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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, excercise

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  • $\begingroup$ sorry but where is the system of Hypothesis to verify? $\endgroup$
    – tommik
    Commented Feb 11, 2021 at 13:07
  • $\begingroup$ 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 $\endgroup$
    – jon smar
    Commented Feb 11, 2021 at 13:15
  • $\begingroup$ you can well understand that if, by the way, a person wants to respond he would know the problem? $\endgroup$
    – tommik
    Commented Feb 11, 2021 at 13:30
  • $\begingroup$ You're right my apologies, any way I could make things clearer? $\endgroup$
    – jon smar
    Commented Feb 11, 2021 at 13:33

2 Answers 2

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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

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  • $\begingroup$ 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. $\endgroup$
    – jon smar
    Commented Feb 11, 2021 at 14:00
  • $\begingroup$ 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. $\endgroup$
    – jon smar
    Commented Feb 11, 2021 at 14:01
  • $\begingroup$ @jonsmar : I see but it is not correct. Type II error is defined (as you stated) to accept H_0 GIVEN $H_1$ $\endgroup$
    – tommik
    Commented Feb 11, 2021 at 14:01
  • $\begingroup$ 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. $\endgroup$
    – jon smar
    Commented Feb 11, 2021 at 14:02
  • $\begingroup$ 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? $\endgroup$
    – jon smar
    Commented Feb 11, 2021 at 14:03
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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

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  • $\begingroup$ 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. $\endgroup$
    – jon smar
    Commented Feb 11, 2021 at 19:12

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