$\chi^2$ distribution: hypothesis testing

Question

This is actually a statistic question from my econometrics class. It's not very difficult but I am having a lot of trouble with its wording.

Suppose we have a sample of 10 independent observations, drawn from a $\chi^2$ distributions. Denote the observations in our sample $\{X_1, \ldots, X_{10}\} \sim \chi^2_k$. We want to test the null hypothesis that $k = 1$ versus the alternative hypothesis that $k > 1.$ Formally, our null and alternative hypotheses are: $H_0:k=1$ and $H_1:k>1$.

So far so good, but I am a bit confused about the samples. Is each individual $X_i$ drawn from a $\chi^2_k$ distribution?

Continuing

Suppose the sample mean is $\bar{X}=1.5$. Recall that if $X \sim \chi^2_k$, then $E(X)=k \\$.

No problems here.

a. Propose an estimator for $k$.

My attempt: $\hat{k}=\frac{1}{N}\sum_{i=1}^N X_i$, in this case $N=10$. Now where the confusion begins:

b. Propose a test statistic to test the null hypothesis. Denote this estimator $T_N$. For now, just provide the formula (not the realized value). What is the distribution of $T_N$ under the null hypothesis?

Ummm.. I am not sure what my instructor means by distribution of $T_N$ under the null. I don't even know what my $T_N$ would be, although my guess is $$T_N=\frac{\hat{k}-k}{\frac{\sigma^2}{N}}$$ is that right?

What is the realized value of your test statistic in this sample?

There are more routine computation questions, but if I can get the wording of the estimator question right I believe I can answer them. Basically, it would be very helpful if someone could guide me through part b. Thank you.

Answer 1

Questions like this are always open to interpretation, and you actually do have some freedom in picking a test statistic. Choosing $$T_N = \frac{\hat{k}-1}{\sigma^2/N}$$ is fine, but a much simpler choice would be $$N\hat{k} = \sum_{i=1}^N X_i$$ In fact any test based on rejecting large values of $T_n$ will be equivalent to rejecting large values of $\sum_{i=1}^N X_i$. The reason for choosing $\sum_{i=1}^N X_i$ is, that it is much easier to compute it's distribution, since it is a sum of indepedent $\chi^2_{k}$- variables and is therefore $\chi^2_{kN}$ distributed (and simply $\chi^2_N$ under the null hypothesis).

Answer 2

The sum of independent chi-square random variables is itself a chi-square random variable. So $$ X_1 + \cdots + X_{10} \sim \chi^2_{10k}\,. \tag 1 $$ Since this has expected value $10k$, the sample mean is an unbiased estimator of $k.$

You can use the sum $(1)$ as a test statistic, since if $k>1,$ the probability of that sum being larger than any proposed critical value is larger than it would be if $k=1.$

If $\text{H}_0$ is true, then that sum is distributed as $\chi^2_{10}.$ Find the value of $c$ for which $\Pr(\chi^2_{10}>c)$ is equal to the level of your test, and reject $\text{H}_0$ if the sum $(1)$ exceeds that.