Hypothesis Testing

H. Kemal Ilter
2020

## Concepts

1. The null hypothesis $H_0$ is a claim about the value of a population parameter. The alternate hypothesis $H_1$ is a claim opposite to $H_0$.
2. A test of hypothesis is a method for using sample data to decide whether to reject $H_0$. $H_0$ will be assumed to be true until the sample evidence suggest otherwise.
3. A test statistic is a function of the sample data on which the decision is to be based.
4. A rejection region is the set of all values of a test statistic for which $H_0$ is rejected.
5. Type I error: you reject $H_0$ when $H_0$ is true. $P(\text{Type I error}) = P(\text{reject }H_0 \mid H_0\text{ true}) = \alpha$. The resulting $\alpha$ is called the significance level of the test and the corresponding test is called a level $\alpha$ test. We will use test procedures that give $\alpha$ less than a specified level (0.05 or 0.01).

## A Problem[1]

I believe that dogs are as smart as people. Assume IQ of a dog follows $X_i \sim N(\mu,10^2)$. IQ of 10 dogs are measured: 30, 25, 70, 110, 40, 80, 50, 60, 100, 60. We want to test if dogs are as smart as people by testing

$H_0 : \mu = 100 \text{ vs. } H_1 : \mu \lt 100$.

One reasonable thing one may try is to see how high the sample mean is.

> x<-c(30, 25, 70, 110, 40, 80, 50, 60, 100, 60)
> mean(x)
[1] 62.5


Since the average IQ of 10 dogs are lower than 100, one would be inclined to reject $H_0$.

Let $\bar{X}$ be a test statistic and $R = (−∞,90]$ to be a rejection region. Let’s compute the probability of making Type I error based on this testing procedure. Under the assumption $H_0$ is true,

$X_i \sim N(100,10^2)$

Under this condition, $\bar{X} \sim N(100, 10)$ and

$\alpha = P(\bar{X} \leq 90)$

> pnorm(90,100,sqrt(10))
[1] 0.0007827011


By using this test procedure, it is highly unlikely to make Type I error. Let’s see what happens when we change the rejection region.

When $R = (−∞,95], \alpha = P(\bar{X} \leq 95)$.

> pnorm(95,100,sqrt(10))
[1] 0.05692315


When $R = (−∞,99], \alpha = P(\bar{X} \leq 99)$.

> pnorm(99,100,sqrt(10))
[1] 0.3759148


The test procedure based on rejecting $H_0 \text{ if } \bar{X} \leq 99$ will produce huge Type I error.

References

1. M. K. Chung's lecture notes, 2003.