1. ## Type 1 and 2 errors

In the problem I am given :
$
H_0 : u=5 ; H_1:u\neq5$

n = 8 (sample size)

The random variable follows a normal law with a standard deviation of 0.25
The problem also states that we need to reject $$H_0$$ if $\overline{x}<4.85$ or $\overline{x}>5.15.$

I need to find the probability of a type I error.

What I've tried so far: (alpha represents my Type I error)
$\alpha=P(reject H_0|H_0 true)
=P(\overline{x}<4.85 or \overline{x}>5.15 | u=5)$

So I tried to start and find the probability of $\overline{x}$ not in that interval.
$\overline{x} \not \in [u-0.15;u+0.15]$
$[u-\frac{s*Z_{\alpha/2}}{\sqrt{n}},u+\frac{s*Z_{\alpha/2}}{\sqrt{n}}]$
which gives me $\frac{s*Z_{\alpha/2}}{\sqrt{n}} = 0.15$
when I plug my numbers and isolate $Z_{\alpha/2}$, i get $Z_{\alpha/2}=0.5*\sqrt{8}/0.25^2=22.62
$

It seems to me that this value of Z can't be right since my table only goes to a maximum of 3.09.
As for the rest of the problem, I could use a tip or two about how to find P(u=5) and how to calculate the conditionnal probability.

Thanks

2. alpha is the PROBABILTY of the type 1 error, it's not the error.

$P(\bar X>5.15)=P(Z>{5.15-5\over .25/\sqrt{8}}) = P(Z>{3\sqrt{8}\over 5})$

Likewise
$P(\bar X<4.85)= P(Z<-{3\sqrt{8}\over 5})$, which is the same as above.

I am assuming that .25 is the POPULATION st deviation. If it's the sample st dev., we will have a t distribution with 7 degrees of freedom.

3. Ok wow I had it wrong, thanks!

(you are correct, it's the standard deviation for the population)
So I end up with P(Z< -1,697) or P(Z>1.697)
which is a probability of 0.0448. Now I guess I add those 2 probabilities and I get alpha? Which is 0.0896 or 8,96%, right?

4. yes, via a table
you can always get a better approximation of these distributions via an online calculator...
Free Two Tailed Area Under the Standard Normal Curve Calculator
They give me... .0896966

5. Ok that's great.

For the same problem I also need to find the probability of a type 2 error if $\mu$ is 5.1 (I'll use B to represent the probability of this error)

So I know that B(5.1)=P{accept H_0|H_0 is false}

$P(Z>\frac{4.85-5.1}{0.25/\sqrt{8}})$ and $P(Z<\frac{5.15-5.1}{0.25/\sqrt{8}})$

so I end up with P(Z>-2.828) and P(Z<0.567)
that's

(the regions -2.828 to 0 and 0 to 0.567 in the graph of the std normal law)
so P=0.9977-0.5 and 0.7146-0.5
gives me : P=0.4977+0.2146 = 0.7123

This probability seems very high (71.23%) so I'm wondering if I did everything correctly?

6. tex errors

Originally Posted by Lok
Ok that's great.

For the same problem I also need to find the probability of a type 2 error if mu is 5.1 (I'll use B to represent the probability of this error)

So I know that B(5.1)=P{accept H_0|H_0 is false}

$P(Z>{4.85-5.1\over 0.25/\sqrt{8}}) + P(Z<{5.15-5.1\over 0.25/\sqrt{8}})$

so I end up with P(Z>-2.828) and P(Z<0.567)
that's

(-2.828 to 0 and 0 to 0.567)
so 0.9977-0.5 and 0.7146-0.5
gives me : 0.4977+0.2146 = 0.7123

This probability seems very high (71.23%) so I'm wondering if I did everything correctly?

7. Use $P(4.85<\bar X<5.15)=P\biggl({4.85-5.1\over 0.25/\sqrt{8}} instead.

8. Yea, that's what I have (I think).
Is it normal that the probability to make an error of type 2 be that high? (71.23%, I think my numbers are correct)

9. it's high because 5.1 is near 5
the closer you get to 5 the closer beta will be to 1-alpha
so the beta at 5.00001 will be near .91
GOOD NIGHT!!! it's 3am

10. Ok! Thanks a lot (:

Good night.