BoxCox Transformation
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 This topic has 5 replies, 4 voices, and was last updated 18 years ago by DD.

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September 22, 2004 at 7:09 pm #36968
HendersonParticipant@Melissa Include @Melissa in your post and this person will
be notified via email.The data I collected was nonnormal. I performed the BoxCox transformation with a lambda = 0.225 which yielded normality. After performing basic staitstics (mean and standard deviation), I wanted to convert those tranformations back to understandable numbers. However, when I try to transform the numbers back to a ‘real’ number, the results are totally not representative of my data set. I want to report the mean, standard deviation and 3s range. My data is below; can someone please help me transform data back?!
Descriptive Statistics: Transformed WLR
Variable N N* Mean SE Mean StDev Minimum Q1 Median
Transformed 16 0 0.9149 0.0503 0.2014 0.5780 0.7286 0.9529
Variable Q3 Maximum
Transformed 1.0836 1.1690
WLR Transformed WLR11.4 0.578054.6 0.709148.3 0.620882.9 0.786795.7 0.675710.7 1.083640.7 1.083640.8 1.051541.1 0.978762.2 0.837302.4 0.821051.1 0.978760.5 1.168950.5 1.168951.4 0.927020.5 1.168950September 22, 2004 at 7:20 pm #107768
Chris SeiderParticipant@cseider Include @cseider in your post and this person will
be notified via email.Dear Melissa,
I think you can find details in Minitab help but just take your mean, s.d., etc to the (1/0.225) power. Doesn’t work if lambda is zero, but then you would take the inverse log base 10.
If you lambda was 0.333, you would take transformed mean and cube it.
I would caution you reporting the mean, s.g., etc. after you reverse the transformation–they really mean nothing to the casual observer. You distribution obviously changes so the numbers won’t “look” like your real mean, etc.0September 22, 2004 at 7:31 pm #107772
HendersonParticipant@Melissa Include @Melissa in your post and this person will
be notified via email.Can the s.d also be transformed in the same manner? If so, then why is the s.d so unrepresentative of my data set.
0September 22, 2004 at 7:39 pm #107774Melissa I would caution you against the used of backtransformed parameters directly. If you data is not normal on the first place, What would be the interest of describing it with parameters typicaly used for normal distributed populations?
Having said that, you can calculate the mean and median directly with your data, without transforming. Be aware that for not normally distributed data the median may be a more meaninful descriptor that the mean. Regarding SD, I rather suggest to used another descriptor of error, such as variance.
BoxCox is useful to transfor the data, get a “normalized” SD and calculate control limits. then you can backtransform those control limits to your original data.0September 22, 2004 at 7:39 pm #107775
Chris SeiderParticipant@cseider Include @cseider in your post and this person will
be notified via email.Melissa,
Yes, you can apply the same reverse transformation to your standard deviation as your mean. However, remember that the reversed transformed s.d. does NOT represent the standard deviation of the original data. That’s why I caution even reporting the reverse transformed data.
People use Box Cox transformations to make a better description of their process. After a change has occurred, hopefully an improvement you made, then they will use the same transform on the new data and see if the transformed mean or standard deviation has changed–using the statistics found with normal distributions.0September 23, 2004 at 9:17 am #107816Hi
Once you have decided to transform the data after thoughroughly undrestanding your process characteristics(assuming there is no other way of getting the normal distUse a Lambda of 0.5 as it gives you a better transform into a Normal dist(see below) I have used Minitab14 to get this
Descriptive Statistics
N N* Mean StDev Median Minimum Maximum Skewness Kurtosis
16 0 2.8 3.19186 1.25 0.5 11.4 1.77312 2.65146
BoxCox transformation for Normal distribution: Lambda = 0.5
Goodness of Fit Test
Distribution AD P LRT P
Normal 1.575 (<) 0.005
Normal (After Transformation) 0.376 0.369
Lognormal 0.440 0.255
3Parameter Lognormal 0.816 * 0.001
Exponential 0.646 0.304
2Parameter Exponential 1.169 0.032 0.013
Weibull 0.633 0.088
3Parameter Weibull 0.377 0.432 0.000
Smallest Extreme Value 1.956 (<) 0.010
Largest Extreme Value 1.224 (<) 0.010
Gamma 0.723 0.074
3Parameter Gamma 0.260 * 0.000
Logistic 1.321 (<) 0.005
Loglogistic 0.449 0.217
3Parameter Loglogistic 0.650 * 0.001
ML Estimates of Distribution Parameters
Distribution Location Shape Scale Threshold
Normal* 2.80000 3.19186
Normal (After Transformation) 0.87116 0.39277
Lognormal* 0.50351 1.03606
3Parameter Lognormal 0.73721 2.48225 0.49500
Exponential 2.80000
2Parameter Exponential 2.30500 0.49500
Weibull 0.99052 2.78697
3Parameter Weibull 0.54634 1.47157 0.49500
Smallest Extreme Value 4.54748 3.90589
Largest Extreme Value 1.55563 1.76479
Gamma 1.08674 2.57651
3Parameter Gamma 0.41591 5.54211 0.49500
Logistic 2.18928 1.55931
Loglogistic 0.43032 0.60239
3Parameter Loglogistic 0.42644 1.40205 0.49500
* Scale: Adjusted ML estimate
Here is the transformed Data
WLR Transformed Data Back Transform11.4 0.29617 11.44.6 0.46625 4.68.3 0.34711 8.32.9 0.58722 2.95.7 0.41885 5.70.7 1.19523 0.70.7 1.19523 0.70.8 1.11803 0.81.1 0.95346 1.12.2 0.67420 2.22.4 0.64550 2.41.1 0.95346 1.10.5 1.41421 0.50.5 1.41421 0.51.4 0.84515 1.40.5 1.41421 0.5
To Transform it back I have used the calculator functionof minitab using the formula ANTILOG( – 2*LOGT(‘Transformed Data’))
Hope this helps
DD0 
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