How to find the standard error of the mean?

Question

Is there any command to find the standard error of the mean in R?

Cristinecristiona

Answer 1

There's the plotrix package with has a built-in function for this: std.error

Shirelyshirey

Answer 2

The standard error is just the standard deviation divided by the square root of the sample size. So you can easily make your own function:

> std <- function(x) sd(x)/sqrt(length(x))
> std(c(1,2,3,4))
[1] 0.6454972

Bohol

Answer 3

The standard error (SE) is just the standard deviation of the sampling distribution. The variance of the sampling distribution is the variance of the data divided by N and the SE is the square root of that. Going from that understanding one can see that it is more efficient to use variance in the SE calculation. The sd function in R already does one square root (code for sd is in R and revealed by just typing "sd"). Therefore, the following is most efficient.

se <- function(x) sqrt(var(x)/length(x))

Bangle

Interestingly, your function and Ian's are nearly identically fast. I tested them both 1000 times against 10^6 million rnorm draws (not enough power to push them harder than that). Conversely, plotrix's function was always slower than even the slowest runs of those two functions - but it also has a lot more going on under the hood. – Oren

That is likely because the advantage of John's function avoids one function call and calculating the square root of a single number. If there is any advantage at all, it is at very small n. – Bohol

Not just likely but true... I did timing runs of rnorm(5) but did 5e4 executions of the se command. My version takes about 2.5s while the one with two sqrts takes about 2.8. A small speed up indeed. (I did several runs and am reporting a rounded average) – Bangle

Note that stderr is a function name in base. – Wince

That's a very good point. I typically use se. I have changed this answer to reflect that. – Bangle

Tom, NO stderr does NOT calculate standard error it displays display aspects. of connection – Brassiere

@Brassiere Tom didn't say stderr calculates the standard error, he was warning that this name is used in base, and John originally named his function stderr (check the edit history...). – Shrift

I believe the second version is not a good idea because for na.rm = TRUE it gives incorrect results as length(x) does not account for removed NAs. – Florentinaflorentine

Correct @Quasimodo. I've removed the erroneous function. Now the answer is just back to it's old simple self. The arandomlypickedname answer is correct for a more general case. – Bangle

Answer 4

A version of John's answer above that removes the pesky NA's:

stderr <- function(x, na.rm=FALSE) {
  if (na.rm) x <- na.omit(x)
  sqrt(var(x)/length(x))
}

Quotha

Note that there is an existing function called stderr in the base package that does something else, so it might be better to chose another name for this one, e.g. se – Ayacucho

Answer 5

The sciplot package has the built-in function se(x)

Silicle

Answer 6

As I'm going back to this question every now and then and because this question is old, I'm posting a benchmark for the most voted answers.

Note, that for @Ian's and @John's answers I created another version. Instead of using length(x), I used sum(!is.na(x)) (to avoid NAs). I used a vector of 10^6, with 1,000 repetitions.

library(microbenchmark)

set.seed(123)
myVec <- rnorm(10^6)

IanStd <- function(x) sd(x)/sqrt(length(x))

JohnSe <- function(x) sqrt(var(x)/length(x))

IanStdisNA <- function(x) sd(x)/sqrt(sum(!is.na(x)))

JohnSeisNA <- function(x) sqrt(var(x)/sum(!is.na(x)))

AranStderr <- function(x, na.rm=FALSE) {
  if (na.rm) x <- na.omit(x)
  sqrt(var(x)/length(x))
}

mbm <- microbenchmark(
  "plotrix" = {plotrix::std.error(myVec)},
  "IanStd" = {IanStd(myVec)},
  "JohnSe" = {JohnSe(myVec)},
  "IanStdisNA" = {IanStdisNA(myVec)},
  "JohnSeisNA" = {JohnSeisNA(myVec)},
  "AranStderr" = {AranStderr(myVec)}, 
  times = 1000)

mbm

Results:

Unit: milliseconds
       expr     min       lq      mean   median       uq      max neval cld
    plotrix 10.3033 10.89360 13.869947 11.36050 15.89165 125.8733  1000   c
     IanStd  4.3132  4.41730  4.618690  4.47425  4.63185   8.4388  1000 a  
     JohnSe  4.3324  4.41875  4.640725  4.48330  4.64935   9.4435  1000 a  
 IanStdisNA  8.4976  8.99980 11.278352  9.34315 12.62075 120.8937  1000  b 
 JohnSeisNA  8.5138  8.96600 11.127796  9.35725 12.63630 118.4796  1000  b 
 AranStderr  4.3324  4.41995  4.634949  4.47440  4.62620  14.3511  1000 a  


library(ggplot2)
autoplot(mbm)

Stool

Answer 7

Remembering that the mean can also by obtained using a linear model, regressing the variable against a single intercept, you can use also the lm(x~1) function for this!

Advantages are:

• You obtain immediately confidence intervals with confint()
• You can use tests for various hypothesis about the mean, using for example car::linear.hypothesis()
• You can use more sophisticated estimates of the standard deviation, in case you have some heteroskedasticity, clustered-data, spatial-data etc, see package sandwich

## generate data
x <- rnorm(1000)

## estimate reg
reg <- lm(x~1)
coef(summary(reg))[,"Std. Error"]
#> [1] 0.03237811

## conpare with simple formula
all.equal(sd(x)/sqrt(length(x)),
          coef(summary(reg))[,"Std. Error"])
#> [1] TRUE

## extract confidence interval
confint(reg)
#>                   2.5 %    97.5 %
#> (Intercept) -0.06457031 0.0625035

^{Created on 2020-10-06 by the reprex package (v0.3.0)}

Learn

Answer 8

You can use the function stat.desc from pastec package.

library(pastec)
stat.desc(x, BASIC =TRUE, NORMAL =TRUE)

you can find more about it from here: https://www.rdocumentation.org/packages/pastecs/versions/1.3.21/topics/stat.desc

Tammitammie