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Question: In one faculty the distribution of the heights of the students is N(180,25) In the second faculty it's N(160,20)

We take some students from both faculties and put them together in a class. There are 2 hypothesis regarding how they are distributed:

$H_0$: $\frac 14$ of them are from faculty I, and $\frac 34$ are from faculty II.

$H_1$: $\frac 14$ of them are in faculty II, and $\frac 34$ are from faculty I.

If we look at a random student- If his height is bigger than 168 we reject $H_0$.

What is the rejection zone?

What I did: To try to formalize it I rephrased this in math terms:

X is a Bernulli RV representing a single random student, s.t. p is the probability of a student to be in faculty I.

Therefore $H_0: p=0.75$

I think the statistic S(X) is the height of a single random student, so I want to find out the prob. under $H_0$ that his height is bigger than 168. I did it this way : $0.75P(\frac {S(X)-180}{25}>\frac {168-180}{25})+0.25 P(\frac{S(X)-160}{20}<\frac{168-160}{20})=0.5994$ (using R) Is this result the alpha used to calculate the rejection zone $C_a$?

(We've been told that $P_{H_0} (S(X)\in C_a)=a$ is how to do a statistic test)

I would love some thorough explanation (maybe some good links too) for this(not only hints) as I feel my understanding of the whole subject is still lacking. Thanks a lot

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The reason why normally one would start with a significance level is in order to find a rejection rule that delivers that significance level.

However, if you begin with a rejection rule, you work out the significance level from the rejection rule that gives.

Your general approach looks okay but you have made a number of errors.

p is the probability of a student to be in faculty I.
Therefore $H_0:p=0.75$

This is wrong already. Look more carefully at your question.

$0.75P(\frac {S(X)-180}{25}>\frac {168-180}{25})+0.25 P(\frac{S(X)-160}{20}<\frac{168-160}{20})=0.5994$

Even after fixing the previous error, this is wrong.

Let S be the height of the student.

$$ \alpha = P(S>168) = P(S>168|F_I)P(F_I) + P(S>168|F_{II})P(F_{II}) \\ =p\,P(S>168|F_I) + (1-p)\,P(S>168|F_{II}) $$

(from the law of total probability)

Both the conditional terms should have their inequality in the same direction:

$$P(S>168|F_i) = P(\frac{S-\mu_i}{\sigma_i}>\frac{168-\mu_i}{\sigma_i}) = P(Z>\frac{168-\mu_i}{\sigma_i})$$

Somehow you flipped one around.

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  • $\begingroup$ Thanks. Okay so now I just want to make sure (after fixing these errors) a=0.429? $\endgroup$
    – jreing
    Jan 1, 2014 at 10:37
  • $\begingroup$ and also: so considering that I found the alpha - can I now claim that that $C_a=[p_0+Z_{1-a}\sqrt{ \frac{p_0(1-p_0)} {n} },\infty)= [0.3255,\infty)?$ $\endgroup$
    – jreing
    Jan 1, 2014 at 10:57
  • $\begingroup$ It looks to me like you have the right probability; just be careful about the direction of your inequalities when you write them down. $\endgroup$
    – Glen_b
    Jan 1, 2014 at 11:10
  • $\begingroup$ On your second query: the rejection region is already specified in the question. You might like to reread the second sentence in my answer. Note that your question defined that $C_\alpha$ was in terms of the height, not the probability (which is useful because height is all you observe). $\endgroup$
    – Glen_b
    Jan 1, 2014 at 11:14
  • $\begingroup$ Can you please elaborate on that sentence? What I understood from it is that the signif. level (=alpha) can be figured from the rejection rule - and that's what I did and how I really found this alpha. Are you saying that the rejection zone is just $C_a=[168,\infty)$ ? If that's true why do I need the alpha in order to answer the question at all? $\endgroup$
    – jreing
    Jan 1, 2014 at 13:55

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