I'm looking for information and guidance to help me understand the outlier test in DHARMa for negative binomial regression. Here is the diagnostic plot from DHARMa using the function simulateResiduals().
First off, The dispersion test is significant in the plot. Using testDispersion() on the model and on the residuals, I get the results of 2.495. Visually, the dots seem to aline pretty well on the QQ line. The developer stated ' If you see a dispersion parameter of 1.01, I would not worry, even if the test is significant. A significant value of 5, however, is clearly a reason to move to a model that accounts for overdispersion.' here I conclude that the deviation is within the acceptable range for the NB regression.
Second, the Outlier test is also significant. I never had this before, and I can't find much information regarding how many outliers is okay vs not okay to have. Following the recommendation of DHARMa's developer, I looked at the magnitude of the outlier to investigate this. reference. Here is the code and output:
ModelNB <- glm.nb(BUD ~ Treatment*YEAR, data=Data_Bud) simulationOutput <- simulateResiduals(fittedModel = ModelNB, plot = T) testOutliers(simulationOutput, type = "binomial")
`
DHARMa outlier test based on exact binomial test with
approximate expectations
data: simulationOutput
outliers at both margin(s) = 12, observations = 576, p-value =
0.00269
alternative hypothesis: true probability of success is not equal to 0.007968127
95 percent confidence interval:
0.01081011 0.03610864
sample estimates:
frequency of outliers (expected: 0.00796812749003984 )
0.02083333
`
**Can someone help me understand this output? ** Is having 12 outliers per 576 observations okay? In statistics classes, I was told that taking out outliers was a big No-No. What does "true probability of success is not equal to 0.007968127" mean? I can't accept H1 and need to accept H0 for the outlier???
Information on my model:
ModelNB <- glm.nb(BUD ~ Treatment*YEAR, data=Data_Bud)
BUD = The number of floral buds on a twig
Treatment = 5 different fertiliser treatment
YEAR = 2 different years (2020 and 2021)
Related
I am using the useful gratia package by Gavin Simpson to extract the difference in two smooths for two different levels of a factor variable. The smooths are generated by the wonderful mgcv package. For example
library(mgcv)
library(gratia)
m1 <- gam(outcome ~ s(dep_var, by = fact_var) + fact_var, data = my.data)
diff1 <- difference_smooths(m1, smooth = "s(dep_var)")
draw(diff1)
This give me a graph of the difference between the two smooths for each level of the "by" variable in the gam() call. The graph has a shaded 95% credible interval (CI) for the difference.
Statistical significance, or areas of statistical significance at the 0.05 level, is assessed by whether or where the y = 0 line crosses the CI, where the y axis represents the difference between the smooths.
Here is an example from Gavin's site where the "by" factor variable had 3 levels.
The differences are clearly statistically significant (at 0.05) over nearly all of the graphs.
Here is another example I have generated using a "by" variable with 2 levels.
The difference in my example is clearly not statistically significant anywhere.
In the mgcv package, an approximate p value is outputted for a smooth fit that tests the null hypothesis that the coefficients are all = 0, based on a chi square test.
My question is, can anyone suggest a way of calculating a p value that similarly assesses the difference between the two smooths instead of solely relying on graphical evidence?
The output from difference_smooths() is a data frame with differences between the smooth functions at 100 points in the range of the smoothed variable, the standard error for the difference and the upper and lower limits of the CI.
Here is a link to the release of gratia 0.4 that explains the difference_smooths() function
enter link description here
but gratia is now at version 0.6
enter link description here
Thanks in advance for taking the time to consider this.
Don
One way of getting a p value for the interaction between the by factor variables is to manipulate the difference_smooths() function by activating the ci_level option. Default is 0.95. The ci_level can be manipulated to find a level where the y = 0 is no longer within the CI bands. If for example this occurred when ci_level = my_level, the p value for testing the hypothesis that the difference is zero everywhere would be 1 - my_level.
