# Standard deviation Problem

Six Sigma – iSixSigma Forums Old Forums General Standard deviation Problem

Viewing 6 posts - 1 through 6 (of 6 total)
• Author
Posts
• #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

³Ò
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
Thanks so much
Victor

0
#164901

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

0
#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
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
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
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
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
Viewing 6 posts - 1 through 6 (of 6 total)

The forum ‘General’ is closed to new topics and replies.