Statistics [12]: Parameter Hypothesis Test

4 minute read

Published: January 12, 2021

Basics of parameter hypothesis test.

Steps of Parameter Hypothesis Test

Establish hypothesis
- Null hypothesis: $H_0: \theta\in \Theta_0$
- Alternate hypothesis: $H_1: \theta\in \Theta_1$
Choose the test statistics and give the form of the rejection region $W$ .
Decide the significance level and rejection region (acceptance region).
- $\alpha = \max\{P(\text{reject} \ \ H_0|H_0 \ \ \text{ is true})\} = \max\{P_{\theta}(x\in W),\theta \in \Theta_0\}$

Example 1

Assume a factory is producing a product obeying the norrmal distribution $N(\mu,2^2)$ , $\mu$ is the average quality standard, $\mu>10$ is considered as high quality. To test the quality of the products, the buyer samples 16 products at random from a batch of products, with the following quality standard:

$9.6, 9.2, 9.3, 9.8, 10.1, 8.5, 9.8, 8.4, 9.2, 9.1, 8.7, 9.3, 8.6, 9.8, 9.2, 10.2$

Test whether the products are in high quality.

Solution. Firstly, establish hypothesis $H_0: \mu \geq 10,\ \ H_1: \mu < 10$

Secondly, choose $\bar{X}$ as test statistic, $\bar{X} = N\left(\mu, 0.5^2\right)$ and the form of the rejection region would be

$\{(x_1,x_2,...,x_n): \bar{x} < c\}$

Thirdly, decide the significance level and rejection region.

$P(\bar{x} < c | \mu = 10) = P\left(\dfrac{\bar{x} - 10}{0.5} < \dfrac{c-10}{0.5}\right) = \alpha$

$\dfrac{c-10}{0.5} = u_{\alpha} = -u_{1-\alpha} \Rightarrow c = 10 - \dfrac{u_{1-\alpha}}{2}$

Hence, the rejection region is

$\{(x_1,x_2,...,x_n): \bar{x} < 10 - \dfrac{u_{1-\alpha}}{2}\}$

When $\alpha = 0.05$ , the rejection region would be $\{(x_1,x_2,...,x_n): \bar{x} < 9.1775\}$ .

Two Types of Errors

There are two types of mistakes in hypothesis testing:

False rejection (type I error): $\alpha(\theta) = \{P(\text{reject} \ \ H_0|H_0 \ \ \text{ is true})\} = \{P_{\theta}(x\in W),\theta \in \Theta_0\}$
False acceptance (type II error): $\beta(\theta) = \{P(\text{accept} \ \ H_0|H_1 \ \ \text{ is true})\} = \{P_{\theta}(x\in \bar{W}),\theta \in \Theta_1\}$

Example 2

A factory is producing screws with tandard length of 35 mm, which obeys normal distribution $N(\mu,3^2)$ . Assume the sample size is 36, $H_0: \mu = 35, H_1: \mu\neq 35$ , rejection region is $W=\{\bar{x}:|\bar{x}-35|>1\}$ , find the type I error, and find type II error when $\mu = 36$ .

Solution. Test statistic is $\bar{X} \sim N\left(\mu,0.5^2\right)$ .

False rejection rate would be

$\alpha = P(|\bar{X} - 35| > 1|\mu = 35) = 1 - P(|\bar{X} - 35| \leq 1|\mu = 35)$

$= 1 - P\left(-2\leq \dfrac{\bar{X}-35}{0.5} \leq 2|\mu = 35\right) = 1 - (\Phi(2)-\Phi(-2)) = 0.0455$

False acceptance rate would be

$\beta = P(|\bar{X} - 35| > 1|\mu = 36) = P(-1\leq \bar{X}\leq 1 |\mu = 36)$

$= P\left(-4\leq \dfrac{\bar{X}-36}{0.5} \leq 0|\mu = 36\right) = \Phi(0)-\Phi(-4) = 0.5$

Three Types of Hypothesis Test Problems

One-tailed test: $H_0:\mu\leq \mu_0, \ \ H_1:\mu > \mu_0$
One-tailed test: $H_0:\mu\geq \mu_0, \ \ H_1:\mu < \mu_0$
Two-tailed test: $H_0:\mu= \mu_0, \ \ H_1:\mu \neq \mu_0$

Example 3

For normal distribution $N(\mu,\sigma^2)$ , $\mu$ is unknown and $\sigma^2$ is known, perform hypothesis to $\mu$ .

Solution. Test statistic $u = \dfrac{\bar{x}-\mu_0}{\sigma/\sqrt{n}} = \dfrac{\sqrt{n}(\bar{x}-\mu_0)}{\sigma}\sim N(0,1)$

$H_0:\mu= \mu_0, \ \ H_1:\mu \neq \mu_0 \Rightarrow \bar{x} > \mu_0 + \dfrac{\sigma}{\sqrt{n}}u_{1-{\alpha}\text{/}{2}}\ \ \text{or}\ \ \bar{x} < \mu_0 - \dfrac{\sigma}{\sqrt{n}}u_{1-{\alpha}\text{/}{2}}$

$H_0:\mu\geq \mu_0, \ \ H_1:\mu < \mu_0 \Rightarrow \bar{x} < \mu_0 - \dfrac{\sigma}{\sqrt{n}}u_{1-{\alpha}}$

$H_0:\mu\leq \mu_0, \ \ H_1:\mu > \mu_0 \Rightarrow \bar{x} > \mu_0 + \dfrac{\sigma}{\sqrt{n}}u_{1-{\alpha}}$

Example 4

For normal distribution $N(\mu,\sigma^2)$ , $\mu$ is unknown and $\sigma^2$ is unknown, perform hypothesis to $\mu$ .