This is not totally satisfactory. For example, it would take a little manual experimentation and it may be difficult to discern accurately when zero drops out of the CI. Although, a function could be written to search the accompanying data frame that is outputted with difference_smooths() as the ci_level is varied. This is not totally satisfactory either because the detection of a non-zero CI would be dependent on the 100 points chosen by difference_smooths() to assess the difference between the two curves. Then again, the standard errors are approximate for a GAM using mgcv, so that shouldn't be too much of a problem.
Here is a graph where the zero first drops out of the CI.
Zero dropped out at ci_level = 0.88 and was still in the interval at ci_level = 0.89. So an approxiamte p value would be 1 - 0.88 = 0.12.
Can anyone think of a better way?
Reply to Gavin Simpson's comments Feb 19
Thanks very much Gavin for taking the time to make your comments.
I am not sure if using the criterion, >= 0 (for negative diffs), is a good way to go. Because of the draws from the posterior, there is likely to be many diffs that meet this criterion. I am interpreting your criterion as sample the posterior distribution and count how many differences meet the criterion, calculate the percentage and that is the p value. Correct me if I have misunderstood. Using this approach, I consistently got p values at around 0.45 - 0.5 for different gam models, even when it was clear the difference in the smooths should be statistically significant, at least at p = 0.05, because the confidence band around the smooth did not contain zero at a number of points.
Instead, I was thinking perhaps it would be better to compare the means of the posterior distribution of each of the diffs. For example
# get coefficients for the by smooths
coeff.level1 <- coef(gam.model1)[31:38]
coeff.level0 <- coef(gam.model1)[23:30]
# these indices are specific to my multi-variable gam.model1
# in my case 8 coefficients per smooth
# get posterior coefficients variances for the by smooths' coefficients
vp_level1 <- gam.model1$Vp[31:38, 31:38]
vp_level0 <- gam.model1$Vp[23:30, 23:30]
#run the simulation to get the distribution of each
#difference coefficient using the joint variance
library(MASS)
no.draws = 1000
sim <- mvrnorm(n = no.draws, (coeff.level1 - coeff.level0),
(vp_level1 + vp_level0))
# sim is a no.draws X no. of coefficients (8 in my case) matrix
# put the results into a data.frame.
y.group <- data.frame(y = as.vector(sim),
group = c(rep(1,no.draws), rep(2,no.draws),
rep(3,no.draws), rep(4,no.draws),
rep(5,no.draws), rep(6,no.draws),
rep(7,no.draws), rep(8,no.draws)) )
# y has the differences sampled from their posterior distributions.
# group is just a grouping name for the 8 sets of differences,
# (one set for each difference in coefficients)
# compare means with a linear regression
lm.test <- lm(y ~ as.factor(group), data = y.group)
summary(lm.test)
# The p value for the F statistic tells you how
# compatible the data are with the null hypothesis that
# all the group means are equal to each other.
# Same F statistic and p value from
anova(lm.test)
One could argue that if all coefficients are not equal to each other then they all can't be equal to zero but that isn't what we want here.
The basis of the smooth tests of fit given by summary(mgcv::gam.model1)
is a joint test of all coefficients == 0. This would be from a type of likelihood ratio test where model fit with and without a term are compared.
I would appreciate some ideas how to do this with the difference between two smooths.
Now that I got this far, I had a rethink of your original suggestion of using the criterion, >= 0 (for negative diffs). I reinterpreted this as meaning for each simulated coefficient difference distribution (in my case 8), count when this occurs and make a table where each row (my case, 8) is for one of these distributions with two columns holding this count and (number of simulation draws minus count), Then on this table run a chi square test. When I did this, I got a very low p value when I believe I shouldn't have as 0 was well within the smooth difference CI across almost all the levels of the exposure. Maybe I am still misunderstanding your suggestion.
Follow up thought Feb 24
In a follow up thought, we could create a variable that represents the interaction between the by factor and continuous variable
library(dplyr)
my.dat <- my.dat %>% mutate(interact.var =
ifelse(factor.2levels == "yes", 1, 0)*cont.var)
Here I am assuming that factor.2levels has the levels ("no", "yes"), and "no" is the reference level. The ifelse function creates a dummy variable which is multiplied by the continuous variable to generate the interactive variable.
Then we place this interactive variable in the GAM and get the usual statistical test for fit, that is, testing all the coefficients == 0.
