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Math Help - Geometric Distribution

  1. #1
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    Geometric Distribution

    Question:


    "The Geometric Distribution and Shorts in NiCad Batteries"

    In their article "A Case Study of the Use of an Experimental Design
    in Preventing Shorts in Nickel-Cadmium Cells"; Ophir, El-Gad, and
    Snyder describe a series of experiments conducted in order to reduce
    the proportion of cells being scrapped by a battery plant because of
    internal shorts.

    The experimental program was successful in reducing the proportion of
    manufactured cells with internal shorts to around 0.03.

    The following procedure is being developed in order to monitor the
    process for increases in the true proportion of manufactured cells
    with internal shorts.

    Suppose that testing of batteries for internal shorts begins
    on a production run in this plant for monitoring/control purposes,

    let B = the number of batteries required until
    the first internal short is discovered.

    [a]
    What is the distribution and expected value of B,
    if the proportion of manufactured cells with internal shorts
    remains at 0.03?

    What is the distribution and expected value of B,
    if the proportion of manufactured cells with internal shorts
    increases to 0.10?

    What happens, in general, to the expected value of B
    as the proportion of manufactured cells with internal shorts
    increases? Choose one: [Larger or Smaller]

    [b]
    Consider using B to test the following hypotheses:

    H_0: % = 0.03 versus H_a: % > 0.03

    where % = true proportion of manufactured cells with internal shorts.

    B has been proposed as the TEST STATISTIC,
    however three Rejection Regions are being considered.

    A choice needs to be made concerning which of the three
    Rejection Regions to implement in the procedure.

    Rejection Region 1: Reject H_0 if B = 1.

    Rejection Region 2: Reject H_0 if B <= 2.

    Rejection Region 3: Reject H_0 if B <= 3.

    [b.i]
    Calculate the probability of a Type I Error for Rejection Region 1.

    Calculate the probability of a Type I Error for Rejection Region 2.

    Calculate the probability of a Type I Error for Rejection Region 3.

    Based upon this information which Rejection Region do you recommend? WHY?

    [b.ii]
    Now suppose, in reality, the true proportion of manufactured cells
    with internal shorts has increased to 0.10

    Calculate the power for Rejection Region 1.

    Calculate the power for Rejection Region 2.

    Calculate the power for Rejection Region 3.

    Based upon this information which Rejection Region do you recommend? WHY?

    [b.iii]
    For each Rejection Region determine an equation/formula/function that
    relates power to %, true proportion of manufactured cells w/ internal shorts.

    For Rejection Region 1: Power = ??function of %??

    For Rejection Region 2: Power = ??function of %??

    For Rejection Region 3: Power = ??function of %??

    [b.iv]
    Graph each of the above power functions on the same set of axes
    use EXCEL, MATLAB, etc. to get a "nice" graph.

    Place % on the x-axis and power on the y-axis.
    Note: the range for both % and power is 0 to 1.

    Now make a final recommendation concerning the Rejection Region
    you would recommend. Justify your answer by giving clear reasoning
    why you selected the Rejection Region you did.

    [c]
    Now suppose the experiment has been conducted and the observed value of B = 5.
    Calculate the p-value.

    My Answer so far:

    (a) Given that, in batteries, an experimental process successfully reduced the proportion of manufactured cells with internal shorts to around 0.03. Let B = the number of batteries required until the first internal short is discovered. Given, the proportion of manufactured cells with internal shorts to around 0.03.

    p = 0.03
    q = 1 - p = 1 - 0.03 = 0.97

    B ~ Geometric (p = 0.03)

    P(X = x) = (0.03)*(0.97)^x for x = 1, 2,...

    E(x) = q/p = 0.97/0.03 = 32.333

    so, the expected value of B is 32


    I'm wondering if my attempt at (a) is correct and if anyone has any helpful hints about how I should start part (b)
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  2. #2
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    Re: Geometric Distribution

    wouldn't $P(X=x) = (0.03)(0.97)^{x-1}$ ?

    you had $x-1$ good batteries before you found the defective one at try $n$
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  3. #3
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    Re: Geometric Distribution

    oh, yes, you're right.

    For the second part of part (a) would I use that same formula, except this time instead of using p = 0.03, I would use p = 0.10?
    Last edited by AwesomeHedgehog; April 27th 2014 at 07:04 PM.
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  4. #4
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    Re: Geometric Distribution

    yes
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  5. #5
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    Re: Geometric Distribution

    okay, so to finish up part (a) this is what I did:

    p = 0.10

    q = 1 - 0.10 = 0.90

    P(X = x) = (0.10)*(0.90)^(x - 1) for x = 1, 2, ...

