Finding the hypothesis test of $\sigma^2$

Question

I have been working on a problem and this is what is given.

$$X_1, ... X_n \sim_{iid} N(\mu_0,\sigma ^2 )$$

I was able to find the MLE of $\sigma ^2 $ which is

$$\hat{\sigma^2}_{MLE} = \frac{1}{n} \sum_{i = 1}^{n}(X_i-\mu_0)^2 $$

I was also able to find that

$$\sim \chi^2_n$$

How would I be able to construct a hypothesis testing where

$$H_0: \sigma^2=\sigma^2_0 \quad \text{vs} \quad H_1:\sigma^2> \sigma^2_0$$

?

I want to say that I start with

$$Pr\left[\frac{n \hat{\sigma^2}_{MLE}}{\sigma^2} >k \quad| \quad \sigma^2= \sigma^2_0 \right] = \alpha$$

but I am not sure how to work from there and also find the power function . . .

I would really appreciate your help.

Answer 1

I will try to get you started on this.

To keep notation simple, let $V = \frac 1 n \sum_i(X_i - \mu_0)^2,$ for $\mu_0$ known. Then $$\frac{nV}{\sigma^2} = \sum_{i=1}^n \left(\frac{X_i - \mu_0}{\sigma}\right)^2 = \sum_{i=1}^nZ_i^2 \sim \mathsf{Chisq}(n),$$ where $Z_i$ are IID standard normal.

To test $H_0: \sigma^2 = \sigma_0^2$ against $H_a: \sigma^2 > \sigma_0^2,$ you want to reject for large values of $V,$ say when $V > c.$ Specifically, you will reject at the 5% level when $nV/\sigma_0^2 > U$, where $U$ cuts 5% from the upper tail of $\mathsf{Chisq}(n)$ or when $V >\sigma_0^2U/n = c.$

To find the power against a particular $\sigma_a^2 > \sigma_0^2,$ you need to find $P(V > c\,|\,\sigma^2 = \sigma_a^2).$

---

Perhaps it is easier to think about significance levels and power if you look at a couple of particular datasets. Here are two generated using R statistical software.

Dataset 1: In testing $H_0: \sigma^2 = 100$ vs $H_a: \sigma^2 > 100$ you might hope not to reject $H_0$ at the 5% level because data are randomly sampled from a population with $\sigma^2 = 100.$
Here $V = 87.72, U = 37.65.$ (You could also find $U$ from printed tables of the chi-squared distribution with 25 degrees of freedom.) Begin the test by finding $c.$

(Results are from R statistical software. R uses the SD parameter $\sigma = 10$ in rnorm. You can ignore numbers in square brackets [ ], which give the index of the first number in a line of output.)

set.seed(1234)  # for reproducibility
> x = round(rnorm(25, 100, 10), 2); x
 [1]  87.93 102.77 110.84  76.54 104.29 105.06  94.25
 [8]  94.53  94.36  91.10  95.23  90.02  92.24 100.64
[15] 109.59  98.90  94.89  90.89  91.63 124.16 101.34
[22]  95.09  95.59 104.60  93.06
v = sum((x-100)^2)/25;  v
## 87.71786
U = qchisq(.95, 25); U
## 37.65248

Dataset 2: Again test $H_0: \sigma^2 = 100$ vs $H_a: \sigma^2 > 100,$ but this time you might hope to reject because data are sampled from a normal population with $\sigma^2 = 20^2 = 400.$

set.seed(4321)  # for reproducibility
x = round(rnorm(25, 100, 20), 2); x
 [1]  91.46  95.53 114.35 116.83  97.43 132.19  94.06
 [8] 103.92 124.81  85.63  98.66 106.89  74.78 122.79
[15]  75.56 131.47 101.47  76.50  68.23  85.05 109.67
[22]  99.94  99.82 111.87  98.02
v = sum((x-100)^2)/25;  v
## 292.622
U = qchisq(.95, 25); U
## 37.65248

Finally, as you consider how to find your power function, you might ask for the probability of rejecting $H_0: \sigma^2 = 100$ when the actual $\sigma^2 = 400.$ as in Dataset 2.