# Calculate the confidence interval of parameter of exponential distribution with summarySE in R?

I am trying to calculate the confidence interval for a set of data with the assumption they follow Exp dist. To achieve this, I am merging this with this in R, but does not work as I am not very efficient in R.

 summarySE <- function(data=NULL, measurevar, groupvars=NULL, na.rm=FALSE,
conf.interval=.95, .drop=TRUE) {
library(plyr)

# New version of length which can handle NA's: if na.rm==T, don't count them
length2 <- function (x, na.rm=FALSE) {
if (na.rm) sum(!is.na(x))
else       length(x)
}

# This does the summary. For each group's data frame, return a vector with
# N, mean, and sd
datac <- ddply(data, groupvars, .drop=.drop,
.fun = function(xx, col) {
c(N    = length2(xx[[col]], na.rm=na.rm),
mean = mean   (xx[[col]], na.rm=na.rm),
sd   = sd     (xx[[col]], na.rm=na.rm)
)
},
measurevar
)

# Rename the "mean" column
datac <- rename(datac, c("mean" = measurevar))

#   datac$$se <- datac$$sd / sqrt(datac$N) # Calculate standard error of the mean # Confidence interval multiplier for standard error # Calculate t-statistic for confidence interval: # e.g., if conf.interval is .95, use .975 (above/below), and use df=N-1 # ciMult <- qt(conf.interval/2 + .5, datac$$N-1) datac$$ci <- qgamma(c(.025,.975), 20, 20)/ mean(datac$measurevar )

return(datac)
}

mean <- 25
dataframe <- data.frame(rnd=rexp(1000, rate =
1/mean),Description=c(rep('A',500),rep('B',500)),dayname= c(rep(
'Sunday',500),rep( 'Monday',500)))

tgc= summarySE(dataframe, measurevar="rnd", groupvars=c("Description"
,"dayname"))