    E(x) = q/p = 0.90/0.10 = 9

    So, in general, as the proportion of manufactured cells with internal shorts
    increases, the expected value of B decreases.

    Then for (bi)

    P[Type I Error for Rejection Region 1] = P[ b = 1 | % = 0.03 ] = ??

    this part I'm stuck at because I'm not sure if I would use an integral going from 0 to 0.03 of the geometric distribution or another method to calculate the probability. My teacher uses the computer software to figure out the probability, but doesn't show us how to figure it out by hand.
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    Re: Geometric Distribution

    Would the above be correct?

    Also, for (bi) would this be correct:

    P[Type I Error for Rejection Region 1] = P[ b = 1 | % = 0.03 ] = the integral from 0 to 1 of (0.03)*(0.97)^(x-1) dx = 0.0304616

    And then I'd use that integral again for the other rejection regions, except I would have the integral going from 0 to 2 for Rejection Region 2 and then have the integral going from 0 to 3 for Rejection Region 3
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  7. #7
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    Re: Geometric Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    okay, so to finish up part (a) this is what I did:

    p = 0.10

    q = 1 - 0.10 = 0.90

    P(X = x) = (0.10)*(0.90)^(x - 1) for x = 1, 2, ...

    E(x) = q/p = 0.90/0.10 = 9
    this isn't quite right. Since we never choose 0 batteries. B has a support of {1,2,3, ....}
    For this support $E[B] = \dfrac 1 p$ not $\dfrac {1-p} p$ as you've used. This last is the formula for the mean when 0 is a possible value.

    So, in general, as the proportion of manufactured cells with internal shorts
    increases, the expected value of B decreases.
    yes as it better had. If it's more likely the batteries are bad I'd expect to see a bad one faster.

    Then for (bi)

    P[Type I Error for Rejection Region 1] = P[ b = 1 | % = 0.03 ] = ??

    this part I'm stuck at because I'm not sure if I would use an integral going from 0 to 0.03 of the geometric distribution or another method to calculate the probability. My teacher uses the computer software to figure out the probability, but doesn't show us how to figure it out by hand.
    no.

    Consider the experiment for Region 1. We grab a battery and test it. It either passes or not with probability p=0.03, thus

    Pr[Type I error] = Pr[H0=True | rejected at B=1] = Pr[we get a bad battery first try | p=0.03] = 0.03

    For region 2, we sample a battery, if it fails we reject, if it passes, sample again and pass if it passes else reject. In other words we fail on {fail} or {pass, fail}

    The probability of failing is thus p+(1-p)p = 0.059. Similarly for Region 3. We fail on {fail}, {pass, fail}, {pass, pass, fail}

    figure out Pr[Type I error] in region 3
    Thanks from AwesomeHedgehog
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  8. #8
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    Re: Geometric Distribution

    Okay here's what I have:

    The completed part of part (a):

    When p = 0.03

    q = 1 - p = 1 - 0.03 = 0.97

    B ~ Geometric (p = 0.03)

    P(X = x) = (0.03)*(0.97)^(x - 1) for x = 1, 2,...

    E(x) = q/p = 0.97/0.03 = 32.333

    so, the expected value of B is 32

    Then when p = 0.10

    q = 1 - 0.10 = 0.90

    P(X = x) = (0.10)*(0.90)^(x - 1) for x = 1, 2, ...

    E(x) = 1/p = 1/0.10 = 10

    So, in general, as the proportion of manufactured cells with internal shorts increases, the expected value of B decreases.

    For part (bi):

    You said that the probability of a type I error for region one would be equal to 0.03. I used this formula and got the same answer:

    P[Type I Error for Rejection Region 1] = P[ b = 1 | % = 0.03 ] = the integral from 0 to 1 of (0.03)*(0.97)^(x-1) dx = 0.0304616

    I used the geometric distribution in the integral to show my work as for how I got 0.03.

    But for the Rejection Region 2 and 3, the integral wouldn't be going from 0 to 1 anymore, that part would change.

    So, P[Type I Error for Rejection Region 2] = P[ b <= 2 | % = 0.03 ] = the integral from 0 to 2 of (0.03)*(0.97)^(x-1) dx = 0.06

    which is the same answer that you got, except I calculated it differently.

    And then, P[Type I Error for Rejection Region 3] = P[ b <= 3 | % = 0.03 ] = the integral from 0 to 3 of (0.03)*(0.97)^(x-1) dx = 0.089
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    Re: Geometric Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    Okay here's what I have:

    The completed part of part (a):

    When p = 0.03

    q = 1 - p = 1 - 0.03 = 0.97

    B ~ Geometric (p = 0.03)

    P(X = x) = (0.03)*(0.97)^(x - 1) for x = 1, 2,...

