Write a function that will allow the user
to input a vector of numerical values,
with no missing values "the data", and
a vector of 1's and 2's, representing
two different groups that you want to compare.
"the treatments". The number of 1's and 2's does not need to be equal. You may assume
for now, that treatment 2 has a higher mean
than treatment 1.
The function will create the randomization
distribution of differences, and plot them
in a histogram. It will use the distribution
to calculate the p-value -- the chance that
the observed difference (or higher) could have
occurred by chance. It will print the observed
difference and the p-value, both
rounded to 4 digits, using text:
"The observed difference is xxxx and the
p-value is xxxx"
Using these two vectors I have determined how to get the differences but do not know how to put it into a function and implement a randomization test.
dat<- c(1,4,2,5,2,4,8,6,9,7)
trt <- c(1,1,1,1,1,2,2,2,2,2)
How to find the observed difference:
obsdiff <- mean(dat[trt == 2]) - mean(dat[trt == 1])
How to 'shuffle the treatments':
trtsh <- sample(trt, size = length(trt))
How to find a difference simulated
under the null hypothesis,
i.e., difference in means for shuffled
treatment 2 minus treatment 1:
simdiff <- mean(dat[trtsh == 2]) - mean(dat[trtsh == 1])
The p-value using these vectors should be .011
dat<- c(1,4,2,5,2,4,8,6,9,7)
trt <- c(1,1,1,1,1,2,2,2,2,2)
In general, it is a good idea to coerce all of your data into a data frame, a la
data.frame(dat, trt) -> mydata
Now you can calculate your obsdiff as
obsdiff <- mean(mydata$dat[mydata$trt == 2]) - mean(mydata$dat[mydata$trt == 1])
Here's one way you can shuffle your treatment values using a for-loop:
simdiff <- vector(length=10000)
for(j in 1:10000){
cat(paste(j, '\n') )
trtsh <- sample(trt)
mydatash <- data.frame(dat, trtsh)
simdiff[j] <- mean(mydatash$dat[mydatash$trt == 2]) - mean(mydatash$dat[mydatash$trt == 1])
}
For help with plotting, see ?hist (e.g. hist(simdiff)).
Now, you just need to wrap the pieces above into a function that calculates the quantile of simdiff where obsdiff >= simdiff and outputs the text.
I am trying to simulate data (Y) from an AR(1) model with rho=0.7. Then I will use this data to run a regression of Y on an intercept ( by so doing the parameter estimate becomes the mean of Y), then test the null hypothesis of the coefficient being less than or equal to zero ( alternative is greater than 0) using robust standard errors.
I want to run a Monte Carlo simulation of this hypothesis using 2000 replications for different lag values. the purpose is to show the finite sample performance of the Newey West estimator as the lag changes. so this is how I began
A<-array(0, dim=c(2000,1))
for(i in 1:2000){
y_new<-arima.sim(model=list(ar=0.7), n=50, mean=0,sd=1)
reg<-lm(y_new~1)
ad<-coeftest(reg, alternative="greater", vcov=NeweyWest(reg, lag=1, prewhite=FALSE))
A[i]<-ad[,3]
}
My question: is the code above the right way of doing this kind of simulation? And if it is, how can I get a code to repeat this process for different lag values in the HAC test. I want to run the test each time increasing the lag by 1, thus I will be doing this 50 times for lags 1,2,3,4......,50, each time storing the 2000 simulated test statistics in a vector with different names. calculate rejection probabilities for the test statistic (sig. level =0,05, using the critical value of 1.645) for each case and plot them(rejection probabilities) against the various lag values.
Please help
Because you didn't mention the possible purpose of the simulation, it is hard to tell whether it is the right way.
You save a lot of time by computing 50 test statistics for each simulated sample, instead of repeating the simulation 2000 times for each lag (that is, the number of simulation is 2000*50).