#GavinSimpson actually posted a method of how to get the difference between two smooths and assess its statistical significance here in 2017. Thanks to Matteo Fasiolo for pointing me in that direction.
In that approach, the by variable is converted to an ordered categorical variable which causes mgcv::gam to produce difference smooths in comparison to the reference level. Statistical significance for the difference smooths is then tested in the usual way with the summary command for the gam model.
However, and correct me if I have misunderstood, the ordered factor approach causes the smooth for the main effect to now be the smooth for the reference level of the ordered factor.
The approach I suggested, see the main post under the heading, Follow up thought Feb 24, where the interaction variable is created, gives an almost identical result for the p value for the difference smooth but does not change the smooth for the main effect. It also does not change the intercept and the linear term for the by categorical variable which also both changed with the ordered variable approach.
A variation of this question has been asked, but certain items remain unanswered -
I am modeling the proportion of mortality (Prop) using a single continuous predictor variable which is temperature (Temp). I have three questions.
1.) Should I be using meanBEINF for my model predicted estimates of the response? If so, how would I extract the associated standard errors? You think the way I have it specified currently would give you the response estimates, however, running predict(beinf_mod, type = "response", what = "mu") yields the same results which has me questioning.
2.) If I exponentiate the predictor variable coefficient (contained within the mu parameter) does this give me the odds between (0,1)? nu and tau currently don't have predictor variable coefficients so I'm not sure if those are to be worked in to get odds for the total domain [0,1].
3.) Is my interpretation of the odds correct in this scenario? I am familiar with a regular beta regression or logistic model, but, the uncertainties in question 2 make me wonder if this is appropriate.
Thanks in advance for the help, and it is greatly appreciated.
# generate DB
DB <- data.frame(Prop = c(0.688888889, 0.519230769, 0.378294574, 0.253644315, 0.234200744, 0.156626506,
0.191011236, 0.0625, 0.064516129, 0, 0, 0),
Temp = c(62.90857143, 62.75428571, 60.05428571, 60.23428571, 59.64285714, 57.94571429,
57.71428571, 57.14857143, 54.39714286, 51.87714286, 50.38571429, 49.1))
# beta inflated model. I understand na.omit works on the data, and that na.exclude is not really useful.
# I removed the NA's for this reproduction of the problem
beinf_mod <- gamlss(Prop ~ cs(Temp),
family=BEINF,data=na.omit(DB),na.action=na.exclude)
# obtain predictions for the estimated/expected value of y
predict(beinf_mod, type="response", se.fit=TRUE)
# get the odds of the explanatory variable. exponentiation gets us 1.47,
# so a one unit increase in temperature results in a 47% increase in the odds of mortality within the domain (0,1)
exp(coef(beinf_mod)[2])
allow me to answer my own questions
1.) yes, meanBEINF provides the estimated response values and bootstrapping would get you the errors.
2.) yes. Nu and Tau do not get worked in because one can only calculate the odds for the mu parameter (0,1) because one cannot calculate the odds for 0 and 1. You cannot divide by 0 if the odds are 1 and the odds of 0 prob to any other prob is 0.
3.) yes.
I was reading this topic on Rbloggers about the use of the Wilcoxon rank sum test: https://www.r-bloggers.com/wilcoxon-mann-whitney-rank-sum-test-or-test-u/
Especially this part, here I quote:
"We can finally compare the intervals tabulated on the tables of Wilcoxon for independent samples. The tabulated interval for two groups of 6 samples each is (26, 52)".
How can I get these "tabulated" values ?
I understand they used a table where the values are reported following the size of each samples, but I was wondering if there was a way to get them in R.
It is important because as I can understand the post, once you have a p-value > 0.05 and so cannot reject the null hypothesis H0, you can actually confirm H0 by comparing "computed" and "tabulated" intervals.
So what I would need is the tabulated intervals, using R.
tl;dr
You can get confidence intervals for a Mann-Whitney-Wilcoxon test by specifying conf.int=TRUE.
Don't believe everything you read on the internet ...
