The four parts of a hypothesis test concerning a population mean from a normal population are shown below.
$H_0: μ=50$
$H_a: μ>50$
TS: $Z=\dfrac{(\bar{X}−μ_0)}{(σ/√n)}$
RR: $Z ≥ z_α$
Assume the sample size is $n=25$; $σ=7.5$ and $α=0.01$
a) Find the probability of a type II error for the alternative mean $μ_a=54$ ; that is, find $β(54)$
b) Find $β(55)$ and $β(56)$
c) Repeat parts (a) and (b) for $α=0.025$
The first step is to quantify the rejection region, aka find $z_\alpha$. The easiest way to do this is to consult a table, or use some software. In R, the appropriate command is
qnorm(1-0.001)which gives us $z_\alpha \approx 2.33$. So, the rejection region solves:$\frac{(X - 50)}{\sigma/\sqrt{n}} > z_\alpha \rightarrow X > z_\alpha \sigma / \sqrt{n} + 50$. Plugging in the given parameters, we get:
$X > 53.495$.
We want to know: If the true mean is 54, what is the probability that the test statistic will be less than $53.495$ and we will fail to reject? Well, we know that in such a case the sample mean $X$ is distributed symmetrically around $54$ with a standard deviation of $\sigma/\sqrt{n} = 1.5$. So, we want to find:
$\Phi(\frac{53.495 - 54}{1.5}) = \Phi(-0.3366667)$ where $\Phi$ is the CDF of the normal distribution. You can find this using the same table as before (but you start by looking for the appropriate z-score, rather than the appropriate probability), although I used the R command
pnorm(-0.3366667)= 0.3681841. This is $\beta(54)$.Note that when you look for $\beta(55)$ and $\beta(56)$, the rejection region does not change, because the rejection region is a function of the null hypothesis rather than what the mean actually is. For part c of the problem, you will have to recalculate the rejection region.