Finding Type I error

Question

Suppose the sample size $n=16$ is drawn from a normal distribution with mean $\mu$ and standard deviation $\sigma = 4$. Consider the testing hypothesis $H_o:\mu = 0$ vs $H_a:\mu \ne 0$. Let the rejection region be $|\bar{x}| > 2$, where $|\bar{x}|$ is the sample mean. Find the type I error.

$$P(\text{Type I error}) = P(|\bar{x}| > 2 \mid \mu = 0)$$

Since we know that the sample follows a normal distribution, then we know that $\bar{x}$ also follows a normal distribution. When $\mu = 0$, then the variance is $\frac{\sigma^2}{n} = 1$.

$$P(|\bar{x}| > 2 \mid \mu = 0)\\ =P(\bar{x} < -2 \mid \mu = 0) + P(\bar{x} > 2 \mid \mu = 0)\\ =2P(\bar{x} > 2 \mid \mu = 0)$$

At this point, I'm not sure how to get the probability of $P(\bar{x} > 2 \mid \mu = 0)$

Answer 1

Under $H_0$, $\bar{x}$ follows $N(0,\frac{\sigma^2}{n})=N(0,1)$. So $$P(\bar{x}>2)=1-\Phi(2)$$ where $\Phi$ is the cdf of a standard normal. There is no explicit way to compute $\Phi(2)$, so you have to look it up on a table.

Answer 2

Arithmetic mean of sample size follows Normal Distribution with $\mu{}, \sigma{}^2/n$.

Calculate the probability of $\bar{x} > 2$ given that $X \sim Normal(\mu = \mu, \sigma^2 = \sigma^2/n)$ (by standardizing it to standard normal).