Standard error of sampling distribution of means is generally defined as (population SD/sqrt(n)) where n=sample size. But where population parameters are not known I have seen sample SD being used in the numerator in place of population SD. But I believe this will result in a SE that is drastically different.Since obviously a sample SD will be much different from population SD. Can you explain this intuitively how this works? It would make more sense for me if, in the absence of population parameters which are unknown, we draw a large sample (with n>>>30), and then use the SD of that sample inthe numerator. But sample size will have to be really high(how high I do not know quantitatively). But even with this approach your SE is only an approximate value, not an accurate one

  • $\begingroup$ Can you be more specific about exactly what you are doing? Are you testing a hypothesis or making a confidence interval? Is this a one-sample problem about estimating a population proportion, a population mean, or a population variance? Or are you asking whether two populations are the same. $\endgroup$
    – BruceET
    Commented Feb 24, 2021 at 16:18
  • $\begingroup$ I thought my question was fairly simple. I am talking about the formula for standard error of means where in the absence of information on population variance, the formula used is standard error = (sample standard deviation)/sqrt(n) where n=sample size. My question is how can a single sample's standard deviation be a substitute for the population standard deviation originally used in the numerator. The standard error that we get is going to be highly inaccurate. so, how is this acceptable? I am trying to find the standard error of means when my population parameters are unknown. $\endgroup$
    – daraj
    Commented Feb 24, 2021 at 17:34

1 Answer 1


If you have a random sample of size $n$ from a normal population with known standard deviation $\sigma,$ then the standard error of the sample mean, $\bar X =\frac 1n\sum_{i=1}^n X_i,$ is $\sigma/\sqrt{n}.$

Then a 95% z confidence interval for $\mu$ is of the form $\bar X \pm 1.96\frac{\sigma}{\sqrt{n}},$ where $\pm 1.96$ cut probability from the upper and lower tails, respectively, of a standard normal distribution.

If $\sigma$ is unknown and estimated by the sample standard deviation $S=\sqrt{\frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2},$ then the (estimated) standard error of $\bar X$ is $S/\sqrt{n}.$ [Because, upon seeing $S$ in place of $\sigma,$ one knows we're dealing with an estimate, the word estimated is often omitted.]

Then a 95% t confidence interval for $\mu$ is of the form $\bar X \pm t^*\frac{\sigma}{\sqrt{n}},$ where $\pm t^*$ cut probability from the upper and lower tails, respectively, of Student's t distribution with degrees of freedom $\nu = n-1.$

In practice $t^* > 1.96,$ making the t confidence interval somewhat longer than the "corresponding" z CI. This extra length takes into account that the sample standard deviation $S$ is only approximately equal to $\sigma.$

For a small sample size, such as $n = 10,$ we have degrees of freedom $\nu = n-1 = 9$ and $t^* = 2.262 > 1.96.$ For samples of moderate size, such as $n = 30,$ we have $\nu = 29$ and $t^* = 2.04,$ close to $1.96.$ For large sample, such as $n = 200,$ we have $t^* = 1.972 \approx 1.96.$ [You can get these values of $t^*$ from a printed table or by using software, such as R as below.]

qt(.975, 9)
[1] 2.262157
qt(.975, 29)
[1] 2.04523
qt(.975, 199)
[1] 1.971957

As $n$ increases the sample standard deviation $S$ gets closer to the population standard deviation $\sigma,$ the estimated standard error of $\bar X$ gets more accurate, and the length of the 95% t CI gets closer to the length of a 95% z CI.

