## 6.4 Asymptotic tests

As evidenced in Section 5.4, another primary source of inferential tools beyond normal populations are asymptotic results. Assume that we want to test the hypotheses

1. $$H_0:\theta=\theta_0$$ vs. $$H_1:\theta>\theta_0;$$
2. $$H_0:\theta=\theta_0$$ vs. $$H_1:\theta<\theta_0;$$
3. $$H_0:\theta=\theta_0$$ vs. $$H_1:\theta\neq \theta_0.$$

If we know a test statistic that, under $$H_0,$$ has an asymptotic normal distribution, that is

\begin{align} Z=\frac{\hat{\theta}-\theta_0}{\hat{\sigma}(\theta_0)}\stackrel{d}{\longrightarrow}\mathcal{N}(0,1), \tag{6.4} \end{align}

then the asymptotic critical regions are given by

\begin{align*} C_a=\{Z>z_{\alpha}\}, \quad C_b=\{Z<-z_{\alpha}\}, \quad C_c=\{|Z|>z_{\alpha/2}\}. \end{align*}

Note that despite not knowing the exact distribution of $$Z$$ under $$H_0,$$ $$Z$$ is a test statistic because its distribution is known asymptotically.

An especially relevant instance of (6.4) is given by likelihood theory (Theorem 4.1 under $$H_0$$):

\begin{align*} Z=\frac{\hat{\theta}_{\mathrm{MLE}}-\theta_0}{1\big/\sqrt{n\mathcal{I}(\theta_0)}}=\sqrt{n\mathcal{I}(\theta_0)}\left(\hat{\theta}_{\mathrm{MLE}}-\theta_0\right)\stackrel{d}{\longrightarrow}\mathcal{N}(0,1). \end{align*}

Due to Corollary 4.3, another asymptotic pivot is

\begin{align*} Z=\sqrt{n\hat{\mathcal{I}}(\theta_0)}\left(\hat{\theta}_{\mathrm{MLE}}-\theta_0\right)\stackrel{d}{\longrightarrow}\mathcal{N}(0,1),\quad \hat{\mathcal{I}}(\theta_0)=\frac{1}{n}\sum_{i=1}^n \left(\left.\frac{\partial\log f(X_i;\theta)}{\partial\theta}\right\vert_{\theta=\theta_0}\right)^2, \end{align*}

which is always straightforward to compute from the srs $$(X_1,\ldots,X_n)$$ from $$X\sim f(\cdot;\theta)$$ (analogous if $$X$$ is discrete).

Other test statistics can be obtained from non-normal asymptotic distributions (see, e.g., the forthcoming Theorem 6.2).

Let us see some examples of asymptotic tests.

Example 6.12 Let $$(X_1,\ldots,X_n)$$ be a srs of a rv $$X$$ with mean $$\mu$$ and variance $$\sigma^2,$$ both unknown. We want to test:

1. $$H_0:\mu=\mu_0$$ vs. $$\mu>\mu_0;$$
2. $$H_0:\mu=\mu_0$$ vs. $$\mu<\mu_0;$$
3. $$H_0:\mu=\mu_0$$ vs. $$\mu\neq\mu_0.$$

For that, employing the CLT (Theorem 2.5) we know that under $$H_0:\mu=\mu_0,$$

\begin{align*} Z=\frac{\bar{X}-\mu_0}{S'/\sqrt{n}}\stackrel{d}{\longrightarrow} \mathcal{N}(0,1). \end{align*}

Therefore, $$Z$$ is a test statistic and $$H_0$$ is rejected if the observed value of $$Z$$ belongs to the corresponding critical region ($$C_a,$$ $$C_b,$$ or $$C_c$$).

Example 6.13 Let $$(X_1,\ldots,X_n)$$ be a srs of a rv $$\Gamma(k,1/\theta)$$ with $$k$$ known and $$\theta$$ unknown. We want to test:

1. $$H_0:\theta=\theta_0$$ vs. $$H_1:\theta>\theta_0;$$
2. $$H_0:\theta=\theta_0$$ vs. $$H_1:\theta<\theta_0;$$
3. $$H_0:\theta=\theta_0$$ vs. $$H_1:\theta\neq\theta_0.$$

We use that, by Example 4.12, $$\mathcal{I}(\theta)=k/\theta^2.$$ Then, we know that under $$H_0:\theta=\theta_0,$$

\begin{align*} Z=\frac{\hat{\theta}_{\mathrm{MLE}}-\theta_0}{\theta_0/\sqrt{nk}}\stackrel{d}{\longrightarrow} \mathcal{N}(0,1). \end{align*}

Therefore, $$Z$$ is a test statistic and $$H_0$$ is rejected if the observed value of $$Z$$ belongs to the corresponding critical region ($$C_a,$$ $$C_b,$$ or $$C_c$$).

Example 6.14 A certain machine has to be repaired if more than $$10\%$$ of the items that it produces per day are defective. A srs of $$n=100$$ items of the daily production contains $$15$$ that are defective and the foreman decides that the machine has to be repaired. Is the sample supporting his decision at a significance level $$\alpha=0.01$$?

Let $$Y$$ be the number of defective items that were found. Then $$Y\sim\mathrm{Bin}(n,p).$$ We want to test

\begin{align*} H_0: p=0.10\quad \text{vs.}\quad H_1:p>0.10. \end{align*}

Because of the CLT (Theorem 2.5), $$Y$$ has a normal asymptotic distribution, so under $$H_0:p=p_0$$ it follows that

\begin{align*} Z=\frac{\hat{p}-p_0}{\sqrt{p_0(1- p_0)/n}}\stackrel{d}{\longrightarrow}\mathcal{N}(0,1). \end{align*}

Therefore, $$Z$$ is a test statistic with observed value

\begin{align*} Z=\frac{0.15-0.10}{\sqrt{0.1\times 0.9/100}}=5/3. \end{align*}

The rejection region is

\begin{align*} C=\{Z>z_{0.01}\approx2.33\}. \end{align*}

Since $$Z=5/3\approx1.67<2.33,$$ the sample does not provide enough evidence supporting the foreman decision, that is, that the actual percentage of defective items the machine is producing is above $$10\%.$$ The machine should not be repaired and the larger proportion of defective items in the batch can be attributed to chance.