart.
GGRAPH
  /GRAPHDATASET NAME="graphdataset" VARIABLES=Prop N Low90 High90 Low95 High95 Low99 High99
  /GRAPHSPEC SOURCE=INLINE.
BEGIN GPL
  SOURCE: s=userSource(id("graphdataset"))
  DATA: N=col(source(s), name("N"))
  DATA: Prop=col(source(s), name("Prop"))
  DATA: Low90=col(source(s), name("Low90"))
  DATA: High90=col(source(s), name("High90"))
  DATA: Low95=col(source(s), name("Low95"))
  DATA: High95=col(source(s), name("High95"))
  DATA: Low99=col(source(s), name("Low99"))
  DATA: High99=col(source(s), name("High99"))
  DATA: N=col(source(s), name("N"))
  GUIDE: text.title(label("Base Rate 50%"))
  GUIDE: axis(dim(1), label("Sample Size"))
  GUIDE: axis(dim(2), delta(0.05))
  SCALE: linear(dim(1), min(50), max(3000))
  ELEMENT: area.difference(position(region.spread.range(N*(Low99+High99))), color.interior(color.grey), 
           transparency.interior(transparency."0.6"))
  ELEMENT: area.difference(position(region.spread.range(N*(Low95+High95))), color.interior(color.grey), 
           transparency.interior(transparency."0.6"))
  ELEMENT: area.difference(position(region.spread.range(N*(Low90+High90))), color.interior(color.grey), 
           transparency.interior(transparency."0.6"))
  ELEMENT: line(position(N*Prop))
END GPL.

So here the lightest gray area is the 99% CI, the second lightest is the 95%, and the darkest area is the 90% CI. So here you can make value trade offs, is it worth it to get an extra precision of under 10% by going from 500 samples to 1000 samples? Going from 1500 to 3000 you don’t gain as much precision as going from 500 to 1000 (diminishing returns).

It is easier to see the progression of the CI’s when plotting the sample size on a log scale (or sometimes square root), although often times costs of sampling are on the linear scale.

You can then change the macro arguments if you know your baseline is likely to not be 50%, but say here 15%:

!PrecisionProp Prop=0.15 MinN=50 MaxN=3000.

So you can see in this graph that the confidence intervals are smaller in width than the 50% – a baseline of 50% results in the largest variance estimates for binary 0/1 data, so anything close to 0% or 100% will result in smaller confidence intervals. So this means if you know you want a precision of 5%, but only have a rough guess as to the overall proportion, planning for 50% baseline is the worst case scenario.

Also note that when going away from 50%, the confidence intervals are asymmetric. The interval above 15% is larger than the interval below.

A few times at work I have had pilots that look something like “historical hit rates for these audits are 10%, I want the new machine learning model to increase that to 15%”.

So here I can say, if we can only easily use a sample of 500 cases for the new machine learning model, the precision I expect to be around 11% to 19%. So cutting a bit close to be sure it is actually above the historical baseline rate of 10%. To get a more comfortable margin we need sample sizes of 1000+ in this scenario.

Spatial sample size suggestions for SPPT analysis

I’ve reviewed several papers recently that use Martin Andresen’s Spatial Point Pattern Test (Andresen, 2016). I have been critical of these papers, as I think they are using too small of samples to be reasonable. So here in this blog post I will lay out spatial sample size recommendations. Or more specifically if you have N crimes, advice about how you can conduct the SPPT test in S spatial units of analysis.

Long story short, if you have N crimes, I think you should either use 0.85*N = S spatial units of analysis at the high end, but can only detect very large changes. To be well powered to detect smaller changes between the two distributions, use 0.45*N = S. That is, if you have two crime samples you want to compare, and the smaller sample has 1000 crimes, the largest spatial sample size I would recommend is 850 units, but I think 450 units is better.

For those not familiar with the SPPT technique, it compares the proportion of events falling inside a common area (e.g. police beats, census block groups, etc.) between two patterns. So for example in my work I compared the proportion of violent crime and the proportion of SQF in New York City (Wheeler et al., 2018). I think it makes sense as a gross monitoring metric for PDs this way (say for those doing DDACTS, swap out pedestrian stops with traffic stops), so you can say things like area A had a much lower proportion of crimes than stops, so we should emphasize people do fewer stops in A overall.

If you are a PD, you may already know the spatial units you want to use for monitoring purposes (say for each police sector or precinct). In that case, you want the power analysis to help guide you for how large a sample you need to effectively know how often you can update the estimates (e.g. you may only have enough traffic stops and violent crimes to do the estimates on a quarterly basis, not a monthly one) Many academic papers though are just generally theory testing, so don’t have an a priori spatial unit of analysis chosen. (But they do have two samples, e.g. a historical sample of 2000 shootings and a current sample of 1000 shootings.) See Martin’s site for a list of prior papers using the SPPT to see it in action.

