Hypothesis testing variance using sample mean

Question

I know how to test hypotheses for variance using methods like the chi-square test. However, this problem is asking me to use a rejection region construction in terms of the sum of the sample values (I assume that could be interpreted as n*(sample mean)) $$R = \{x\mid x_1 + x_2 + x_3 + x_4 >\gamma\}$$

The problem asks me to find the value of the scalar γ for which the significance level of the rejection region above is 0.05, and then find the probability of type II error.

Original formulation of the problem:

I need some help. How do I approach this problem? Thank you!

Answer 1

This is not the usual way to distinguish between two variances, but as an exercise in hypothesis testing it might be interesting.

If $H_0: \sigma^2 = 16$ is true, then the sum of the observations has $S \sim \mathsf{Norm}(\mu = 80, \sigma = 8).$

By contrast, if $H_1: \sigma^2 = 25$ is true, then the sum $S \sim \mathsf{Norm}(\mu=80, \sigma = 10).$

In the sketch below, the blue curve is for $H_0$ and the red curve for $H_1.$ Roughly speaking, a total much above 90 may be slightly more favorable for rejecting $H_0$ in favor of $H_1.$ I will leave a formal comparison of the likelihoods to you.

Under $H_0,$ we have $P(S > 90) \approx 0.106$ and under $H_1, P(X > 90) \approx 0.159).$ [Computations in R statistical software.]

1 - pnorm(90, 80, 8)
## 0.1056498
1 - pnorm(90, 80, 10)
## 0.1586553

Such a test at the 5% level would have critical value $c \approx 3.16 $, and power only about $0.094.$

c = qnorm(.95, 80, 8);  c  
## 93.15883 
1 - pnorm(c, 80, 10)
## 0.09410667

Note: I'm wondering if the test statistic isn't also supposed to use the sum of squares of the four observations. Properly stated, that could lead to a better test using $Q = \sum_i(X_i - 20)^2/\sigma^2 \sim \mathsf{Chisq}(\nu = 4)$ as the test statistic.