If by "confirm" you mean "make sure that the computation is true", you don't need to double-check by consulting the original tables; the p-value should be enough to decide whether you can reject H0 or not. You can trust R for standard, widely used statistical methods. (I also show below how to repeat the computation with a different implementation from the coin package, which is a nearly independent check.)
if by "confirm" you mean "accept the null hypothesis", please don't do this; this is a fundamental violation of frequentist statistical theory, which says that you can reject a null hypothesis, but that you can never accept the null. Wide confidence intervals and p-values greater than a given threshold are evidence that the conclusion is uncertain (we can't be sure whether the null or the alternative is true), not that the null is true. The concluding text of the blog post referred to ("we conclude by accepting the hypothesis H0 of equality of means") is statistically incorrect.
A better way to interpret the uncertainty is to look at the confidence intervals. You can compute these for the Wilcoxon test: from ?wilcox.test:
... (if argument ‘conf.int’ is true [and a two-sample test is being performed]), a nonparametric
confidence interval and an estimator for ... the difference of the location parameters
‘x-y’ is computed.
> a = c(6, 8, 2, 4, 4, 5)
> b = c(7, 10, 4, 3, 5, 6)
> wilcox.test(b,a, conf.int=TRUE, correct=FALSE)
data: b and a
W = 22, p-value = 0.5174
alternative hypothesis: true location shift is not equal to 0
95 percent confidence interval:
-1.999975 4.000016
sample estimates:
difference in location
0.9999395
The high p-value (0.5174) says that we really can't tell whether the values in a or b have signicantly different ranks. The difference in location gives us the estimated difference between the median ranks, and the confidence interval gives the confidence interval on this difference. In this case, for a sample size of 12, the estimated difference in ranks is 1 (group b has slightly higher ranks than group a), and the confidence interval is (-2, 4) (the data are consistent with group b having slightly lower or much higher ranks than group a). It is admittedly rather difficult to interpret the substantive meaning of these values - that's one of the disadvantages of rank-based nonparametric tests ...
You can assume that the p-value computed by wilcox.test() is a reasonable summary of the evidence against the null hypothesis; there's no need to look up ranges in the tables. If you're worried about wilcox.test() in base R, you can try wilcox_test() from the coin package:
dd <- data.frame(f=rep(c("a","b"),each=6),x=c(a,b))
wilcox_test(x~f,data=dd,conf.int=TRUE) ## asymptotic test
which gives nearly identical results to wilcox.test(), and
wilcox_test(x~f,data=dd,conf.int=TRUE, distribution="exact")
which gives a slightly different p-value, but essentially the same confidence intervals.
of historical interest only
As for the tables: I found them on Google books, by doing a Google Scholar search with author:katti author:wilcox. There you can read the description of how they were computed; this wouldn't be impossible to replicate, but it seems unnecessary since p-values and confidence intervals are available via other methods. Digging through you find this:
The number 0.0206 in the red box indicates that the interval (26,52) corresponds to a one-tail p-value of 0.0206 (2-tailed = 0.0412); that's the closest you can get with a discrete range. The next closest range is given in the line below [(27,51), one-tailed p=0.0325, two-tailed=0.065]. In the 21st century you should never have to do this procedure.
My data is in the following format and includes a particular statistic
site LRStat
1 3.580728
2 2.978038
3 5.058644
4 3.699278
5 4.349046
This is just a sample of the data.
I then obtained the null LR distribution as well by permuting random pairs of data. I used this to plot a histogram with frequency in the y-axes and LR statistic in the x-axes. How is it possible to determine the critical p-value cut-off points based on the null distribution (as shown in the below figure)?
You now have a sampling distribution of LR values. The quantile function in R will give you an estimate of whatever "critical value" you prefer. If, for instance, you decided you wanted the conventional 0.05 "p-value" you could take your dataframe, named LR_df for illustration, and issue this command:
quantile( LR_df[ , 'LRStat'] , 0.95)
If you wanted all of those "probabilities" on the figure, you would use a vector of values complementary to unity. The following code gives you the LSstat values at which a given proportion of the sample are higher than that value.
quantile( LR_df[ , 'LRStat'] , c(0.9, 0.95, 0.99, 0.999, 0.9999) )
The p-values are just a sampling distribution of a test statistic under a null hypothesis. Your null hypothesis in this case is that the LRstats are uniformly distributed. (I know it sounds strange to put it that way, but if you want to argue with the statisticians then get a copy of http://amstat.tandfonline.com/doi/pdf/10.1198/000313008X332421 .) The choice of p-value for cutoff will depend on scientific or business setting. If you were assessing an investment opportunity the cutoff might be 0.15 but if you are trying to find new scientific knowledge, I think it should be smaller (more stringent test). The field of molecular genetics has a lot of junk (i.e. fails to reproduce results) in their literature because they were not strict enough in the statistical methods.