I’ve reviewed several papers that examine these proportion changes using the SPPT at very tiny spatial units of analysis, such as street segments. They also happen to have very tiny numbers of overall crimes, and then break the crimes into subsequent subsets. For example reviewed a paper that had around 100 crimes in each subset of interest, and had around 20,000 street segments. I totally get wanting to examine micro place crime patterns – but the SPPT is not well suited for this I am afraid.

Ultimately if you chunk up the total number of crimes into smaller and smaller areas, you will have less statistical power to uncover differences. With very tiny total crime counts, you will be basically only identifying differences between areas that go from 0 to 1 or 0 to 2 etc. It also becomes much more important to control for multiple comparisons when using a large number of spatial units. In general this technique is not going to work out well for micro units of analysis, it will only really work out for larger spatial units IMO. But here I will give my best advice about how small you can reasonably go for the analysis.

Power analysis logic

There are quite a few different ways people have suggested to determine the spatial sample for areas when conducting quadrat analysis (e.g. when you make your own spatial areas). So one rule of thumb is to use 2*A/N, where A is the area of the study and N is the total number of events (Paez, 2021).

Using the SPPT test itself, Malleson et al. (2019) identify the area at which the spatial pattern exhibits the highest similarity index with itself using a resampling approach. Ramos et al. (2021) look at the smallest spatial unit at which the crime patterns within that unit show spatial randomness.

So those later two take an error metric based approach (the spatial unit of analysis likely to result in the miminal amount of error, with error defined different ways). I take a different approach here – power analysis. We want to compare to spatial point patterns for proportional differences, how can we construct the test to be reasonably powered to identify differences we want to detect?

I do not have a perfect way to do this power analysis, but here is my logic. Crime patterns are often slightly overdispersed, so here I assume if you split up say 1000 crimes into 600 areas, it will have an NB2 distribution with a mean of 1000/600 = 1.67 and an overdispersion parameter of 2. (I assume this parameter to be 2 for various reasons, based on prior analysis of crime patterns, and that 2 tends to be in the general ballpark for the amount of overdispersion.) So now we want to see what it would take to go from a hot spot of crime, say the 98th percentile of this distribution to the median 50th percentile.

So in R code, to translate the NB2 mean/dispersion to N & P notation results in N & P parameters of 1.0416667 and 0.3846154 respectively:

trans_np <- function(mu,disp){
    a <- disp
    x <- mu^2/(1 - mu + a*mu)
    p <- x/(x + mu)
    n <- (mu*p)/(1-p)
    return(c(n,p))
}

# Mean 1000/600 and dispersion of 2
nb_dis <- trans_np(1000/600,2)

Now we want to see what the counts are to go from the 98th to the 50th percentile of this distribution:

crime_counts <- qnbinom(c(0.98,0.5), size=nb_dis[1], prob=nb_dis[2])

And this gives us a result of [1] 8 1 in the crime_counts object. So a hot spot place in this scenario will have around 8 crimes, and the median will be around 1 crime in our hypothetical areas. So we can translate these to percentages, and then feed them into R’s power.prop.test function:

crime_prop <- crime_counts/1000
power.prop.test(n = 1000, p1 = crime_prop[1], p2 = crime_prop[2])

And this gives us a result of:

     Two-sample comparison of proportions power calculation 

              n = 1000
             p1 = 0.008
             p2 = 0.001
      sig.level = 0.05
          power = 0.6477139
    alternative = two.sided

NOTE: n is number in *each* group

Note that this is for one N estimate, and assumes that N will be the same for each proportion. In practice for the SPPT test this is not true, oftentimes we have two crime samples (or crime vs police actions like stops), which have very different total baseline N’s. (It is part of the reason the test is useful, it doesn’t make so much sense in that case to compare densities as it does proportions.) So subsequently when we do these estimates, we should either take the average of the total number of crimes we have in our two point patterns for SPPT (if they are close to the same size), or the minimum number of events if they are very disparate. So if you have in sample A 1000 crimes, and sample B 2000 crimes, I think you should treat the N in this scenario as 1500. If you have 5000 crimes vs 1000000 crimes, you should treat N here as 5000.

So that estimate above is for one set of crimes (1000), and one set of areas (600). But what if we vary the number of areas? At what number of areas do we have the maximum power?

So I provide functions below to generate the power estimate curve, given these assumptions about the underlying crime distribution (which will generally be in the ballpark for many crime patterns, but not perfect), for varying numbers of spatial units. Typically we know the total number of crimes, so we are saying given I have N crimes, how finely can a split them up to check for differences with the SPPT test.

Both the Malleson and Ramos article place their recommendations in terms of area instead of total number of units. But it would not surprise me if our different procedures end up resulting in similar recommendations based on the observed outputs of each of the papers. (The 2A/N quadrat analysis suggestion translates to N*0.5 total number of areas, pretty close to my 0.45*N suggestion for example.)

