Calculating Variance: Sample vs Population

[CAH]

See the photo of the question+ mark scheme!
I don't understand a question on finding an expression for the variance of something...

Attempt at solution: also I worked out c as (3/2) previously, which is correct
Var(U) = (3/2)^2 x Var(Xbar) = 9/4n^2 x a^2/18

I'll attach a photo of this too if it's easier to read, my problem is that I thought var(Xbar) = var(x/n) = 1/n^2 x var(x)

... But they have done var(Xbar) = 1/n x var(x) ...?

 

[statdad]

Think about this:
<br /> Var\left(\bar X\right) = Var\left(\frac 1 n \sum_{i=1}^n X_i\right) = \frac 1 {N^2} \sum_{i=1}^n Var(X_i)<br />

What happens when you simplify the sum?

 

[CAH]

Ah I see so it's just sigma^2/n, is this the same for all cases as in, calculating the variance of a sample and estimating a population variance?.. In other words, kinda, will it ever (a level standard) be 1/n^2 x var(x)

Probably a stupid question, just checking

 

[statdad]

The only time you would have n^2 in the denominator is if you were (for some reason) considering mathematically a single observation X_1 and calculate
<br /> Var\left( \dfrac{X_1}{n}\right) = \dfrac{Var(X)}{n^2}<br />

"is this the same for all cases as in, calculating the variance of a sample and estimating a population variance?"
I'm not exactly sure what you mean by this, so if my response is off-target that's why.

If you've talked about sampling distributions for the sample mean, the expression \frac{\sigma^2}n is the population variance for that sampling distribution. It will never have denominator n^2, since, as long as the distribution being sampled has a variance, the steps shown above apply.

The (sample) variance of a sample is a different beast. Essentially
* if the population variance is \sigma^2, then the sample variance
<br /> s^2 = \dfrac 1 {n-1} \sum_{i=1}^n \, \left(x_i - \bar x\right)^2<br />
is an unbiased estimator of the population variance

* If you refer to the variance of the sampling distribution of \bar x - which is given above - then to estimate that you have two options
a) If you have a single sample, use the sample variance s^2 to estimate the sampling distribution's variance by calculating
<br /> \dfrac{s^2}{n}<br />

b) If you have a large number of samples, all the same sample size, from the same population, then calculate each sample mean and treat those sample means as a new sample. The sample variance of those (call it s_{\bar x}^2) is the estimate of \dfrac{\sigma^2}{n}

So there are several subtleties to wade through, but in none but the one unusual and unrealistic comment I made at the start will \frac{\sigma^2}{n^2} play a role.