I'm new to R and trying to learn stats..
Here is one practice question that I'm trying to figure out
How should I use R code to create a function based on this math equation?
I have a dataframe like this
the "exposed" column from the df contains two groups, one is called"Test Group (Exposed)" the other one is called "Control Group". So the math function is referring to these two groups.
In another practice I have these codes here to calculate the confidence interval
# sample size
# OK for non normal data if n > 30
n <- 150
# calculate the mean & standard deviation
will_mean <- mean(will_sample)
will_s <- sd(will_sample)
# normal quantile function, assuming mean has a normal distribution:
qnorm(p=0.975, mean=0, sd=1) # 97.5th percentile for a N(0,1) distribution
# a.k.a. Z = 1.96 from the standard normal distribution
# calculate standard error of the mean
# standard error of the mean = mean +/- critical value x (s/sqrt(n))
# "q" functions in r give the value of the statistic at a given quantile
critical_value <- qt(p=0.975, df=n-1)
error <- critical_value * will_s/sqrt(n)
# confidence inverval
will_mean - error
will_mean + error
but I'm not sure how to do the exposed 2 groups
Don't worry it's quite easy if you have experience in at least one programming language, R is quite trivial.
The only remarkable difference between R and most of other programming languanges is that R was developed for statistical purposes.
You can compute what is the quantile for a certain significance level α (reminds to divide it by 2 for your formula) by using the function qnorm(). By default it is set up for standardized normal distribution, like in your case, but you can get more details using the documentation, reachable by the command ?qnorm().
Actually in the exercise you are not required to compute it, since you have to pass it as argument, but in reality you need to.
The code should be something like:
conf <- function(p1,p2,n1,n2,z){
part = z*(p1*(1-p1)/n1+p2*(1-p2)/n2)**(1/2)
return(c(p1-p2-part,
p1-p2+part))
}
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.
Is it possible to/how can I generate a beta-binomial distribution from an existing vector?
My ultimate goal is to generate a beta-binomial distribution from the below data and then obtain the 95% confidence interval for this distribution.
My data are body condition scores recorded by a veterinarian. The values of body condition range from 0-5 in increments of 0.5. It has been suggested to me here that my data follow a beta-binomial distribution, discrete values with a restricted range.
set1 <- as.data.frame(c(3,3,2.5,2.5,4.5,3,2,4,3,3.5,3.5,2.5,3,3,3.5,3,3,4,3.5,3.5,4,3.5,3.5,4,3.5))
colnames(set1) <- "numbers"
I see that there are multiple functions which appear to be able to do this, betabinomial() in VGAM and rbetabinom() in emdbook, but my stats and coding knowledge is not yet sufficient to be able to understand and implement the instructions provided on the function help pages, at least not in a way that has been helpful for my intended purpose yet.
We can look at the distribution of your variables, y-axis is the probability:
x1 = set1$numbers*2
h = hist(x1,breaks=seq(0,10))
bp = barplot(h$counts/length(x1),names.arg=(h$mids+0.5)/2,ylim=c(0,0.35))
You can try to fit it, but you have too little data points to estimate the 3 parameters need for a beta binomial. Hence I fix the probability so that the mean is the mean of your scores, and looking at the distribution above it seems ok:
library(bbmle)
library(emdbook)
library(MASS)
mtmp <- function(prob,size,theta) {
-sum(dbetabinom(x1,prob,size,theta,log=TRUE))
}
m0 <- mle2(mtmp,start=list(theta=100),
data=list(size=10,prob=mean(x1)/10),control=list(maxit=1000))
THETA=coef(m0)[1]
We can also use a normal distribution:
normal_fit = fitdistr(x1,"normal")
MEAN=normal_fit$estimate[1]
SD=normal_fit$estimate[2]
Plot both of them:
lines(bp[,1],dbetabinom(1:10,size=10,prob=mean(x1)/10,theta=THETA),
col="blue",lwd=2)
lines(bp[,1],dnorm(1:10,MEAN,SD),col="orange",lwd=2)
legend("topleft",c("normal","betabinomial"),fill=c("orange","blue"))
I think you are actually ok with using a normal estimation and in this case it will be:
normal_fit$estimate
mean sd
6.560000 1.134196
For an analysis I need to perform a "F-pseudosigma", also called the "pseudo standard deviation". I tried to look if it's in any R package, but can't find it myself.
There isn't much info on it to begin with.
Does any of you know a package that holds it, or if it is calculated in a function from a package?
I have to admit that I haven't heard about F-pseudo sigma (or pseudo sigma) before; but a bit of research suggests that it is simply defined as the scaled difference between the third and first quartile.
