For a normal distribution of $N(30,10)$ and sagnificence of 5% (α=0.05),
the sample mean is 32.5 and the population size is 64.
H0: $μ=30$
H1: $μ>30$
A. What is the probability for a second type error?(β)
What I did was, $=P( accept H_0 | H_0 false)=P(X ̅<30 | μ=32.5)=Φ((30-32.5)/(10/√64))=1-Φ(2)=0.0227$
B. What is the sample size needed for a $90$% discovery chance of the second type error?
what I did was,
$n≥((Z_(1-α)+Z_(1-β) )^2 σ^2)/(μ_1-μ_0 )^2 =((1.65+1.285)^2 ×10^2)/(32.5-30)^2 =137.83$
My problem is,
In part B, the probability for a second type error is $0.1$, and the sample size I get is 138, BUT in part A the probability for a second type error I get is 0.027 and the sample size is 64.
How is that possible that I get a bigger sample size for a higher probability for an error?
The full excercise,
After reading the original text, it is very different from you initial summary.
First, looking at your calculations, I assume that with $N(30;10)$ you mean a Gaussian with st dev =10 but usually this notation is used as a variance =10. Using a variance =10 the calculations makes no sense thus I correct the solution I wrote before considering $N(30;10^2)$
A:
Verifying the system of hypothesis (one tail) at 5% you get the following critical region
$$\frac{\overline{X}_{64}-30}{10}8 \geq 1.64$$
That is $\overline{X}_{64}\geq 32.06$
Thus we reject the null hypothesis as the sample mean is $\overline{X}_{64}=33>32.06$
The manager's claim is correct.
B: the minimal $\alpha$ is
$$P\left(\overline{X}_{64}>33|H_0\right)=\dots=P(Z>2.4)=0.82\%$$
(this is called p-value of the test)
C: "if really the mean moved to 32.5" means that 32.5 is the new alternative Hypothesis, thus
$$\beta=P(\overline{X}_{64}<32.06|\mu=32.5)=P\left(Z<\frac{32.06-32.5}{10}8\right)=\Phi(-0.355)=0.36$$
Now I think you can proceed by yourself