Much better format of doing simulation is
library(AER)
library(dplyr)
lags <- 1:50
nreps <- 2000
sim <- function (){
ynew <- arima.sim(model = list(ar=0.7), n=50, mean=0, sd=1)
reg <- lm(ynew ~ 1 )
s <- rep(NA, 50)
for(i in lags){
ad <- coeftest(reg, alternative="greater", vcov=NeweyWest(reg, lag = i, prewhite=FALSE))
s[i] <- ad[ ,4]
}
s
}
Following code stores simulation results in a data.frame
result <- lapply(1:nreps, function(i)data.frame(simulation = i, lag = lags, pvalues = sim())) %>%
rbind_all
From your vague description, I extrapolate what you want looks something like
library(ggplot2)
result %>%
group_by(lag) %>%
summarize(rejectfreq = mean(pvalues > 0.05)) %>%
ggplot(., aes(lag, rejectfreq)) + geom_line()+
coord_cartesian(ylim = c(0,1)) +
scale_y_continuous(breaks=seq(0, 1, by=0.1))
Although the figure was created using only 100 simulations, it is evident that the choice of the lags in Newey-West wouldn't matter much when the disturbance terms are i.i.d.
Ok straight to the question. I have a database with lots and lots of categorical variable.
Sample database with a few variables as below
gender <- as.factor(sample( letters[6:7], 100, replace=TRUE, prob=c(0.2, 0.8) ))
smoking <- as.factor(sample(c(0,1),size=100,replace=T,prob=c(0.6,0.4)))
alcohol <- as.factor(sample(c(0,1),size=100,replace=T,prob=c(0.3,0.7)))
htn <- as.factor(sample(c(0,1),size=100,replace=T,prob=c(0.2,0.8)))
tertile <- as.factor(sample(c(1,2,3),size=100,replace=T,prob=c(0.3,0.3,0.4)))
df <- as.data.frame(cbind(gender,smoking,alcohol,htn,tertile))
I want to test the hypothesis, using a chi square test, that there is a difference in the portion of smokers, alcohol use, hypertension (htn) etc by tertile (3 factors). I then want to extract the p values for each variable.
Now i know i can test each individual variable using a 2 by 3 cross tabulation but is there a more efficient code to derive the test statistic and p-value across all variables in one go and extract the p value across each variable
Thanks in advance
Anoop
If you want to do all the comparisons in one statement, you can do
mapply(function(x, y) chisq.test(x, y)$p.value, df[, -5], MoreArgs=list(df[,5]))
# gender smoking alcohol htn
# 0.4967724 0.8251178 0.5008898 0.3775083
Of course doing tests this way is somewhat statistically inefficient since you are doing multiple tests here so some correction is required to maintain an appropriate type 1 error rate.
You can run the following code chunk if you want to get the test result in details:
lapply(df[,-5], function(x) chisq.test(table(x,df$tertile), simulate.p.value = TRUE))
You can get just p-values:
lapply(df[,-5], function(x) chisq.test(table(x,df$tertile), simulate.p.value = TRUE)$p.value)
This is to get the p-values in the data frame:
data.frame(lapply(df[,-5], function(x) chisq.test(table(x,df$tertile), simulate.p.value = TRUE)$p.value))
Thanks to RPub for inspiring.
http://www.rpubs.com/kaz_yos/1204
I am working with the cumulative emergence of flies over time (taken at irregular intervals) over many summers (though first I am just trying to make one year work). The cumulative emergence follows a sigmoid pattern and I want to create a maximum likelihood estimation of a 3-parameter Weibull cumulative distribution function. The three-parameter models I've been trying to use in the fitdistrplus package keep giving me an error. I think this must have something to do with how my data is structured, but I cannot figure it out. Obviously I want it to read each point as an x (degree days) and a y (emergence) value, but it seems to be unable to read two columns. The main error I'm getting says "Non-numeric argument to mathematical function" or (with slightly different code) "data must be a numeric vector of length greater than 1". Below is my code including added columns in the df_dd_em dataframe for cumulative emergence and percent emergence in case that is useful.
