calculating the standard error of the mean?

Question

The mean of a random sample of size $n = 35$ is going to be used to estimate the mean of a finite population of $N = 400$. Given that the population standard deviation is thought to be 9.355, what is the standard error of the mean? Ok so I know the the equation for calculating the standard error of the mean is
$\sigma M = (\sigma)/(\sqrt{N})$
$\sigma M =$ standard error of the mean
$\sigma =$ the standard deviation of the original distribution
$N =$ the sample size
$\sqrt{N} =$ Root of the sample size but I am not too sure on if I just use $400$ as $N$ or there is a trick to the question. Thank you in advance!

Answer 1

Technically, 'standard error of the mean' is just alternate terminology for $SD(\bar X).$ Because $Var(\bar X) = \sigma^2/n,$ that gives $SD(\bar X) = \sigma/\sqrt{n}.$

Often in applications, the population variance $\sigma^2$ is unknown and is estimated by the sample variance $S^2.$ In that one often uses 'standard error of the mean' as shorthand for '$estimated$ standard deviation of the mean'.

For example, this gives rise to the 95% t confidence interval for the population mean $\mu$: $$ \bar X \pm t^* S/\sqrt{n},$$ where $t^*$ cuts probability 2.5% from the upper tail of Student's t distribution with $n - 1$ degrees of freedom (and by symmetry $-t^*$ cuts 2.5% from the lower tail). Using tables of the t distribution, you need the 'parameter' $DF = n-1$, to look up the appropriate value $t^*$.

For $n > 30$ you will see that $t^* \approx 2.0.$ This gives rise to the rough and informal statement that the CI extends about 'two standard errors on either side of the sample mean'.

Maybe the 'trick' you are thinking about involves the distinction between sample size $n$ and degrees of freedom $n - 1.$ Always use $n$ when computing the standard error, and always use DF when looking up values in a table of Student's t distribution.