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If we want to test a mean and are lucky enough to know the population variance, we can use a z-test. Even if our population is not normal, for a sufficiently large sample, we can appeal to the central limit theorem and say that our test statistic has an approximately normal distribution. (Throughout this whole post, I mean under the null hypothesis.)

$$\dfrac{\bar{x}-\mu_0}{\sigma/\sqrt{n}}\sim N(0,1),\text{ at least asymptotically}$$

When we are not so lucky that we know the population variance, we would use the t-test and have our t-stat follow a t-distribution.

$$\dfrac{\bar{x}-\mu_0}{s/\sqrt{n}}\sim t_{df}$$

Playing with the central limit theorem, we get that the numerator has, asymptotically, a normal distribution.

$$\dfrac{\bar{x}-\mu_0}{\sigma/\sqrt{n}}\sim N(0,1) \implies \bar{x} - \mu_0 \sim N\bigg(0, \dfrac{\sigma^2}{n} \bigg)$$

The denominator, however, also has a distribution, unlike when we know $\sigma$.

But then there's Slutsky's theorem saying that $X_n/Y_n \overset{d}{\rightarrow}X/c$ if $X_n \overset{d}{\rightarrow} X$ and $Y_n \overset{d}{\rightarrow} c$.

We have $\bar{x}-\mu_0 \overset{d}{\rightarrow} N\bigg(0, \dfrac{\sigma^2}{n}\bigg)$, so we have convergence in the numerator. In the denominator, the usual $s$ is a consistent estimator of $\sigma$.

I have been naughty in two places, however!

  1. Converging to $N\bigg(0, \dfrac{\sigma^2}{n}\bigg)$ means converging to $N(0,0)$.

  2. While $s^2$ is (I'm pretty sure) a consistent estimator of $\sigma^2$, I do not know that $s$ is consistent for $\sigma$. (I do know that $s$ is biased.)

This leaves me with three questions.

  1. How do I resolve the two places where I've been naughty?

  2. What would be a situation where we would not assume that Slutsky's theorem rescues us from the issue of having a random variable in the denominator?

  3. If we lack normality but then appeal to the central limit theorem to say that our large sample size means that we're "close enough", why do a t-test instead of a z-test?

Thanks!

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The statement $$\bar{x}-\mu \to N \left(0, \frac{\sigma^2}{n}\right)$$ is somewhat nonsensical. The right hand side still depends on $n$, so what is the convergence here?

What you should be writing here is $$\sqrt{n}(\bar{x}-\mu) \to N(0,\sigma^2).$$ The point is that after taking the limit $n \to \infty$, the right hand side should no longer be a function of $n$.

Secondly, we know that from the law of large numbers $$s^2 \overset{p}{\to} \sigma^2$$

By the continuous mapping theorem, this implies that $$1/s \overset{d}{\to} 1/\sigma.$$

Now use Slutsky's to combine the above statements to obtain: $$\frac{(\bar{x}-\mu)}{s/\sqrt{n}} = \frac{\sqrt{n}(\bar{x}-\mu)}{s} \overset{d}{\to} N(0,1).$$


2) What would be a situation where we would not assume that Slutsky's theorem rescues us from the issue of having a random variable in the denominator?

Slutsky's theorem works so long as the assumptions hold, which can be found here.

3) If we lack normality but then appeal to the central limit theorem to say that our large sample size means that we're "close enough", why do a t-test instead of a z-test?

$t$-tests are normally used when the sample size is not large. i.e., when that appeal to the central limit theorem is not valid.

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  • $\begingroup$ Okay, (1) makes sense. For (2), Slutsky is an asymptotic property, so much like how the CLT might only result in convergence too slow to save us from lacking normality in the population, what would be the equivalent for Slutsky? For (3), then t-testing makes no sense to me when we don't have normal data. Either the sample size is so large that we use asymptotic arguments to justify a z-test, or the sample size is too small for the asymptotics to kick in for a t-test, and we have to resort to some other form of testing. $\endgroup$ – Dave Jun 9 at 21:56

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