degree_days <- c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94)
emergence <- c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0)
cum_em <- cumsum(emergence)
df_dd_em <- data.frame (degree_days, emergence, cum_em)
df_dd_em$percent <- ave(df_dd_em$emergence, FUN = function(df_dd_em) 100*(df_dd_em)/46)
df_dd_em$cum_per <- ave(df_dd_em$cum_em, FUN = function(df_dd_em) 100*(df_dd_em)/46)
x <- pweibull(df_dd_em[c(1,3)],shape=5)
dframe2.mle <- fitdist(x, "weibull",method='mle')
Here's my best guess at what you're after:
Set up data:
dd <- data.frame(degree_days=c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94),
emergence=c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0))
dd <- transform(dd,cum_em=cumsum(emergence))
We're actually going to fit to an "interval-censored" distribution (i.e. probability of emergence between successive degree day observations: this version assumes that the first observation refers to observations before the first degree-day observation, you could change it to refer to observations after the last observation).
library(bbmle)
## y*log(p) allowing for 0/0 occurrences:
y_log_p <- function(y,p) ifelse(y==0 & p==0,0,y*log(p))
NLLfun <- function(scale,shape,x=dd$degree_days,y=dd$emergence) {
prob <- pmax(diff(pweibull(c(-Inf,x), ## or (c(x,Inf))
shape=shape,scale=scale)),1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
library(bbmle)
I should probably have used something more systematic like the method of moments (i.e. matching the mean and variance of a Weibull distribution with the mean and variance of the data), but I just hacked around a bit to find plausible starting values:
## preliminary look (method of moments would be better)
scvec <- 10^(seq(0,4,length=101))
plot(scvec,sapply(scvec,NLLfun,shape=1))
It's important to use parscale to let R know that the parameters are on very different scales:
startvals <- list(scale=1000,shape=1)
m1 <- mle2(NLLfun,start=startvals,
control=list(parscale=unlist(startvals)))
Now try with a three-parameter Weibull (as originally requested) -- requires only a slight modification of what we already have:
library(FAdist)
NLLfun2 <- function(scale,shape,thres,
x=dd$degree_days,y=dd$emergence) {
prob <- pmax(diff(pweibull3(c(-Inf,x),shape=shape,scale=scale,thres)),
1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
startvals2 <- list(scale=1000,shape=1,thres=100)
m2 <- mle2(NLLfun2,start=startvals2,
control=list(parscale=unlist(startvals2)))
Looks like the three-parameter fit is much better:
library(emdbook)
AICtab(m1,m2)
## dAIC df
## m2 0.0 3
## m1 21.7 2
And here's the graphical summary:
with(dd,plot(cum_em~degree_days,cex=3))
with(as.list(coef(m1)),curve(sum(dd$emergence)*
pweibull(x,shape=shape,scale=scale),col=2,
add=TRUE))
with(as.list(coef(m2)),curve(sum(dd$emergence)*
pweibull3(x,shape=shape,
scale=scale,thres=thres),col=4,
add=TRUE))
(could also do this more elegantly with ggplot2 ...)
These don't seem like spectacularly good fits, but they're sane. (You could in principle do a chi-squared goodness-of-fit test based on the expected number of emergences per interval, and accounting for the fact that you've fitted a three-parameter model, although the values might be a bit low ...)
Confidence intervals on the fit are a bit of a nuisance; your choices are (1) bootstrapping; (2) parametric bootstrapping (resample parameters assuming a multivariate normal distribution of the data); (3) delta method.
