4.3 Best critical regions

Let $X_1,X_2,\mathrm{...},X_n$ be a random sample from a normal distribution $N\left(\mu ,36\right)$ . We shall find the best critical region for testing the simple hypothesis $H_0:\mu =50$ against the simple alternative hypothesis $H_1:\mu =55$ . Using the ratio of the likelihood functions, namely $L\left(50\right)/L\left(55\right)$ , we shall find those points in the sample space for which this ratio is less than or equal to some constant k .

That is, we shall solve the following inequality: $$\begin{array}{l}\frac{L\left(50\right)}{L\left(55\right)}=\frac{{\left(72\pi \right)}^{-n/2}\exp \left[-\left(\frac{1}{72}\right){\displaystyle \sum _1^n{\left(x_i-50\right)}^2}\right]}{{\left(72\pi \right)}^{-n/2}\exp \left[-\left(\frac{1}{72}\right){\displaystyle \sum _1^n{\left(x_i-55\right)}^2}\right]}\\ =\exp \left[-\left(\frac{1}{72}\right)\left(10{\displaystyle \sum _1^n{x}_i}+n{50}^2-n{55}^2\right)\right]\le k.\end{array}$$

If we take the natural logarithm of each member of the inequality, we find that $$-10{\displaystyle \sum _1^n{x}_i}-n{50}^2+n{55}^2\le \left(72\right)\ln k.$$

Thus, $$\frac{1}{n}{\displaystyle \sum _1^n{x}_i}\ge -\frac{1}{10n}\left[n{50}^2-n{55}^2+\left(72\right)\ln k\right]$$ Or equivalently, $\overline{x}\ge c,$ where $c=-\frac{1}{10n}\left[n{50}^2-n{55}^2+\left(72\right)\ln k\right].$

Thus $L\left(50\right)/L\left(55\right)\le k$ is equivalent to $\overline{x}\ge c$ .

A best critical region is, according to the Neyman-Pearson lemma, $$C=\left\{\left(x_1,x_2,\mathrm{...},x_n\right):\overline{x}\ge c\right\},$$ where c is selected so that the size of the critical region is $\alpha$ . Say n =16 and c =53. Since $\overline{X}$ is $N\left(50,36/16\right)$ under ${\text{H}}_{\text{0}}$ we have

$$\alpha =P\left(\overline{X}\ge 53;\mu =50\right)=P\left[\frac{\overline{X}-50}{6/4}\ge \frac{3}{6/4};\mu =50\right]=1-\Phi \left(2\right)=\mathrm{0.0228.}$$

Let $X_1,X_2,\mathrm{...},X_n$ denote a random sample of size n from a Poisson distribution with mean $\lambda$ . A best critical region for testing $H_0:\lambda =2$ against $H_1:\lambda =5$ is given by $$\frac{L\left(2\right)}{L\left(5\right)}=\frac{2^{{\displaystyle \sum x_i}}{e}^{-2n}}{x_1!x_2!···x_n!}\frac{x_1!x_2!···x_n!}{5^{{\displaystyle \sum x_i}}{e}^{-5n}}\le k.$$

The inequality is equivalent to ${\left(\frac{2}{5}\right)}^{{\displaystyle \sum x_i}}{e}^{3n}\le k$ and $\left({\displaystyle \sum x_i}\right)\ln \left(\frac{2}{5}\right)+3n\le \ln k.$

Since $\ln \left(2/5\right)<0$ , this is the same as $${\displaystyle \sum _{i=1}^n{x}_i\ge \frac{\ln k-3n}{\ln \left(2/5\right)}=c.}$$

If n =4 and c =13, then $$\alpha =P\left({\displaystyle \sum _{i=1}^4{X}_i\ge 13;\lambda =2}\right)=1-0.936=0.064,$$ from the tables, since ${\displaystyle \sum _{i=1}^4{X}_i}$ has a Poisson distribution with mean 8 when $\lambda$ =2.

Let $X_1,X_2,\mathrm{...},X_n$ be a random sample from $N\left(\mu ,36\right)$ . We have seen that when testing $H_0:\mu =50$ against $H_1:\mu =55$ , a best critical region C is defined by $$C=\left\{\left(x_1,x_2,\mathrm{...},x_n\right):\overline{x}\ge c\right\},$$ where c is selected so that the significance level is $\alpha$ . Now consider testing $H_0:\mu =50$ against the one-sided composite alternative hypothesis $H_1:\mu >50$ . For each simple hypothesis in $H_1$ , say $\mu ={\mu }_1$ the quotient of the likelihood functions is

$$\begin{array}{l}\frac{L\left(50\right)}{L\left({\mu }_1\right)}=\frac{{\left(72\pi \right)}^{-n/2}\exp \left[-\left(\frac{1}{72}\right){\displaystyle \sum _1^n{\left(x_i-50\right)}^2}\right]}{{\left(72\pi \right)}^{-n/2}\exp \left[-\left(\frac{1}{72}\right){\displaystyle \sum _1^n{\left(x_i-{\mu }_1\right)}^2}\right]}\\ =\exp \left[-\left(\frac{1}{72}\right)\left\{2\left({\mu }_1-50\right){\displaystyle \sum _1^n{x}_i}+n\left({50}^2-{\mu }_1^2\right)\right\}\right].\end{array}$$

Now $L\left(50\right)/L\left({\mu }_1\right)\le k$ if and only if $$\overline{x}\ge \frac{\left(-72\right)\ln \left(k\right)}{2n\left({\mu }_1-50\right)}+\frac{50+{\mu }_1}{2}=c.$$

Thus the best critical region of size $\alpha$ for testing $H_0:\mu =50$ against $H_1:\mu ={\mu }_1$ , where ${\mu }_1>50$ , is given by

$$C=\left\{\left(x_1,x_2,\mathrm{...},x_n\right):\overline{x}\ge c\right\},$$ where is selected such that $$P\left(\overline{X}\ge c;H_0:\mu =50\right)=\alpha .$$