Ref: https://onlinecourses.science.psu.edu/stat414/node/167

Let *X*_{1}, *X*_{2}, … , *X*_{n}_{ }be a random sample of size *n *from a distribution (population) with mean *μ*_{ }and variance *σ*^{2}. What is the variance of X¯? Var(X¯) named sample mean variance.

**Solution.** Starting with the definition of the sample mean, we have:

Var(X¯)=Var(X1+X2+⋯+Xnn)Var(X¯)=Var(X1+X2+⋯+Xnn)

Rewriting the term on the right so that it is clear that we have a linear combination of *X _{i}*‘s, we get:

Var(X¯)=Var(1nX1+1nX2+⋯+1nXn)Var(X¯)=Var(1nX1+1nX2+⋯+1nXn)

Then, applying the theorem on the last page, we get:

Var(X¯)=1n2Var(X1)+1n2Var(X2)+⋯+1n2Var(Xn)Var(X¯)=1n2Var(X1)+1n2Var(X2)+⋯+1n2Var(Xn)

Now, the *X _{i}* are identically distributed, which means they have the same variance

*σ*

^{2}. Therefore, replacing

*Var*(

*X*) with the alternative notation

_{i}*σ*

^{2}, we get:

Var(X¯)=1n2[σ2+σ2+⋯+σ2]Var(X¯)=1n2[σ2+σ2+⋯+σ2]

Now, because there are *n* *σ*^{2}‘s in the above formula, we can rewrite the expected value as:

Var(X¯)=1n2[nσ2]=σ2nVar(X¯)=1n2[nσ2]=σ2n

Our result indicates that as the sample size *n *increases, the variance of the sample mean decreases. That suggests that on the previous page, if the instructor had taken larger samples of students, she would have seen less variability in the sample means that she was obtaining. This is a good thing, but of course, in general, the costs of research studies no doubt increase as the sample size* n *increases. There is always a trade-off!