Using bbmle::mle2 makes it easy to do things like get profile confidence intervals:
confint(m1)
## 2.5 % 97.5 %
## scale 1576.685652 1777.437283
## shape 4.223867 6.318481
dd <- data.frame(degree_days=c(998.08,1039.66,1111.29,1165.89,1236.53,1293.71,
1347.66,1387.76,1445.47,1493.44,1553.23,1601.97,
1670.28,1737.29,1791.94,1849.20,1920.91,1967.25,
2036.64,2091.85,2152.89,2199.13,2199.13,2263.09,
2297.94,2352.39,2384.03,2442.44,2541.28,2663.90,
2707.36,2773.82,2816.39,2863.94),
emergence=c(0,0,0,1,1,0,2,3,17,10,0,0,0,2,0,3,0,0,1,5,0,0,0,0,
0,0,0,0,1,0,0,0,0,0))
dd$cum_em <- cumsum(dd$emergence)
dd$percent <- ave(dd$emergence, FUN = function(dd) 100*(dd)/46)
dd$cum_per <- ave(dd$cum_em, FUN = function(dd) 100*(dd)/46)
dd <- transform(dd)
#start 3 parameter model
library(FAdist)
## y*log(p) allowing for 0/0 occurrences:
y_log_p <- function(y,p) ifelse(y==0 & p==0,0,y*log(p))
NLLfun2 <- function(scale,shape,thres,
x=dd$degree_days,y=dd$percent) {
prob <- pmax(diff(pweibull3(c(-Inf,x),shape=shape,scale=scale,thres)),
1e-6)
## multinomial probability
-sum(y_log_p(y,prob))
}
startvals2 <- list(scale=1000,shape=1,thres=100)
m2 <- mle2(NLLfun2,start=startvals2,
control=list(parscale=unlist(startvals2)))
summary(m2)
#graphical summary
windows(5,5)
with(dd,plot(cum_per~degree_days,cex=3))
with(as.list(coef(m2)),curve(sum(dd$percent)*
pweibull3(x,shape=shape,
scale=scale,thres=thres),col=4,
add=TRUE))
In the following code I use bootstrapping to calculate the C.I. and the p-value under the null hypothesis that two different fertilizers applied to tomato plants have no effect in plants yields (and the alternative being that the "improved" fertilizer is better). The first random sample (x) comes from plants where a standard fertilizer has been used, while an "improved" one has been used in the plants where the second sample (y) comes from.
x <- c(11.4,25.3,29.9,16.5,21.1)
y <- c(23.7,26.6,28.5,14.2,17.9,24.3)
total <- c(x,y)
library(boot)
diff <- function(x,i) mean(x[i[6:11]]) - mean(x[i[1:5]])
b <- boot(total, diff, R = 10000)
ci <- boot.ci(b)
p.value <- sum(b$t>=b$t0)/b$R
What I don't like about the code above is that resampling is done as if there was only one sample of 11 values (separating the first 5 as belonging to sample x leaving the rest to sample y).
Could you show me how this code should be modified in order to draw resamples of size 5 with replacement from the first sample and separate resamples of size 6 from the second sample, so that bootstrap resampling would mimic the “separate samples” design that produced the original data?
EDIT2 :
Hack deleted as it was a wrong solution. Instead one has to use the argument strata of the boot function :
total <- c(x,y)
id <- as.factor(c(rep("x",length(x)),rep("y",length(y))))
b <- boot(total, diff, strata=id, R = 10000)
...
Be aware you're not going to get even close to a correct estimate of your p.value :
x <- c(1.4,2.3,2.9,1.5,1.1)
y <- c(23.7,26.6,28.5,14.2,17.9,24.3)
total <- c(x,y)
b <- boot(total, diff, strata=id, R = 10000)
ci <- boot.ci(b)
p.value <- sum(b$t>=b$t0)/b$R
> p.value
[1] 0.5162
How would you explain a p-value of 0.51 for two samples where all values of the second are higher than the highest value of the first?
The above code is fine to get a -biased- estimate of the confidence interval, but the significance testing about the difference should be done by permutation over the complete dataset.
Following John, I think the appropriate way to use bootstrap to test if the sums of these two different populations are significantly different is as follows:
x <- c(1.4,2.3,2.9,1.5,1.1)
y <- c(23.7,26.6,28.5,14.2,17.9,24.3)
b_x <- boot(x, sum, R = 10000)
b_y <- boot(y, sum, R = 10000)
z<-(b_x$t0-b_y$t0)/sqrt(var(b_x$t[,1])+var(b_y$t[,1]))
pnorm(z)
So we can clearly reject the null that they are the same population. I may have missed a degree of freedom adjustment, I am not sure how bootstrapping works in that regard, but such an adjustment will not change your results drastically.
While the actual soil beds could be considered a stratified variable in some instances this is not one of them. You only have the one manipulation, between the groups of plants. Therefore, your null hypothesis is that they really do come from the exact same population. Treating the items as if they're from a single set of 11 samples is the correct way to bootstrap in this case.
If you have two plots, and in each plot tried the different fertilizers over different seasons in a counterbalanced fashion then the plots would be statified samples and you'd want to treat them as such. But that isn't the case here.