Standard deviation Problem
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 This topic has 5 replies, 4 voices, and was last updated 14 years, 2 months ago by Eric Maass.

AuthorPosts

November 16, 2007 at 12:18 pm #48689
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
account015
68
36
68
57163
23
31
22
543160
15
22
5
66190
5
23
1
091120
1
5
2
0121150
8
8
0
0³Ò
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
Victor0November 16, 2007 at 12:36 pm #164901
Fake Stan AlertParticipant@FakeStanAlert Include @FakeStanAlert in your post and this person will
be notified via email.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 opinion0November 16, 2007 at 1:11 pm #164903What was the logic behind having the first two “buckets” (015,1630)in 15 minute ranges and all the remaining “buckets” (3160, 6190…)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.0November 17, 2007 at 9:21 am #164961Thanks 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 ) 015, 1630, 3160…since the beginnig I disagree with my professor about his because there no continuity of the ranges..It should be 015, 1630, 3145, 4660….This was my opinion but I could not do it nothing…What can I do to solve this problem out ?
Thanks a lot0November 18, 2007 at 12:57 pm #165000
Eric MaassParticipant@poetengineer Include @poetengineer in your post and this person will
be notified via email.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 AndersonDarling (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
GoodnessofFit
AndersonDarling Correlation
Distribution (adj) Coefficient
Weibull 1.362 0.986
Lognormal 1.349 0.995
3Parameter Weibull 1.362 0.986
3Parameter Lognormal 1.349 0.995
2Parameter 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
LogLikelihood = 642.816
GoodnessofFit
AndersonDarling (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 27.95690November 18, 2007 at 12:57 pm #165001
Eric MaassParticipant@poetengineer Include @poetengineer in your post and this person will
be notified via email.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 AndersonDarling (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
GoodnessofFit
AndersonDarling Correlation
Distribution (adj) Coefficient
Weibull 1.362 0.986
Lognormal 1.349 0.995
3Parameter Weibull 1.362 0.986
3Parameter Lognormal 1.349 0.995
2Parameter 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
LogLikelihood = 642.816
GoodnessofFit
AndersonDarling (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 27.95690 
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