The supplier of a particular brand of vitamin pills claims that the average potency of these pills after a certain exposure to heat and humidity is at least $65$. The distributor who wishes to buy these pills is interested in testing the supplier's claim. The distributor considers it a more serious mistake to buy a batch of pills with a mean retention of vitamin potency that is less than $65$ than to reject a batch with a higher mean. It is given that $\sigma$ = $6$ with a sample size of n = $9$.
(a) Formulate the hypotheses about the mean potency of a batch of pills $\mu$ and determine the rejection region of the test with $\alpha$ = $.05.$
(b) If $\mu$ = $67$ what is the probability of a type II error?
(c) Compute the power of the test at a few values of $\mu$ that range between $63$ and $70$ and plot the power curve.
Attemped Solution:
(a) I'm most concerned that I'm mixing up $H_0$ and $H_1$. I'm having trouble distinguishing them from the question, but this is my guess.
$H_0$: $\mu < 65$
$H_1$: $\mu \geq 65$
$t_{8,.05}$ = $1.8595$
$\Rightarrow$ $65 + 1.8595(6)$ = $76.157$ so reject when $\mu_0$ > $76.157$
(b) P(Type II error) = P($\mu$ < $65$ | $\mu$ = $67$)
$\frac{65-67}{6\over\sqrt9}$ = $-1$ $\Rightarrow$ P(Type II error) = $.173$
(c)
Without answering everything, I'll try to get you started on the right track.
Your null hypothesis $H_0: \mu < 65,$ cannot be right. First, it is formally incorrect for any null hypothesis because a null hypothesis must always contain an equal sign, whether in the form $=, \le,$ or $\ge.$
Second, Because $\sigma = 6$ is known, you should be doing a z-test instead of a t-test. (That is important, because finding Type II error probabilities for a t-test is probably beyond what you're studying in this course.)
You are wondering whether the direction of your null hypothesis is correct. To help with that, you might read this page where $\alpha$ and 'Type I and II errors' are explained in terms of 'Producer's Risk` and 'Consumer's Risk'.
Suppose we frame $H_0$ and $H_1$ so that rejecting the null hypothesis corresponds to rejecting the lot. Then we want $H_0: \mu \ge 65$ and $H_1: \mu < 65.$
Then the test statistic is
$$Z = \frac{\bar X - \mu_0}{\sigma/\sqrt{n}} = \frac{\bar X - 65}{2},$$ where $\bar X$ is the mean potency of $n = 9$ randomly chosen pills. Under $H_0$ (meaning 'assuming the null hypothesis to be true') you have $Z \sim \mathsf{Norm}(0,1)$ and you reject $H_0$ at level $\alpha = 0.05$ if $Z < - 1.645,$ chosen because $P(Z < - 1.645) = 0.05.$ In terms of $\bar X,$ you reject $H_0$ if $\bar X < 61.71.$
Notice that if $\bar X = 64.5,$ just a little below 65, we don't reject $H_0$ because a slight deficiency in the sample of 9 might be due to chance. But if $\bar X$ is significantly below 65, we do reject $H_0.$
If the true potency is $65,$ you would reject $H_0,$ and the lot of pills, with probability 5%. (That would be unfavorable to the 'producers' because they are losing a sale.)
Now the distributor may worry if a lot of pills in which $\bar X$ is a little under 65 may still be of undesirably low potency. That is where Type II error comes in. Type II error is $P(\text{Accept}|\text{Low potency}).$ But there are many ways for potency to be low. So $P(\text{Type II Err}|\mu) = \beta(\mu)$ is a not a single value, but a function of $\mu.$ This corresponds to the distributor's or 'consumer's' risk, because (for $\mu <65$) it involves accepting a batch of pills that are below the required potency.
Also $$\text{Power}(\mu) = \gamma(\mu) = 1 - \beta(\mu) = P(\text{Reject}|\mu) = P(\bar X < 61.71|\mu).$$
I will show you what I get for a power curve for some values of $\mu$ between 55 and 70. Based on what you show or your work, I think you should be able to do the computations to make a similar curve on your own. The value at $\mu=65$ shows the significance level 0.05. (Graph made using R statistical software.)
It seems to me that $n=9$ may be too few sample pills to give a good power for batches that are just a little below 65 in potency. The dotted red curve on a similar plot made using Minitab, shows better performance for $n = 16.$ (The horizontal axis shows differences from $\mu=65;$ the solid blue curve is essentially the same as mine.)