Suppose $x_1,x_2,...,x_{20}$ is a random sample from a normal population with mean = 0 and variance $ \sigma ^2 $. I want to test the hypothesis $H_0: \sigma ^2 \geq 4$ against the alternative $H_1: \sigma ^2 < 4$ at level $0.05$. How can I can find the distribution of the test statistic?
I think it involves a t distribution with 19 degrees of freedom.
Student's t distribution with 19 "degrees of freedom" would be correct for testing a hypothesis about the mean of normal (or nearly normal) data, where the standard deviation (or variance) is unknown. Your situation is to test a hypothesis about the variance of a distribution where the mean is known, which is quite different.
Let us suppose your 20 observations $x_i$ are from a normal population with known mean 0 and unknown variance $\sigma^2.$ Let $Q = \sum_{i = 1}^{20} x_i^2$. Then $Q/\sigma^2 \sim CHISQ(n),$ the chi-squared distribution with $n$ degrees of freedom (where $n = 20$ in your problem).
Specifically, suppose you have data (rounded to 2 places) as follows:
0.45, 8.46, 6.85, -1.98, 5.80, -0.39, -4.83, 0.80, 2.04, -0.88, -5.96, 1.64, -2.26, 2.88, 6.93, 2.15, 0.86, 0.56, 5.63, -3.34.
For these data, $Q = 333.4803$. An unbiased point estimate of the population variance $\sigma ^2$ is $T = Q/20 = 16.64.$
A test at the 5% level might use lower and upper critical values 9.5908 and 34.1696. respectively. These values cut 2.5% from the lower and upper tails, respectively, of $CHISQ(20)$. Because $Q/16 = Q/\sigma_0 ^2 = 20.84$ lies between the two critical values, we do not reject the null hypothesis.
Notes: (1) The critical values can be obtained using software or printed tables of the chi-squared distribution. (2) It is customary to use a "probability symmetric" test that cuts the same probability from each tail of the relevant chi-squared distribution, even though a slightly better test at the 5% level might be achieved by cutting a little less than 2.5% from one tail and a little more from the other. (3) My data above were simulated from a normal distribution with mean 0 and variance 25, so we might have hoped to reject the null hypothesis. But this test with only 20 observations does not have enough power to distinguish between $\sigma = 4$ and $\sigma = 5$.
Caution: A more common problem deals with $n$ observations from a normal population for which neither the mean $\mu$ nor the variance $\sigma ^ 2$ is known, and $S^2 = \sum_{i=1}^n (x_i - \bar x)^2/(n-1)$ is an unbiased estimate of $\sigma^2.$ Then $(n-1)S^2/\sigma^2 \sim CHISQ(n - 1)$. This situation is routinely covered in statistics texts at a variety of levels, and I will not elaborate here.