# Hypothesis testing variance using sample mean

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!

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.