Suppose $\bar X$ is based on a normal population with $\sigma= 5$ of size $n=10.$ Then the margin of error of a 95% z confidence interval for unknown $\mu$ is $M = 1.96(5)/\sqrt{10} = 3.099$ and the width of the CI is $2M= 6.198.$ If $\sigma$ is unknown and we use the (highly variable) sample standard deviation $S$ to make a t confidence interval, then the margin of error will vary and will tend to be longer than $3.1$--on average $3.48,$ as shown in the R simulation below.

n = 10;  t.10 = qt(.975, 9)
E = replicate(10^5, t.10*sd(rnorm(n,50,5))/sqrt(n))
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.6693  2.8994  3.4508  3.4812  4.0224  7.5266 

For $n = 30,$ the expected value of the margin of error for a 95% t CI is $E(M) = 1.852.$ And for $n = 200,$ we have $E(M) = 0.696.$ [The exact values of $M$ in a 95% z CI for $n = 30$ and $200$ are $M=1.789$ and $M=0.693,$ respectively]

n = 30;  t.30 = qt(.975, n-1)
E = replicate(10^5, t.30*sd(rnorm(n,50,5))/sqrt(n))
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.8887  1.6838  1.8464  1.8520  2.0140  2.9576 

n = 200;  t.200 = qt(.975, n-1)
E = replicate(10^5, t.200*sd(rnorm(n,50,5))/sqrt(n))
    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.5443  0.6726  0.6963  0.6964  0.7198  0.8516 

For each sample size, the 95% z CIs and the 95% t CIs will cover (include) the population mean $\mu$ for 95% of samples, over the long run. For each sample size, the 'probability factors' $t^*$ are chosen to be sufficiently larger than $1.96$ to compensate for the variability of $S$ as an estimate of $\sigma,$ when $\sigma$ is not known.

The R program below illustrates that 95% of 100,000 z and t confidence intervals based on $n = 25$ normal observations cover the population mean $\mu.$ On average, the t CIs are a little longer.

n = 25;  mu = 50;  sg = 5;  t.25 = qt(.975,24)
m = 10^5; LL.t = UL.t = LL.z = UL.z = numeric(m)
for(i in 1:m) {
 x = rnorm(n,mu,sg); a=mean(x); s=sd(x)
 LL.z[i] = a - 1.96*sg/5
 UL.z[i] = a + 1.96*sg/5
 LL.t[i] = a - t.64*s/5
 UL.t[i] = a + t.64*s/5

mean(mu > LL.z & mu < UL.z)
[1] 0.95046       # aprx 95%
mean(mu > LL.t & mu < UL.t)
[1] 0.95023       # aprx 95%

2 * 1.96*sg/5
[1] 3.92          # exact length of z CIs
mean(UL.t - LL.t)
[1] 4.086751      # average length of t CIs
  • $\begingroup$ Thanks. I did not know about the t* and the t-distribution related math behind using the sample variance as substitute. Is there theory published somewhere on how they arrived at this? (ie the t* values which you use as multiplier to increase the confidence interval to account for the fact that sample SD is only an estimate of population SD)? I want to understand the following better: "Then a 95% t confidence interval for μ is of the form X¯±t∗σn√, where ±t∗ cut probability from the upper and lower tails, respectively, of Student's t distribution with degrees of freedom ν=n−1." $\endgroup$
    – daraj
    Commented Feb 24, 2021 at 20:40
  • $\begingroup$ Yes. Famous episode in statistical history. William S. Gossett writing under the pseudonym 'A. Student' (perhaps to preserve his employer's confidential affairs), introduced the t distribution. You can google various sites, perhaps starting with Wikipedia. $\endgroup$
    – BruceET
    Commented Feb 24, 2021 at 20:53
  • $\begingroup$ Thanks. Looks like the math involved is a bit more difficult than what I had anticipated when i had asked this question :-( Now I need to understand student's t distribution in order to get your answer fully starting from "degrees of freedom" which I have not encountered much before. I was hoping that there would be a simpler intuitive answer without involving much math. $\endgroup$
    – daraj
    Commented Feb 24, 2021 at 20:58
  • $\begingroup$ If it were simple and intuitive maybe Gossett wouldn't be famous. I seem to recall he collaborated with others on some mathematical details of 'Student's t distribution.' $\endgroup$
    – BruceET
    Commented Feb 24, 2021 at 21:00
  • $\begingroup$ Okay, at the very least, the question I asked has led me to a totally new distribution altogether. So it is revealing in that sense :-) Thanks $\endgroup$
    – daraj
    Commented Feb 24, 2021 at 21:04

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