I am trying to find a structural break in the mean of a time-series that is skewed, fat-tailed, and heteroskedastic. I apply the Andrews(1993) supF-test via the strucchange package. My understanding is that this is valid even with my nonspherical disturbances. But I would like to confirm this via bootstrapping. I would like to estimate the max t-stat from a difference in mean test at each possible breakpoint (just like the Andrews F-stat) and then bootstrap the critical value. In other words, I want to find my max t-stat in the time-ordered data. Then scramble the data and find the max t-stat in the scrambled data, 10,000 times. Then compare the max t-stat from the time-ordered data to a critical value given by the rank 9,500 max t-stat from the unordered data. Below I generate example data and apply the Andrews supF-test. Is there any way to "correct" the Andrews test for nonspherical disturbances? Is there any way to do the bootstrap I am trying to do?
library(strucchange)
Thames <- ts(matrix(c(rlnorm(120, 0, 1), rlnorm(120, 2, 2), rlnorm(120, 4, 1)), ncol = 1), frequency = 12, start = c(1985, 1))
fs.thames <- Fstats(Thames ~ 1)
sctest(fs.thames)
I'm adding a second answer to analyze the simulated Thames data provided.
Regarding the points from my first general methodological answer: (1) In this case, a log() transformation is clearly appropriate to deal with the extreme skewness of the observations. (2) As the data are heteroscedastic, the inference should be based on HC or HAC covariances. Below I employ the Newey-West HAC estimator, although the data are just heteroscedastic but not autocorrelated. The HAC-corrected inference affects the supF test and the confidence intervals for the breakpoint estimates. The breakpoints themselves and the corresponding segment-specific intercepts are estimated by OLS, i.e., treating the heteroscedasticity as a nuisance term. (3) I did not add any bootstrap or permutation inference as the asymptotic inference appears to be convincing enough in this case.
First, we simulate the data using a particular seed. (Note that other seeds may not lead to such clear-cut breakpoint estimates when analyzing the series in levels.)
library("strucchange")
set.seed(12)
Thames <- ts(c(rlnorm(120, 0, 1), rlnorm(120, 2, 2), rlnorm(120, 4, 1)),
frequency = 12, start = c(1985, 1))
Then we compute the sequence of HAC-corrected Wald/F statistics and estimate the optimal breakpoints (for m = 1, 2, 3, ... breaks) via OLS. To illustrate how much better this works for the series in logs rather than in levels, both versions are shown.
fs_lev <- Fstats(Thames ~ 1, vcov = NeweyWest)
fs_log <- Fstats(log(Thames) ~ 1, vcov = NeweyWest)
bp_lev <- breakpoints(Thames ~ 1)
bp_log <- breakpoints(log(Thames) ~ 1)
The visualization below shows the time series with the fitted intercepts in the first row, the sequence of Wald/F statistics with the 5% critical value of the supF test in the second row, and the residual sum of squares and BIC for the selection of the number of breakpoints in the last row. The code to replicate the graphic is at the end of this answer.
Both supF tests are clearly significant but in levels (sctest(fs_lev)) the test statistic is "only" 82.79 while in logs (sctest(fs_log)) it is 282.46. Also, the two peaks pertaining to the two breakpoints can be seen much better when analyzing the data in logs.
