Page 2 of 2 FirstFirst 12
Results 16 to 26 of 26
Like Tree1Thanks

Math Help - Geometric Distribution

  1. #16
    Member
    Joined
    Nov 2013
    From
    Philadelphia
    Posts
    187

    Re: Geometric Distribution

    ????
    Follow Math Help Forum on Facebook and Google+

  2. #17
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,776
    Thanks
    1141

    Re: Geometric Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    ????
    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
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Member
    Joined
    Nov 2013
    From
    Philadelphia
    Posts
    187

    Re: Geometric Distribution

    Quote Originally Posted by romsek View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Member
    Joined
    Nov 2013
    From
    Philadelphia
    Posts
    187

    Re: Geometric Distribution

    Quote Originally Posted by romsek View Post
    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
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #20
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,776
    Thanks
    1141

    Re: Geometric Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    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?
    regions 2 and 3 yes,

    region 1 seems wrong
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Member
    Joined
    Nov 2013
    From
    Philadelphia
    Posts
    187

    Re: Geometric Distribution

    Quote Originally Posted by romsek View Post
    regions 2 and 3 yes,

    region 1 seems wrong
    Oops, I redid Region 1 and now got 0.55

    So, that would be it for part (bii)

    Then for part (biii) where it says to make a formula for power, wouldn't that just be the same formula I just used? I guess I'm just not understanding the wording of the question.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Member
    Joined
    Nov 2013
    From
    Philadelphia
    Posts
    187

    Re: Geometric Distribution

    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
    Follow Math Help Forum on Facebook and Google+

  8. #23
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,776
    Thanks
    1141

    Re: Geometric Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #24
    Member
    Joined
    Nov 2013
    From
    Philadelphia
    Posts
    187

    Re: Geometric Distribution

    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
    Last edited by AwesomeHedgehog; May 1st 2014 at 10:41 AM.
    Follow Math Help Forum on Facebook and Google+

  10. #25
    Member
    Joined
    Nov 2013
    From
    Philadelphia
    Posts
    187

    Re: Geometric Distribution

    Quote Originally Posted by romsek View Post
    Look at post #19

    see how I first defined $power_p$ ? Then that was averaged over the range of $p$ to get the final formula for the test power. Now they don't want you to average. They want to know what the power of the test is given that the internal short probability is actually some number p.

    You derived the Type I error in terms of $p$ for each region in post #10 (for region 1 it's $p$ in general)

    You've got a formula now for the power of the test in terms of $p$.

    Plot one against the other. This is an example of the ROC curve I mentioned earlier and gave you a graph of.
    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
    Last edited by AwesomeHedgehog; May 1st 2014 at 12:04 PM.
    Follow Math Help Forum on Facebook and Google+

  11. #26
    Member
    Joined
    Nov 2013
    From
    Philadelphia
    Posts
    187

    Re: Geometric Distribution

    For (biv) would I want to recommend rejection region 2 since it has the a low type I error and it's type II error isn't too bad, it's in between both of the others.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Is this a geometric distribution?
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 21st 2013, 06:13 AM
  2. Geometric distribution
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: July 28th 2011, 07:24 AM
  3. geometric distribution...p.g.f
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 26th 2011, 06:45 AM
  4. Geometric Distribution HELP!
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 15th 2009, 11:16 AM
  5. Geometric (p) Distribution on (0,1,2,...)
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 7th 2009, 07:32 PM

Search Tags


/mathhelpforum @mathhelpforum