Question

Suppose a large population has mean μ and standard deviation σ, and a simple random sample of size n is taken. The sampling distribution of the sample mean has mean and variance respectively equal to

Answer 1

mean \(\mu\) and variance \(\sigma^2/n\). This can easily be shown:
Say you have \(X_1,\ldots,X_n\) being a random sample n random variables. Since it's a random sample, they are iid. From the problem, we have \(E(X_i)=\mu\) and \(StDev(X_i)=\sqrt{Var(X_i)}=\sigma\)\(\implies\) \(Var(X_i)=\sigma^2\).

Now, let \[\bar{X}=\frac{1}{n}\sum_{i=1}^n X_i\] denote the sample mean random variable.

Thus, we need to find \(E(\bar{X})\) and \(Var(\bar{X})\).

\[E(\bar{X})=E\left(\frac{1}{n}\sum_{i=1}^n X_i\right)=\frac{1}{n}E\left(\sum_{i=1}^nX_i\right)=\frac{1}{n}\sum_{i=1}^nE(X_i)=\frac{1}{n}\sum_{i=1}^n\mu=\frac{1}{n}n\mu=\mu\]

\[Var(\bar{X})=Var\left(\frac{1}{n}\sum_{i=1}^n X_i\right)=\frac{1}{n^2}Var\left(\sum_{i=1}^nX_i\right)=\frac{1}{n^2}\sum_{i=1}^nVar(X_i),\text{by independence}\]
\[=\frac{1}{n^2}\sum_{i=1}^n\sigma^2=\frac{1}{n^2}n\sigma^2=\sigma^2/n\]

Answer 2

It is important to take into account the independence of the random variable for the variance, because otherwise there would be covariance terms.

Answer 3

I hope not too much of my stuff got cut off your screen.. the word "independence" got slightly cut off my screen but that is it.

Answer 4

Oh wait I just noticed this is a simple random sampling. Hm well the mean is the same, but the variance has the finite population correction factor, and is thus
\[\large \left(1-\frac{n}{N}\right)\frac{\sigma^2}{n}\]