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Thread: Probability: Sampling to find defects

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    Probability: Sampling to find defects




    Hi, I'm new to the site.

    We discovered that a process is resulting in a defect rate of 14% for some batches.
    Each batch of product contains 2,500 units.

    I want to sample each batch to determine if the problem exists in that batch or not, to a 95% confidence level. I need to calculate the sample size required that will give me a 95% chance of detecting the problem.

    I have performed a calculation using probability where we sample without replacement and based on this it gives me a sample size of 20 per batch, which shows we should be 95% confident of identifying a defective unit. Data below.

    I also considered solving this problem using the Binomial distribution with Minitab, but this seems to give me a different answer.

    Is this a sound approach or can you suggest an alternative?

    Unit No. good Total post sample Prob. Good unit Prob. next pump good
    1 2150 2500 0.86
    2 2149 2499 0.860 0.740
    3 2148 2498 0.860 0.636
    4 2147 2497 0.860 0.547
    5 2146 2496 0.860 0.470
    6 2145 2495 0.860 0.404
    7 2144 2494 0.860 0.347
    8 2143 2493 0.860 0.299
    9 2142 2492 0.860 0.257
    10 2141 2491 0.859 0.221
    11 2140 2490 0.859 0.190
    12 2139 2489 0.859 0.163
    13 2138 2488 0.859 0.140
    14 2137 2487 0.859 0.120
    15 2136 2486 0.859 0.103
    16 2135 2485 0.859 0.089
    17 2134 2484 0.859 0.076
    18 2133 2483 0.859 0.066
    19 2132 2482 0.859 0.056
    20 2131 2481 0.859 0.048
    21 2130 2480 0.859 0.042


    Thank you for reading,

    Joe

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    Re: Probability: Sampling to find defects

    Quote Originally Posted by Joerice View Post
    Hi, I'm new to the site.
    Welcome to Talkstats forum.

    Quote Originally Posted by Joerice View Post
    I also considered solving this problem using the Binomial distribution with Minitab, but this seems to give me a different answer.
    As per Binomial also the answer is same, the sample size as 20. What was the 'different answer' are talking here?
    In the long run, we're all dead.

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    Re: Probability: Sampling to find defects

    Vinux,
    firstly thank so much you for your reply, I'm really pleased that this is the correct sample size, I was starting to doubt myself!

    Using the Binomial in Minitab I get the following:

    "Welcome to Minitab, press F1 for help.

    Cumulative Distribution Function

    Binomial with n = 20 and p = 0.86

    x P(*X*≤*x*)
    1 0.00000
    2 0.00000
    3 0.00000
    4 0.00000
    5 0.00000
    6 0.00000
    7 0.00000
    8 0.00000
    9 0.00002
    10 0.00014
    11 0.00080
    12 0.00384
    13 0.01534
    14 0.05067
    15 0.13748
    16 0.30412
    17 0.54498
    18 0.79157
    19 0.95103
    20 1.00000"

    Which suggests a sample size of 19. Perhaps I am not using the correct function in minitab.

    Any suggestions would be welcome.

    Joe

  4. #4
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    Re: Probability: Sampling to find defects


    Quote Originally Posted by Joerice View Post
    Vinux,
    firstly thank so much you for your reply, I'm really pleased that this is the correct sample size, I was starting to doubt myself!

    Using the Binomial in Minitab I get the following:

    "Welcome to Minitab, press F1 for help.

    Cumulative Distribution Function

    Binomial with n = 20 and p = 0.86

    x P(*X*≤*x*)
    1 0.00000
    2 0.00000
    3 0.00000
    4 0.00000
    5 0.00000
    6 0.00000
    7 0.00000
    8 0.00000
    9 0.00002
    10 0.00014
    11 0.00080
    12 0.00384
    13 0.01534
    14 0.05067
    15 0.13748
    16 0.30412
    17 0.54498
    18 0.79157
    19 0.95103
    20 1.00000"

    Which suggests a sample size of 19. Perhaps I am not using the correct function in minitab.

    Any suggestions would be welcome.

    Joe
    This is not the way to calculate the sample size.
    Hint: In your calculation n is fixed (x is varying). You need to calculate the sample size required that will give me a 95% chance of detecting the problem. So, sample size is the variable.
    In the long run, we're all dead.

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