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Sample Size Determination for Variance Reduction

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  • #52348

    Travis
    Member

    I’m trying to locate a formula that can be used to determine the sample size requirements to detect a reduction in variance.  All literature and discussion threads I’ve found only discuss sample size determinations for differences in the mean.
     
    Basically, I want to determine how many samples must be tested in order to detect a 10% reduction in variance at 95% confidence.  The current sample standard deviation is 0.0008. 
     
    I would greatly appreciate any feedback.

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    #184091

    annon
    Participant

    Cant you use the Power and Sample Size function on MTB to accomplish this?

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    #184110

    Allard Munters
    Participant

    No, Minitab’s power and sample size calculation does not support test for equal variance…

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    #184115

    Asleeper
    Participant

    Yes, Travis, there is a simple formula. If I interpret your question correctly, this is a
    one-sample test, that is, you will take one sample
    of parts and test them to see if the population
    standard deviation has changed from a previous
    valueAlso, this is a one-sided test. That is, you are
    not looking for an increase, only a decrease.In this case, here is the sample size formula,
    given as an Excel formula:=1+0.5*(((NORMSINV(Alpha)*Sigma0+NORMSINV(Beta)*Sig
    maBeta)/(Sigma0-SigmaBeta))^2)Where Alpha is the risk of a Type 1 error (0.05?)
    Beta is the risk of a Type 2 error (0.05 to give
    95% confidence of detecting a shift)
    Sigma0 is the old std. dev. (0.0008)
    SigmaBeta is the new std. dev. to be detected
    (0.00072 in your case)With these values, the sample size is 490, after
    rounding up the results of the calculation.This is from p. 413 of my book, Design For Six
    Sigma StatisticsI hope this helps!

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    #184117

    Cone
    Participant

    Your book is wrong.You might want to try a book where the people have an
    appropriate background.Look at the formulas for confidence interval of the standard
    deviation, it is based on a chi squared distribution. You can decide
    how big of a variance reduction you desire to be able to detect and
    solve for n in the equation. The solution is iterative, that’s why
    most stats book stay away from it.There are also graphs in the back of Juran’s Quality Handbook, at
    leas since the second edition that will help you quickly know the
    right answer. The graphs are based on the chi squared distribution.

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    #184119

    Cone
    Participant

    This is also helpful.https://www.isixsigma.com/forum/showmessage.asp?
    messageID=156786

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    #184120

    Asleeper
    Participant

    The formula I offered uses a normal approximation
    to the chi-squared distribution, which is
    reasonably good for large sample sizes. I stand by the formula as being a useful
    approximation.

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    #184133

    Cone
    Participant

    Sloppy stats and it way overestimates the sample size needed.
    Do some simulations using the sample size you have suggested and you will see.

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