    E(x) = q/p = 0.97/0.03 = 32.333

    so, the expected value of B is 32

    Then when p = 0.10

    q = 1 - 0.10 = 0.90

    P(X = x) = (0.10)*(0.90)^(x - 1) for x = 1, 2, ...

    E(x) = 1/p = 1/0.10 = 10

    So, in general, as the proportion of manufactured cells with internal shorts increases, the expected value of B decreases.

    For part (bi):

    You said that the probability of a type I error for region one would be equal to 0.03. I used this formula and got the same answer:

    P[Type I Error for Rejection Region 1] = P[ b = 1 | % = 0.03 ] = the integral from 0 to 1 of (0.03)*(0.97)^(x-1) dx = 0.0304616

    I used the geometric distribution in the integral to show my work as for how I got 0.03.

    But for the Rejection Region 2 and 3, the integral wouldn't be going from 0 to 1 anymore, that part would change.

    So, P[Type I Error for Rejection Region 2] = P[ b <= 2 | % = 0.03 ] = the integral from 0 to 2 of (0.03)*(0.97)^(x-1) dx = 0.06

    which is the same answer that you got, except I calculated it differently.

    And then, P[Type I Error for Rejection Region 3] = P[ b <= 3 | % = 0.03 ] = the integral from 0 to 3 of (0.03)*(0.97)^(x-1) dx = 0.089
    you are messed up because you are including 0 in your support. You never choose 0 batteries. n = {1, 2, 3 ...}
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    Re: Geometric Distribution

    Okay, so I tried it your way and did:

    P[Type I Error for Rejection Region 1 ] = 0.03

    P[Type I Error for Rejection Region 2] = p + (1 - p)*p = 0.0591 = 0.06

    P[Type I Error for Rejection Region 3] = p + (1 - p)*p + (1 - p)*(1 - p)*p = 0.087

    but what I do not understand is how you figured out how to use that to find the probability. I get what you're saying now with how my integrals should not include 0, but I do not get how you figured out how to do it this other way. I wish my teacher would have taught us how to do it by hand instead of using the computer to calculate the values.
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    Re: Geometric Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    Okay, so I tried it your way and did:

    P[Type I Error for Rejection Region 1 ] = 0.03

    P[Type I Error for Rejection Region 2] = p + (1 - p)*p = 0.0591 = 0.06

    P[Type I Error for Rejection Region 3] = p + (1 - p)*p + (1 - p)*(1 - p)*p = 0.087

    but what I do not understand is how you figured out how to use that to find the probability. I get what you're saying now with how my integrals should not include 0, but I do not get how you figured out how to do it this other way. I wish my teacher would have taught us how to do it by hand instead of using the computer to calculate the values.
    I just envisioned what the guy/gal in the battery checking booth had to do.
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    Re: Geometric Distribution

    For the Rejection Region that I'd recommend, would I choose Rejection Region 3 because it makes the rejection region better and it allows for variation to occur?

    and for part (bii)

    to solve for power, that whole thing just confuses me with the geometric distribution. Do you have any hints on how to start that?
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    Re: Geometric Distribution

    For part (bii) I remembered that power = 1 - beta = P[Reject H_0 when it is false] = P[Type II Error]

    Now, I just need to figure out what the formula is in order to calculate the power for each rejection region
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    Re: Geometric Distribution

    Quote Originally Posted by AwesomeHedgehog View Post
    For part (bii) I remembered that power = 1 - beta = P[Reject H_0 when it is false] = P[Type II Error]

    Now, I just need to figure out what the formula is in order to calculate the power for each rejection region
    For part (bii)

    μ = 0.10

    P[Type II Error for Rejection Region 1] = P[ b = 1 | μ = 0.10] = 1 - [the integral from 0 to 1 of (0.10)*(0.90)^(x - 1) dx ] = 0.8945

    I'm guessing that that would be similar for the other Rejection Regions, but I'm not sure.

    Also, (bii) relates to (biii) which really confuses me. Any help would be appreciated.
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  15. #15
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    Re: Geometric Distribution

    So, I talked to my teacher about this and my part (bii) is incorrect.

    I know that power = 1 - beta, so in order to solve for beta, this is what I did:

    beta = P[ x̄ ≠ 1 | H_a: μ > 0.03 ] = integral from 2 to infinity of (0.03)*(0.97)^(x - 1) dx = 0.955

    Then the power = 1 - beta = 0.045

    And that would be for rejection region one. However, the integral doesn't seem correct to me...
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