# Use $\bar{X}^2$ for hypothesis test that $\mu=0$ because faster convergence rate?

Suppose that I have $X_1,\ldots,X_n$ are i.i.d. and I want to do a hypothesis test that $\mu$ is 0. Suppose I have large n and can use Central Limit Theorem. I could also do a test that $\mu^2$ is 0, which should be equivalent to testing that $\mu$ is 0. Further, $n(\bar{X}^2 - 0)$ converges to a chi-squared, where $\sqrt{n}(\bar{X} - 0)$ converges to a normal. Because $\bar{X}^2$ has a faster convergence rate, shouldn't I use that for the test statistic and thus I will get a faster convergence rate and test will be more efficient?

I know this logic is wrong but I have been thinking and searching a long time and cannot figure out why.

• It isn't clear what you are asking. Could you explain in what sense the convergence rate of $\bar X^2$ is "faster" than that of $\bar X$? How are you measuring the rate? What test statistics are you using in the two tests? Clearly these choices can make a difference. – whuber Nov 4 '16 at 16:56
• @whuber thanks for questions. I claim "faster rate" because n is larger than square root of n. Is that intuition incorrect? I have in mind test statistic X-bar or X-bar squared. – Xu Wang Nov 4 '16 at 20:37
• I think you're focusing on the wrong thing. This rate tells you how quickly the sampling distribution approaches the limiting one--either standard Normal or $\chi^2(1)$. Since $n$ is large, its value makes no practical difference--it's irrelevant. The issue concerns the power of each test, not how well approximated the test statistic is to the limiting distribution. – whuber Nov 4 '16 at 23:10
• @whuber thank you for these details. I have been thinking about them but still don't understand. Won't the approximate variance of X-bar^2 eventually be smaller than the approximate variance of X-bar? And isn't that a result of X-bar^2 having a higher rate of convergence than X-bar? I'm sorry for not seeing my fundamental misunderstandings. I know there is something big I am missing and hope to correct such thinking. – Xu Wang Nov 6 '16 at 3:57
• It doesn't matter whether the approximate variance is larger or smaller, because what counts is the distribution of the statistic. To see this, consider a t-test for $\mu = 0$ with $x \sim N(0,1)$ vs $y \sim N(0,10)$. The statistic $\bar{y}$ always has variance 100x that of $\bar{x}$, but the normalization results in both actual test statistics distributed $t(n-1)$. In your case, remember that squaring a $N(0,1)$ variate gives a $\chi^2$ variate. At the limit, this transformation means that the two tests are identical in terms of their power given a specified level. – jbowman Nov 7 '16 at 22:59

Both of the tests you describe are equivalent.

If I have two hypotheses: $$H_0: \mu=0$$ $$H_1: \mu\neq0$$

then they are equivalent to

$$H_0: \mu^2=0$$ $$H_1: \mu^2\gt 0.$$

If the data are known to be normal, then the sample mean $\bar{X}$ will also be Normal with mean $\mu$ and variance $\sigma^2/n$ (which might be known or unknown).

If the data aren't known to be Normal then you can use the central limit theorem and the above will be true asymptotically. You claim that $\bar{X}^2$ will converge to a chi-squared variable "faster" than $\bar{X}$ will converge to a normal one. This is true in the sense that as $n$ tends to infinity,

$$P(|\bar{X} - \mu| > |\bar{X}^2 - \mu^2|) \rightarrow 1$$

but that is not the whole story. We are performing a likelihood ratio test, or at least an approximate one. the ratio will come out the same whether we perform a chi-squared or a normal test. (Recall that the square of a normal random variable follows a chi-squared distribution.) If the sample mean $\bar{X}$ comes out at the 95th percentile of the relevant normal or t-distribution, then the sum-of-squares will be equal to the 95th percentile of the $\chi^2$ distribution (which is not the same number, but that doesn't matter).