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Help with Sample Size Needed

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  • #55037

    John
    Participant

    Hi Everyone,

    I’m new to LSS and am currently taking a course.

    One of the assignments seems very straight-forward, but I simply cannot wrap my head around it. Can anyone help?

    There are 4 parts to this assignment, but I’ve only included the 1st part as well as my response for it.

    Be kind because I am new to the stats world…

    Question Part 1:
    A part is not to exceed a failure rate of 3.4 failures in one million.
    (a) Determine the sample size needed for a hypothesis test of the equality of this failure rate to a 3.4 parts per million failure rate. Use risks of 0.05 and an uncertainty of +/- 10% of the failure rate target.

    Response to Part 1:
    (a) Sample size needed (n) for a failure rate criterion of 3.4 DPMO (3.4 / 1,000,000 = 0.0000034)
    • Risk = a = 0.05 (or 95% confidence level)
     Or Za = 1.645
    • St.Dev. = +/- 10% = 1,000,000 * 0.1 = +/- 100,000
     Or 1,000,000 +/- 100,000
     900,000 <= p <= 1,100,000
    • n = ((Ua * (pa * (1- pa))1/2) + (Ub * (pb * (1-pb)) 1/2))/(pb – pa)
     pa = 3.4 / 1,100,000 = 3.0909090909090909090909090909091e-7
    • pa = 0.0000003091
     pb = 3.4 / 900,000 = 3.74777777777777777777777777777778e-6
    • pb = 0.00000375
     Ua = 1.645 (Table B, Appendix A, Single-Sided Normal Distribution)
     Ub = 1.645 (Table B, Appendix A, Single-Sided Normal Distribution)
     n = ((1.645 * (9.091 * (1 – 0.0000003091))1/2) + (1.645 * (1.111 * (1 – 0.00000375))1/2))/(0.00000375 – 0.0000003091)
     n = ((1.645 * (9.091 * 0.9999996909)1/2) + (1.645 * (1.111 * 0.99999625)1/2))/(0.0000034409)
     n = ((1.645 * (9.0909971899719)1/2) + (1.645 * (1.11099583375)1/2))/(0.0000034409)
     n = ((1.645 * 3.015) + (1.645 * 1.054))/(0.0000034409)
     n = 1.73383/0.0000034409
     n = 503888.52 = 503,889

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    #198319

    John
    Participant

    Sorry, the formatting came out wrong…

    Also, I have access to Minitab 16 but am new to that too…

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    #198324

    Amit Kumar Ojha
    Participant

    Hi John,

    Please find below the solution for the given problem:
    It is given that

    Tolerance = 10
    Alpha = 0.05 and Hence Z = 1.96
    Standard Deviation corresponding to 3.4 parts per million defect rate = 6
    Hence Sample Size = ((1.96*6)/10)^2 = 32.14

    So you need a sample of size = 33 (Rounded Off)

    Good Luck !!!

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    #198326

    John
    Participant

    Thank you AmitOjha,

    How did you determine Z?

    I know 3.4 dpmo is the 6 sigma standard (really its equal to 4.5 sigma).
    And that is used as the st.dev., but are you using reference tables or are these calculated?

    Also, is this related to a type of 1 proportion test

    Regards,
    John

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    #198329

    John
    Participant

    Ok, i realize my last question on Z is silly. I realize the z-table contains my answer.

    But I am being pointed towards a 1 proportion type test to solve this for the assignment.

    Can you help clarify?

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    #198330

    Amit Kumar Ojha
    Participant

    Hi,

    by 1 proportion test for determining sample size what exactly do you imply ? Sorry I dint get that :-(

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    #198335

    Nigel
    Participant

    @John.Lombardo Do you mean a 1 tailed test vs a 2 tailed test? Amit probably knows better but I think that would mean you need to use a Z-score of 1.645 instead of 1.96.

    This article explains the difference between the 2 tests:
    http://www.ats.ucla.edu/stat/mult_pkg/faq/general/tail_tests.htm

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    #198336

    John
    Participant

    Hi,

    So the final solution that I submitted (and was accepted) is below.

    For the 1 proportion test, this is done through minitab as follows:
    Go to: Stat –> Power and sample size –> 1 Proportion …
    Then enter into the dialoge:
    Comparison propotion = 0.00000374
    Power value = 0.95
    Hypothesized proportion: 0.00000306

    Or you can do it the manual way using the following formula:
     n = (((Ua * (pa * (1- pa))^1/2) + (Ub * (pb * (1-pb))^1/2))/(pb – pa))^2
    • pa = 0.0000034 * 1.1 = 0.00000374
    • pb = 0.0000034 * 0.9 = 0.00000306
    • Ua = 1.645 (Table B, Appendix A, Single-Sided Normal Distribution)
    • Ub = 1.645 (Table B, Appendix A, Single-Sided Normal Distribution)
    • n = (((1.645 * (0.00000374 * (1 – 0.00000374))1/2) + (1.645 * (0.00000306 * (1 – 0.00000306))1/2))/(0.00000306 – 0.00000374))^2
    • n = 79,389,227

    In any case, the sample size is deemed unmanageable!

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    #198339

    Amit Kumar Ojha
    Participant

    Hi John,

    See as far as I know, there can be only one point which you need to take care about the formula for sample size calculation i.e. whether you are concerned about one tailed or two tailed test.
    In this context Nigel is correct.

    However, the formula which you used to calculate the sample size is not correct.

    Refer Levin and Rubin – Statistic for Management for sample size calculation. This is best book I have ever known for statistics for beginners.

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    #198391

    John
    Participant

    Thank you Amit, I will certainly take a look at it!

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