????
I think what you have to do here is note the following
a) for a given $n$, and $p$ the power of the tests are given by $power_p=1-(1-p)^n$
b) to get the test power you have to average that across p for the rejection range, in this case either (0.03, 1) or (0.1, 1)
c) for a general actual bad battery probability $p_0$, the formula for this average is given by
$power=\dfrac 1 {1-p_0} \displaystyle{\int_{p_0}^1} 1-(1-p)^n~dp$
$power =1 - \dfrac{(1-p_0)^{n}}{n+1}$
here you are treating p as a uniform$[p_0,1]$ random variable because you don't have any info about it other than it's range.
you can use this last formula to quickly evaluate the power of the the regions corresponding to $n$=1,2,3, with $p_0$=0.03, or 0.1
okay, so for (bii), I would use this formula to calculate the power of each rejection region?
and then for (biii) it tells me to determine a function that relates to %, but wouldn't that just be the exact same formula as the one you just stated? But, I feel like that would be way too simple of an answer.
So I tried using the formula for power to solve (bii) and this is what I got:
I made p_0 = 0.10 since it says that the internal shorts increases to 0.10
Then for Rejection Region 1, I got power = 1 - [(1 - 0.10)^1 / 2 ] = 0.95
Using that same formula for Region 2, I got power = 0.73
and for Region 3, I got power = 0.82
does that seem correct?
I believe that I have parts (a) through (biv) all done.
For part (c), would this be correct?
Our test statistic would be:
[observed - expected] / STD error
but since we don't know the sample size, we cannot compute the STD error.
So, are there any ideas as for how to get the p-value?
I tried getting alpha, but I don't know if that really helps...
alpha = P[Type I Error where B = 5] = the integral from 0 to 5 of (0.03)*(0.97)^(x - 1) dx = 0.143
Any help would be extremely appreciated
0.55 is correct.
i'd like to see your results for biii and if possible the graphs you made for biv
I'm not entirely sure yet what they mean in the last question. Let's make sure you've got iii and iv correct first.
After all this effort you might as well get this question answered perfectly.
For (biii) I didn't really understand what the question was asking. I just took it as the question saying what formula did you use in (bii) to find power? So what I did was rewrite the formula that I used for (bii) in a general form:
power = 1/[1 - p_0] * the integral from p_0 to 1 of 1 - (1 - p)^n dp
n = 1, 2, 3 <-- Rejection Regions
p_0 = 0.10 or 0.30
(biv) This is my graph: https://scontent-b-ord.xx.fbcdn.net/...ff&oe=5364AEB4
So, then for (biii) would it be:
f(x) = 1 - (1 - p)^(n-1) for the general formula without averaging it?
See, I know where I got the formula in post #10 because I understand the Type I Error and how to solve it for any type of distribution, but with power, I am not quite grasping this concept.
I tried plotting one against the other, but it keeps getting screwed up every time I try.
This is what I kept getting when I tried to plot one against the other:
https://scontent-b-ord.xx.fbcdn.net/...ca&oe=53644CA7