Similarly, the breakpoint estimates are somewhat better and the confidence intervals much narrower for the log-transformed data. In levels, we get:
confint(bp_lev, breaks = 2, vcov = NeweyWest)
##
## Confidence intervals for breakpoints
## of optimal 3-segment partition:
##
## Call:
## confint.breakpointsfull(object = bp_lev, breaks = 2, vcov. = NeweyWest)
##
## Breakpoints at observation number:
## 2.5 % breakpoints 97.5 %
## 1 NA 125 NA
## 2 202 242 263
plus an error message and warnings which all reflect that the asymptotic inference is not a useful approximation here. In contrast, the confidence intervals are quite reasonable for the analysis in logs. Due to the increased variance in the middle segment, its start and end are somewhat more uncertain than for the first and last segment:
confint(bp_log, breaks = 2, vcov = NeweyWest)
##
## Confidence intervals for breakpoints
## of optimal 3-segment partition:
##
## Call:
## confint.breakpointsfull(object = bp_log, breaks = 2, vcov. = NeweyWest)
##
## Breakpoints at observation number:
## 2.5 % breakpoints 97.5 %
## 1 107 119 121
## 2 238 240 250
##
## Corresponding to breakdates:
## 2.5 % breakpoints 97.5 %
## 1 1993(11) 1994(11) 1995(1)
## 2 2004(10) 2004(12) 2005(10)
Finally, the replication code for the figure above is included here. The confidence intervals for the breakpoints in levels are cannot added in the graphic due to the error mentioned above. Hence, only the log-transformed series also has the confidence intervals.
par(mfrow = c(3, 2))
plot(Thames, main = "Thames")
lines(fitted(bp_lev, breaks = 2), col = 4, lwd = 2)
plot(log(Thames), main = "log(Thames)")
lines(fitted(bp_log, breaks = 2), col = 4, lwd = 2)
lines(confint(bp_log, breaks = 2, vcov = NeweyWest))
plot(fs_lev, main = "supF test")
plot(fs_log, main = "supF test")
plot(bp_lev)
plot(bp_log)
(1) Skewness and heavy tails. As usual in linear regression models, the asymptotic justification for the inference does not depend on normality and also holds for any other error distribution given zero expectation, homoscedasticity, and lack of correlation (the usual Gauss-Markov assumptions). However, if you have a well-fitting skewed distribution for your data of interest, then you might be able to increase efficiency by basing your inference on the corresponding model. For example, the glogis package provides some functions for structural change testing and dating based on a generalized logistic distribution that allows for heavy tails and skewness. Windberger & Zeileis (2014, Eastern European Economics, 52, 66–88, doi:10.2753/EEE0012-8775520304) used this to track changes in skewness of inflation dynamics over time. (See ?breakpoints.glogisfit for a worked example.) Furthermore, if the skewness itself is not really of interest then a log or sqrt transformation might also be good enough to make the data more "normal".
(2) Heteroscedasticity and autocorrelation. As usual in linear regression models, the standard errors (or more broadly the covariance matrix) is not consistent in the presence of heteroscedasticity and/or autocorrelation. One can either try to include this explicitly in the model (e.g., an AR model) or treat it as a nuisance term and employ heteroscedasticity and autocorrelation consistent (HAC) covariance matrices (e.g., Newey-West or Andrews' quadratic spectral kernal HAC). The function Fstats() in strucchange allows to plug in such estimators, e.g., from the sandwich package. See ?durab for an example using vcovHC().
(3) Bootstrap and permutation p-values. The "scrambling" of the time series you describe above sounds more like applying permutations (i.e., sampling without replacement) rather than bootstrap (i.e., sampling with replacement). The former is feasible if the errors are uncorrelated or exchangeable. If you are regressing just on a constant, then you can employ the function maxstat_test() from the coin package to carry out the supF test. The test statistic is computed in a somewhat different way, however, this can be shown to be equivalent to the supF test in the constant-only case (see Zeileis & Hothorn, 2013, Statistical Papers, 54, 931–954, doi:10.1007/s00362-013-0503-4). If you want to perform the permutation test in a more general model, then you would have to do the permutations "by hand" and simply store the test statistic from each permutation. Alternatively, the bootstrap can be applied, e.g., via the boot package (where you would still need to write your own small function that computes the test statistic from a given bootstrap sample). There are also some R packages (e.g., tseries) that implement bootstrap schemes for dependent series.