# hypothesis testing result problem

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,

• sorry but where is the system of Hypothesis to verify? Commented Feb 11, 2021 at 13:07
• The Hypothesis was verified in the first part of the question that I didn't include for μ=30 or μ>30, but from what I understand doesn't affect this part Commented Feb 11, 2021 at 13:15
• you can well understand that if, by the way, a person wants to respond he would know the problem? Commented Feb 11, 2021 at 13:30
• You're right my apologies, any way I could make things clearer? Commented Feb 11, 2021 at 13:33

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

• Please let me clarify the problem, the mean of the sample is 32. For THAT sample size I calculated the probability for a second type error which came out 0.0227. Commented Feb 11, 2021 at 14:00
• On part B, I tried calculating the sample size needed for a 0.1 probability for second type error, which came out to be 137.83. Commented Feb 11, 2021 at 14:01
• @jonsmar : I see but it is not correct. Type II error is defined (as you stated) to accept H_0 GIVEN $H_1$ Commented Feb 11, 2021 at 14:01
• comparing the 2 results shows me I got something wrong, because I get that for a bigger sample size, I have a bigger probability for a second type error, which makes no sense. Commented Feb 11, 2021 at 14:02
• but calculating $β=P( accept H_0 ┤| H_0 is not true)=P(X ̅<30 | μ=32.5)$ isn't that what I did? how would you do it? Commented Feb 11, 2021 at 14:03

In D they are asking you to calculate $$n$$ with a power of $$90\%$$ thus here $$\beta=10\%$$ is fixed.

This results in

$$P(\overline{X}_n>32.06|\mu=32.5)\geq 90$$

That is

$$\frac{32.5-32.06}{10}\sqrt{n}>1.28$$

$$n=847$$

I did the calculations with paper z-table thus the result can be better approximate

• Why did you choose to not use the formula $n≥((Z(1−α)+Z(1−β))^2σ^2)/(μ1−μ0)^2$? and how come it returns such a different result? Thank you. Commented Feb 11, 2021 at 19:12