That can be easily translated into a custom R function
fpseudosig <- function(x) unname(diff(quantile(x, c(0.25, 0.75)) / 1.35))
For example, let's generate some random data x ~ N(0, 1)
set.seed(2018)
x <- rnorm(100)
Then
fpseudosig(x)
#[1] 0.9703053
References
(in no particular order)
Irwin, Exploratory Data Analysis for Beginners: "Instead of the using the standard deviation in an RSD calculation, one might consider using the sample-data deviation (F-pseudosigma). This is a nonparametric statistic analogous to the standard deviation that is calculated by using the 25th and 75th percentiles in a data set. It is resistant to the effect of extreme outliers."
https://bqs.usgs.gov/srs/SRS_Spr04/statrate.htm: "The F-pseudosigma is calculated by dividing the fourth-spread (analogous to interquartile range) by 1.349; therefore the smaller the F-pseudosigma the more precise the determinations. The 1.349 value is derived from the number of standard deviations that encompasses 50% of the data."
http://mkseo.pe.kr/stats/?p=5: "Simply put, given the first quartile H1 and the third quartile H3, pseudo sigma is (H3-H1)/1.35. Why? It’s because H1= μ – 0.675σ and H3 = μ + 0.675σ if X ∼N. Therefore, H3-H1=1.35σ, resulting in σ = (H3-H1)/1.35. We call H3-H1 as IQR(Inter Quartile Range)."
I want to know whether the sd command in R works accurately when calculating the standard deviation of a binomial distribution.
Take the following example:
coin <- c("heads", "heads", "tails", "heads", "tails", "heads", "heads", "tails")
die <- as.factor(coin)
The standard deviation formula for such a distribution would be:
sd <- sqrt(n*p*(1-p))
where n is the number of trials, and p is the probability of success.
So we would calculate it in R as follows:
sqrt(8*(5/8)*(3/8))
[1] 1.369306
However, when we use the sd command, we get a different answer:
sd(coin)
[1] 0.5175492
Does the sd function in R not take into consideration the fact that the variable is not numeric. That explanation would make sense to me if R returned an error message, but it produces a result. Can you please clarify this for me? Thanks.
The sd function returns the (corrected) sample standard deviation (not the theoretic SD of a Bernoulli random variable). The sample SD is defined as sqrt( sum((x-x_bar)^2)/(N-1)). See ?sd and ?var Your example can be checked:
samp_var_die <- sum((as.numeric(die)-mean(as.numeric(die)))^2)/(length(die)-1)
samp_sd_die <- sqrt(samp_var_die)
samp_sd_die
#[1] 0.5175492
If you are interested in exploring the theoretic aspects of statistical distributions, there is a nice suite of packages devoted to this topic. Check out the distr-package and it's cousins: distrEllipse, distrEx, distrMod, distrRmetrics, distrSim, distrTeach, and RandVar. I found playing with functions and examples from those packages quite educational and entertaining.
By the way, that 1.3+ value you got would be the SD (theoretic sigma) around the estimate of 5 you would have gotten from that series of observations.
I am an statistics student and R beginner (understatement of the year) trying to generate multiple confidence intervals for randomly generated samples of a normal distribution as part of an assignment.
I used the function
data <- replicate(25, rnorm(20, 50, 6))
to generate 25 samples of size n=20 from a N(50, 6^2) distribution (in a double matrix).
My question is, how do I find a 95% confidence interval for each sample of this distribution? I know that I can use colMeans(data) and sd(data) to find the sample mean and sample standard deviation for each sample, but I am having a brain fart trying to think of a function that can generate the confidence intervals for all columns in the double matrix (data).
As of now, my (extremely crude) solution consists of creating the functions
left <- function (x,y){x-(qnorm(0.975)*y/sqrt(20))}
right <- function (x,y){x+(qnorm(0.975)*y/sqrt(20))}
left(colMeans(data), sd(data)
right(colMeans(data), sd(data)
to generate 2 vectors of left and right bounds. Please let me know if there is a better way I can do this.
I suppose you could use the t.test() function. It returns the mean and the 95% confidence interval for a given vector of numbers.
# Create your data
data <- replicate(25, rnorm(20, 50, 6))
data <- as.data.frame(data)
After you make your data, you could apply the t.test() function to all columns using the lapply() function.
# Apply the t.test function and save the results
results <- lapply(data, t.test)
If you only want to see the confidence interval or mean returned, you can call them using the dollar sign operator. For example, for column one of your original data frame, you could type the following:
# Check 95% CI for sample one
results[[1]]$conf.int[1:2]
You could come up with a more eloquent way of saving these data to a results data frame. Remember, you can always see what individual bits of information you can yank from an object by using the str() command. For example:
# Example
example <- t.test(data[,1])
str(example)
Hope this helps. Try this link for more information: Using R to find Confidence Intervals