A normal population has a population standard deviation of $12$. The hypotheses about the population mean, $H_0: \mu = 40$ versus $H_1: \mu > 40$ are to be tested. A random sample of $100$ observations will be selected and the sample mean is denoted by $\bar x$.
If the true population mean is $43$ and the significance level is $5\%$, the probability of a Type II error is closest to:
A. $0.305$
B. $0.273$
C. $0.229$
D. $0.195$
E. $0.187$
I know that the $z$ score is $(43-40)/1.2 = 2.50$, but am lost on what to do afterwards
The Type II error $\beta$ is the probability of failing to reject the null hypothesis $H_0$ when the alternative $H_1 : \mu = 43$ is true. In other words, given that the true sampling distribution of the sample mean is normal with mean $\mu = 43$ and standard error $\sigma/\sqrt{n} = 12/\sqrt{100} = 1.2$, then $H_0$ is not rejected if it is "too small." But what does this mean?
Specifically, suppose we observe a value of the test statistic $$Z \mid H_0 = \frac{\bar x - \mu_0}{\sigma/\sqrt{n}} = \frac{\bar x - 40}{1.2}$$ such that it is less than the critical value for rejection at the $\alpha = 0.05$ level; i.e., $$\frac{\bar x - 40}{1.2} \le z^*_\alpha \approx 1.64485.$$ This means the sample mean would need to be at most $$\bar x \le 41.9738,$$ otherwise you would reject. So we need to compute $$\beta = \Pr[\bar x \le 41.9738 \mid H_1] = \Pr\left[\frac{\bar x - 43}{1.2} \le \frac{41.9738 - 43}{1.2}\right] = \Pr[Z \le -0.855146] \approx 0.196235.$$ Therefore, the Type II error probability is just shy of $20\%$.
The thinking is like this: The test statistic assumes the null is true. Under this assumption, what is the complement of the rejection region; i.e., for what values of the sample mean would we not reject the null? Then, assuming the alternative hypothesis is true, what is the chance of observing a sample mean that lies in the complement of the rejection region? This gives you the probability of committing a Type II error--failure to reject when we should reject.