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Standard deviation Problem

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  • #48689

    Aguilar
    Member

    There Sirs please just help me with this problem.
    I made a survey in Clinic and the result of was this: ( this is what the customers want )

    Time
    recept.
    examin.
    LabTest
    account

    0-15
    68
    36
    68
    57

    16-3
    23
    31
    22
    54

    31-60
    15
    22
    5
    6

    61-90
    5
    23
    1
    0

    91-120
    1
    5
    2
    0

    121-150
    8
    8
    0
    0

    Š³ŽÒ”
    120
    125
    98
    117
    The point is that I know in the reception the waiting time that patients selected was between 0 and 15.  I am performing a Six Sigma Project to try to dimish the waiting time. My teacher told me that I have to calculate the Standard deviation and the media of each section.
    I disagree, because These are LClL = 0 and UCL = 15.  You can check this data.
    I calculate what he told me but this is strange:
    media = 17.43 and standard deviation = 25.94
    This means
    UCL = X + 3 standard deviation = 95.25 minutes
    I know this is wrong but please let know your opinion and advices.
    Thanks so much
    Victor 

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    #164901

    Fake Stan Alert
    Participant

    In  you  case  you  have  to  consider  “smaller  is  better  ”  as  a  major  quality characterstic,that  means  that  you  have  to  drop Cpk (for  the  usl)  and  to consider  only  the  lsl.
    Just  my  opinion

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    #164903

    Ward
    Participant

    What was the logic behind having the first two “buckets” (0-15,16-30)in 15 minute ranges and all the remaining “buckets” (31-60, 61-90…)in 30 minute ranges?
    UCL is not calculated by multiplying the standard deviation by 3!
    Please don’t confuse the UCL for a control chart (based on the moving range) with the 3 standard deviations for a capability study.

    0
    #164961

    Aguilar
    Member

    Thanks Pete for your comment but the big question is How Can I calculate the USL of the data ?.
    The logic behind what I did ( ranges ) 0-15, 16-30, 31-60…since the beginnig I disagree with my professor about his because there no continuity of the ranges..It should be 0-15, 16-30, 31-45, 46-60….This was my opinion but I could not do it nothing…What can I do to solve this problem out ?
    Thanks a lot

    0
    #165000

    Eric Maass
    Participant

    Victor,
    Since the data you have is counts by ranges (like the range of 0 to 15 and the range of 16 to 30), you don’t know the exact numbers of minutes for any response.
    One way to handle it is to apply the counts to the midpoint of each range.
    1) I pulled your data into Minitab 15 and used a different approach (Stat/Reliability/Distribution Analysis (Arbitrary Censoring)/Distribution ID Plot) which allows you to find a good distribution fit and obtain estimates for the mean and standard deviation using that distribution fit. Iused the lower limit of time in each interval as the “Start” and the upper limit in each interval as the “End”
    The Lognormal Distribution looked like a relatively good fit for your data – here is the table of Anderson-Darling (lower is better) and Correlation Coefficients (higher is better; closer to 1).
    2) Then, I used Stat/Reliability/Distribution Analysis (Arbitrary Censoring)/Parametric Distribution Analysis to estimate the mean and standard deviation assuming a fit to a Lognormal distribution.
    Here are the Minitab outputs (I can’t post the graphs too easily):
    1) Distribution Identification: 
    Distribution ID Plot: Start = Low Time and End = High Time
     
    Goodness-of-Fit
    Anderson-Darling Correlation
    Distribution (adj) Coefficient
    Weibull 1.362 0.986
    Lognormal 1.349 0.995
    3-Parameter Weibull 1.362 0.986
    3-Parameter Lognormal 1.349 0.995
    2-Parameter Exponential 1.796 *
    Smallest Extreme Value 1.608 0.918
    Normal 1.691 0.946
    Logistic 1.709 0.963
     
    2) Estimating the mean and standard deviation assuming a normal distribution:
    Distribution Analysis, Start = Low Time and End = High Time
    Variable Start: Low Time End: High Time
    Frequency: Total Count
    Censoring Information Count
    Interval censored value 460
    Estimation Method: Least Squares (failure time(X) on rank(Y))
    Distribution: Lognormal
     
