Assume you are using a significance level of $α=0.05$ to test the claim that $μ<20$ and that your sample is a random sample of 33 values. Find $β$ given that the population actually has a normal distribution with $μ$1 $=$ 15 and $σ=7$.
- These are me workings out up to the point that I got stuck
$H$0 : $μ$ = 20
$H$1 : $μ$ < 20
Firstly, I'm not sure if this is right. To work out $β$, we find P(Do not reject $H$0|$μ$ = 15)
So $\bar X$ = $20 + \frac{7}{33^{0.5}}(-1.96) = 17.61$
And I think I'm supposed to find the probability that $X > 17.61$ given $μ = 15$ and $σ = 7$ but I feel like it's wrong. If so, can anyone please point out where I went wrong?
Thanks in advance.
Edit: Changed the null and alternative hypotheses
First, we determine the rejection criterion for the test. Clearly, we are intending to use the one-sample $Z$-test: $$Z \mid H_0 = \frac{\bar X - \mu_0}{\sigma/\sqrt{n}} = \frac{\bar X - 20}{7/\sqrt{33}}. \tag{2}$$ Then we will reject $H_0$ at level $\alpha$ if $$\Pr[Z < z_\alpha] = \alpha; \tag{3}$$ that is to say, if $Z < -1.64485$ for $\alpha = 0.05$. (Note that your choice of $z_\alpha$ is incorrect because the test is one-sided, not two-sided.) This in turn implies $$\bar X < \frac{(-1.64485)(7)}{\sqrt{33}} + 20 \approx 17.9957, \tag{4}$$ where $\bar X$ is the sample mean.
So if the sample is drawn from a normal distribution with mean $\mu_1 = 15$ and standard deviation $\sigma = 7$, the Type II error is given by $$\beta = \Pr[\bar X > 17.9957 \mid \mu_1 = 15], \tag{5}$$ i.e., it is the probability of failing to meet the rejection criterion under the assumption that the alternative hypothesis is correct. In this case, the alternative hypothesis is the point hypothesis $H_1 : \mu = \mu_1 = 15$.
Hence $$\beta = \Pr\left[\frac{\bar X - \mu_1}{\sigma/\sqrt{n}} > \frac{17.9957 - 15}{7/\sqrt{33}} \mid \mu_1 = 15\right] = \Pr[Z > 2.45841] \approx 0.00697778. \tag{6}$$