R Code

Below I have a nicer function to do the analysis I walked through above, but give a nice power curve and dataframe over various potential spatial sample sizes:

# SPPT Power analysis example
library(ggplot)

# See https://andrewpwheeler.com/2015/01/03/translating-between-the-dispersion-term-in-a-negative-binomial-regression-and-random-variables-in-spss/
trans_np <- function(mu,disp){
    a <- disp
    x <- mu^2/(1 - mu + a*mu)
    p <- x/(x + mu)
    n <- (mu*p)/(1-p)
    return(c(n,p))
}

diff_suggest <- function(total_crimes,areas=round(seq(2,total_crimes*10,length.out=500)),
                         change_quant=c(0.98,0.5), nb_disp=2, alpha = 0.05,
                         plot=TRUE){
         # Figure out mean
         areas <- unique(areas)
         mean_cr <- total_crimes/areas
         # Initialize some vectors to place the results
         n_areas <- length(areas)
         power <- vector("numeric",length=n_areas)
         hign <- power
         lown <- power
         higp <- power
         lowp <- power
         # loop over areas and calculate power
         for (i in 1:length(areas)){
             # Negative binomial parameters
             dp <- trans_np(mean_cr[i],nb_disp)
             hilo <- qnbinom(change_quant, size=dp[1], prob=dp[2])
             hilo_prop <- hilo/total_crimes
             # Power for test
             pow <- power.prop.test(n = total_crimes, p1 = hilo_prop[1], p2 = hilo_prop[2], 
                                    sig.level = alpha)
             # Stuffing results in vector
             power[i] <- pow$power
             hign[i] <- hilo[1]
             lown[i] <- hilo[2]
             higp[i] <- hilo_prop[1]
             lowp[i] <- hilo_prop[2]
         }
         Ncrimes <- rep(total_crimes,n_areas)
         res_df <- data.frame(Ncrimes,areas,mean_cr,power,hign,lown,higp,lowp)
         # replacing missing with 0
         res_df[is.na(res_df)] <- 0
         if (plot) {
            require(ggplot2)
            fmt_cr <- formatC(total_crimes, format="f", big.mark=",", digits=0)
            title_str <- paste0("Power per area for Total number of crimes: ",fmt_cr)
            cap_str <- paste0("NB Dispersion = ",nb_disp,", alpha = ",alpha,
                              ", change quantiles = ",change_quant[1]," to ",change_quant[2])
            p <- ggplot(data=res_df,aes(x=areas,y=power)) + geom_line(size=1.5) +
                 theme_bw() + theme(panel.grid.major = element_line(linetype="dashed")) +
                 labs(x='Number of Areas',y=NULL,title=title_str,caption=cap_str) +
                 scale_x_continuous(minor_breaks=NULL) + scale_y_continuous(minor_breaks=NULL) +
                 theme(text = element_text(size=16), axis.title.y=element_text(margin=margin(0,10,0,0)))
            print(p)
         }
         return(res_df)
}

Once that function is defined, you can make a simple call like below, and it gives you a nice graph of the power given different numbers of grid cells:

diff_suggest(100)

So you can see here we never have very high power over this set of parameters. It is also non-monotonic and volatile at very small numbers of spatial units of analysis (at which the overdispersion assumption likely does not hold, and probably is not of much interest). But once that volatility tamps down we have stepped curve, that ends up happening to step whenever the original NB distribution changes from particular integer values.

So what happens with the power curve if we up the number of crimes to 3000?

diff_suggest(3000)

So those two patterns are quite similar. It happens that when breaking down to the smaller units, the highest power scenario is when the crimes are subdivided into around 0.85 fewer spatial units than total crimes. So if you have 1000 crimes, in this scenario I would suggest to use 850 areas.

Also note the behavior when you break it down into a very large number of spatial units S, where S >> N, you get a progressive decline until around 0 power in this analysis. E.g. if you have 100 times more spatial units than observations, only a handful of locations have any crimes, and the rest are all 0’s. So you need to be able to tell the difference between 1/N and 0, which is tough (and any inferences you do make will just pretty much be indistinguishable from noise).

What about if we change the quantiles we are examining, and instead of looking at the very high crime place to the median, look at the 80th percentile to the 20th percentile:

diff_suggest(3000, change_quant=c(0.8,0.2))

We have a similar step pattern, but here the power is never as high as before, and only maxes out slightly above 0.5. It happens that this 0.5 power is around number of crimes*0.45. So this suggests that to uncover more middling transitions, one would need to have less than half the number of spatial units of crime observed. E.g. if you have 1000 crimes, I would not suggest any more than 450 spatial units of analysis.

So the first scenario, crimes*0.85 you could say something like this is the highest power scenario to detect changes from very high crime locations (aka hot spots), to the middle of the distribution. For the second scenario (my preferred offhand), is to say crimes*0.45 total spatial units results in the highest power scenario to detect more mild changes in the middle of the distribution (and detecting changes from hot spots to cold spots thus have even more power).

For now that is the best advice I can give for determining the spatial sample size for the SPPT test. Will have a follow up blog post on using R to make a grid to conduct the test.

Also I have been wondering about the best way to quantify changes in the overall ranking. I have not come upon a great solution I am happy with though, so will need to think about it some more.

References