    Parameter Estimates
    Standard 95.0% Normal CI
    Parameter Estimate Error Lower Upper
    Location 2.65078 0.0693402 2.51488 2.78668
    Scale 1.15365 0.0697274 1.02477 1.29874
    Log-Likelihood = -642.816
    Goodness-of-Fit
    Anderson-Darling (adjusted) = 1.349
    Correlation Coefficient = 0.995
     
    Characteristics of Distribution
    Standard 95.0% Normal CI
     Estimate Error Lower Upper
    Mean(MTTF) 27.5564 2.07904 23.7685 31.9479
    Standard Deviation 45.9826 7.64497 33.1952 63.6961
    Median 14.1651 0.982209 12.3651 16.2271
    First Quartile(Q1) 6.50555 0.659797 5.33279 7.93621
    Third Quartile(Q3) 30.8428 1.88966 27.3529 34.7781
    Interquartile Range(IQR) 24.3373 1.72171 21.1863 27.9569

    0
    #165001

    Eric Maass
    Participant

    Victor,
    Since the data you have is counts by ranges (like the range of 0 to 15 and the range of 16 to 30), you don’t know the exact numbers of minutes for any response.
    One way to handle it is to apply the counts to the midpoint of each range.
    1) I pulled your data into Minitab 15 and used a different approach (Stat/Reliability/Distribution Analysis (Arbitrary Censoring)/Distribution ID Plot) which allows you to find a good distribution fit and obtain estimates for the mean and standard deviation using that distribution fit. Iused the lower limit of time in each interval as the “Start” and the upper limit in each interval as the “End”
    The Lognormal Distribution looked like a relatively good fit for your data – here is the table of Anderson-Darling (lower is better) and Correlation Coefficients (higher is better; closer to 1).
    2) Then, I used Stat/Reliability/Distribution Analysis (Arbitrary Censoring)/Parametric Distribution Analysis to estimate the mean and standard deviation assuming a fit to a Lognormal distribution.
    Here are the Minitab outputs (I can’t post the graphs too easily):
    1) Distribution Identification: 
    Distribution ID Plot: Start = Low Time and End = High Time
     
    Goodness-of-Fit
    Anderson-Darling Correlation
    Distribution (adj) Coefficient
    Weibull 1.362 0.986
    Lognormal 1.349 0.995
    3-Parameter Weibull 1.362 0.986
    3-Parameter Lognormal 1.349 0.995
    2-Parameter Exponential 1.796 *
    Smallest Extreme Value 1.608 0.918
    Normal 1.691 0.946
    Logistic 1.709 0.963
     
    2) Estimating the mean and standard deviation assuming a normal distribution:
    Distribution Analysis, Start = Low Time and End = High Time
    Variable Start: Low Time End: High Time
    Frequency: Total Count
    Censoring Information Count
    Interval censored value 460
    Estimation Method: Least Squares (failure time(X) on rank(Y))
    Distribution: Lognormal
     
    Parameter Estimates
    Standard 95.0% Normal CI
    Parameter Estimate Error Lower Upper
    Location 2.65078 0.0693402 2.51488 2.78668
    Scale 1.15365 0.0697274 1.02477 1.29874
    Log-Likelihood = -642.816
    Goodness-of-Fit
    Anderson-Darling (adjusted) = 1.349
    Correlation Coefficient = 0.995
     
    Characteristics of Distribution
    Standard 95.0% Normal CI
      Estimate Error Lower Upper
    Mean(MTTF) 27.5564 2.07904 23.7685 31.9479
    Standard Deviation 45.9826 7.64497 33.1952 63.6961
    Median 14.1651 0.982209 12.3651 16.2271
    First Quartile(Q1) 6.50555 0.659797 5.33279 7.93621
    Third Quartile(Q3) 30.8428 1.88966 27.3529 34.7781
    Interquartile Range(IQR) 24.3373 1.72171 21.1863 27.9569

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