Hypothesis testing for variance From a random sample $X_1;...;X_8\sim \mathcal N(3,σ^2)$, we would like to test whether
the variance $σ^2$ can be considered to be at least $5$, against the hypothesis
that it is less than 5. We have recorded $\sum_{i=1}^8 (x_i-3)^2 = 37.5$.
(i) Perform the relevant test at $1\%$ signicance level and conclude.
(ii) In terms of the cumulative distribution function of a known distribution, express the power of the test when (a) $σ^2 = 4$, (b) $σ^2 = 3$ and (c)
$σ^2 = 2$.
(iii) Explain briefly what would change if we did not know the true mean
in the population, but we had recorded the sample mean $x = 3$.
I got the answer for part i,the observed t-statistic is $7.5$ so do not reject the null hypothesis. But I have problem with part ii. I try to use the power method but it seems not working as in the statistic T there is only one variance $σ^2$, can anyone helps me out?
 A: In order to understand the power computation, it will help to write out
the original test in (1). I think it is bad for to call the test statistic $T$
because this has nothing to do with t tests or Student's t distribution; so
I'll use $Q$.
Testing the null hypothesis at level $\alpha.$ 
You want to test $H_0: \sigma^2 \ge 5$ against the alternative 
$H_a: \sigma^2 < 5.$
 Your estimate of the variance is 
$$V = \frac{1}{n}\sum_{i=1}^n (X_i - \mu)^2 
= \frac{1}{8}\sum_{i=1}^8 (X_i - 3)^2 = \frac{37.5}{8} = 4.6875.$$
Thus your estimate $V < 5,$ but the question is whether it is enough
smaller than $5$ to reject $H_0$ at the 1% level of significance.
The test statistic is 
$$Q = \frac{nV}{\sigma_0^2} =\sum_{i=1}^8 \frac{(X_i - 3)^2}{\sigma_0^2} = 
\sum_{i=1}^8 \left(\frac{X_i - 3}{\sigma_0} \right)^2 = \sum_{i=1}^8 Z_i^2,$$
where $Z_i \stackrel{iid}{\sim} \mathsf{Norm}(0,1),$ because $\mu = 3$ is known. Thus $Q \sim \mathsf{Chisq}(df=8).$
We will reject $H_0$ for sufficiently small values of $Q.$ Specifically,
the critical value $q^* = 1.646$ cuts 1% of the area from the lower tail
of $\mathsf{Chisq}(8).$ So we reject $H_0$ at the $\alpha =1\%$ level of
significance against the left-sided alternative if $Q < 1.646.$
 (You can get $q^*$ from a suitable printed table of chi-squared
distributions or by using software; the result from R statistical
software is shown below.)
qchisq(.01, 8)
## 1.646497

In summary, we can write $\alpha = P\{Q = nV/\sigma_0^2 \le q^* = 1.65\; |\; \sigma_0^2 = 5\} = 0.01.$ 
Power against alternative $\sigma_a.$ In finding the critical value $q^*$
we have used the value of $\sigma = \sigma_0 = 5,$ specified by the $=$-sign
in $H_0.$ One can find the power of the tests for any of the values
$\sigma_a$ of $\sigma$ in the alternative hypothesis. Specifically, let's
choose $\sigma_a = 2.$ The power is the probability of rejecting $H_0$
assuming that $\sigma = \sigma_a = 2.$ 
$\pi(\sigma_a=2) = P\{Q = nV/\sigma_0^2 \le q^* = 1.65\; |\; \sigma_0^2 = 2\} = ??.$
The test procedure and significance level $\alpha = .01$ remain
the same. What is different is that the data $X_i$ are now a random sample
from $\mathsf{N}(\mu = 3, \sigma^2 = 2).$  You no longer
have $V = 4.6875.$ You are imagining new data, for which you no longer have
$Q = nV/\sigma_0^2 \sim \mathsf{Chisq}(8).$ Your task in solving this
problem is to decide how to take this change into account. 
[I get $\pi(2) \approx 0.154.$ From simulation and from the R code below:]
 pchisq(qchisq(.01,8)*2.5, 8)
 ## 0.1535139

When $\mu$ is unknown. If $\mu$ is unknown, you would estimate it by
$\bar X.$ Then the estimate of $\sigma^2$ would be
$S^2 = \frac{1}{n-1}\sum_{i=1}^8 (X_i - \bar X)^2$ and
$Q = (n-1)S^2/\sigma^2 \sim \mathsf{Chisq}(n - 1).$
