# Why does a sample of n of a population of n retain degrees of freedom of n?

I understand that if you take a sample from the population that a single data point cannot freely vary if $\bar{x}$ is known and you have the remaining sample items. However, I do not understand why "a sample of size $n$ retains $n$ degrees of freedom if the population mean $μ$ is known". The answer is sort of provided in this paper: Degrees of Freedom (Eisenhauer, J.G.), where it states:

'Note that a sample of size n retains n degrees of freedom if the population mean μ is known, since this does not determine $x_i$ for $i = 1 ... n$ if the other $(n–1)$ values are known. The concept is of importance in statistical inference since it defines the effective size of a sample.'

Which is actually a quote from "A Dictionary of Mathematics".

Why though, does knowing $μ$ "not determine $x_i$ for $i = 1 ... n$ if the other $(n–1)$ values are known"?

Using the same logic, I can calculate the remaining value if I know $n-1$ values and $μ$. So why are the degrees of freedom still $n$?

My intuitive guess is that by knowing $μ$ I must know the entire population of data points. However by the same logic, if I know $\bar{x}$ I must know my entire sample set of data. So I do not understand why the -1 rule does not apply to the population.

The rule seems arbitrary to me, since if the data set is 1000, and I have 999 points, my degrees of freedom are 998, but if my sample is the entire population (1000), then my degrees of freedom are 1000.

My discussion is a bit long winded, so please let me restate my question:

Why does knowing $μ$ "not determine $x_i$ for $i = 1 ... n$ if the other $(n–1)$ values are known"?

• Great example of a well-asked question (+1). – Reviewer – Jim Jun 2 '18 at 21:23

... because the mean of the $n$ values is not $\mu$.
• A sample mean would normally be denoted as $\bar{x}$; it's not the case that $\bar{x}=\mu$ in general. Consider the case where I have 99 tickets labelled 1,2,...,99. The population average of those values, $\mu$, is 50. I draw five of them and get 76, 48, 34, 8, 71, which averages 47.4. If we take the knowledge that the population mean is 50 and the first four sample values were 76, 48, 34, 8 that doesn't really constrain that last value (if we're sampling without replacement there's a few values it can't be, but otherwise it could be any of the other values). – Glen_b Jun 4 '18 at 6:40