Solution. Test statistic $t = \dfrac{\bar{x}-\mu_0}{s/\sqrt{n}} = \dfrac{\sqrt{n}(\bar{x}-\mu_0)}{s}\sim t(n-1)$

$H_0:\mu= \mu_0, \ \ H_1:\mu \neq \mu_0 \Rightarrow \bar{x} > \mu_0 + \dfrac{s}{\sqrt{n}}t_{1-{\alpha}\text{/}{2}}\ \ \text{or}\ \ \bar{x} < \mu_0 - \dfrac{s}{\sqrt{n}}t_{1-{\alpha}\text{/}{2}}$

$H_0:\mu\geq \mu_0, \ \ H_1:\mu < \mu_0 \Rightarrow \bar{x} < \mu_0 - \dfrac{s}{\sqrt{n}}t_{1-{\alpha}}$

$H_0:\mu\leq \mu_0, \ \ H_1:\mu > \mu_0 \Rightarrow \bar{x} > \mu_0 + \dfrac{s}{\sqrt{n}}t_{1-{\alpha}}$

p Value

$p$ value of a test refers to the smallest significance level obtained through the samples at which the null hypothesis would be rejected.

Example 5

Assume the population obeys the normal distribution $N(\mu,3^2)$ and the sample size is 36, $H_0: \mu = 35, H_1: \mu\neq 35$ . Noe suppose the sample value is 36.5 mm, calculate its $p$ value.

Solution.

$p = P(|\bar{X} - 35|\geq 1.5|\mu=35) = 1 - P(|\bar{X}-35| < 1.5 | \mu=35)$

$= 1 - P\left(-3 < \dfrac{\bar{X}-35}{0.5} < 3|\mu=35\right) = 1 - (\Phi(3)-\Phi(-3)) = 0.0027$

Suppliment: Neyman-Pearson Lemma

Let $C$ be a critical regionfor a hypothesis test with significance level $\alpha$ , where the hypothesis are $H_0: \theta = \theta_0$ and $H_1: \theta=\theta_1$ . Then we say that $C$ is a best critical region of size $\alpha$ if whenever $D$ is another critical region with size $\alpha$ , there is $P(C|\theta=\theta_1)\geq P(D|\theta=\theta_1)$ .

What this means is that if the alternative hypothesis is true, then the probability that we reject the null hypothesis is greatest if we use the critical region $C$ .

Neyman-Pearson Lemma

Let $CX_1,X_2,...,X_n$ be a sample from a distribution with pdf $f(x; \theta)$ , and let $L(\theta) {\displaystyle \prod_{i=1}^n f(x_i;\theta)}$ . If there exists a positive constant $k$ and a region $C$ such that (1) $P((X_1,X_2,...,X_n)\in C|\theta=\theta_0) = \alpha$ , (2) $\dfrac{L(\theta_0)}{L(\theta_1)}\leq k$ for $(X_1,X_2,...,X_n)\in C$ , and (3) $\dfrac{L(\theta_0)}{L(\theta_1)}\geq k$ for $(X_1,X_2,...,X_n)\in C^\text{'}$ , then $C$ is the best critical region of size $\alpha$ for testing $H_0: \theta=\theta_0$ against $H_1: \theta=\theta_1$ .

Example 6

Let $CX_1,X_2,...,X_n$ be a random sample from a normal distribution having known variance $\sigma^2$ . Consider two simple hypothesis

$H_0: \mu=\mu_0,\ \ H_1: \mu=\mu_1$

where $\mu_0$ and $\mu_1$ are given constants. Let the significance level be prescribed. The Neyman-Pearson Lemma states that among all tests with significance level $\alpha$ , the tests that rejects for small values of the lieklihood ratio is most powerful. We calculate the likelihood ratio

$\dfrac{f_0(X)}{f_1(X)} ={\displaystyle \dfrac{\exp\left[-\dfrac{1}{2\sigma^2}\sum_{i=1}^n(X_i-\mu_0)^2\right]}{\exp\left[-\dfrac{1}{2\sigma^2}\sum_{i=1}^n(X_i-\mu_1)^2\right]} }$

Ignore the constant term $\dfrac{1}{2\sigma^2}$

$\dfrac{f_0(X)}{f_1(X)} = {\displaystyle \exp\left[ \sum_{i=1}^n(2X_i(\mu_0-\mu_1) + \mu_1^2 - \mu_0^2)\right] = \exp\left[2n\bar{X}(\mu_0-\mu_1) + n\mu_1^2 - n\mu_0^2)\right]}$

If $\mu_0-\mu_1 > 0$ , the likelihood ratio is small if $\bar{X}$ is small; if $\mu_0-\mu_1 < 0$ , the likelihood ratio is small if $\bar{X}$ is large. Now assume the latter case, the Neyman-Pearson lemma thus tells us that the most powerful test rejects for $\bar{X} > x_0$ for some $x_0$ , we choose $x_0$ so as to give the test the desired significance level $\alpha$ . Since