Results 1 to 4 of 4

Math Help - Fisher's LSD test on transformed data: valid backtransformation of the LSD valiue

  1. #1
    Newbie
    Joined
    Oct 2012
    From
    Vienna
    Posts
    2

    Fisher's LSD test on transformed data: valid backtransformation of the LSD valiue

    Hi all,

    To achieve a normal distribution of my experimental data, I did a transformation using the formula: y = square root (x+0,5). On the transformed data I then performed a Fisher's LSD test, giving me the LSD value for the transformed data. Simple backtransformation of that LSD value using the inverse formula x = y^2-0,5 does not seem to work (the backtransfromed LSD is way too low to make any sense at all). I could not find any info on the internet, only that backtransformation of the LSD value using the inverse formula is not valid.

    Anyone knows how to do it?

    Thanks a lot.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,826
    Thanks
    669

    Re: Fisher's LSD test on transformed data: valid backtransformation of the LSD valiue

    Hey jakkoklaak.

    Can you describe exactly what you are trying to do both in terms of the ends and means for your problem?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2012
    From
    Vienna
    Posts
    2

    Re: Fisher's LSD test on transformed data: valid backtransformation of the LSD valiue

    hi chiro,

    ok. it's an agricultural experiment: to control a disease called fusarium head blight, a wheat cultivar was treated with 13 different treatment variants in 5 replications. the disease scores are the data for statistical analysis. However, the data was not normally distributed (a requirement for ANOVA and Fisher's LSD test) and therefore it was previously transformed using the formula y = square root (x+0,5), resulting in a normal distribution . Then, in order to find out what treatment variants produced different disease scores, first an ANOVA and then a Fisher's LSD test was performed (using SAS, glm procedure) on the transformed data. Now I have the treatment variants grouped according to their control efficiency and an LSD value for the transformed data. To present these results, I have to back-transform the data (makes it much easier to interprete in terms of disease control efficiency). Back-transformation is done using the inverse formula x = y^2-0,5. I also want to show the corresponding LSD value, so I backtransform it as well, with the same formula. Here however, I get a backtransformed LSD value that cannot be correct, it is way to low.
    Numerical example:
    transformed treatment mean of the first treatment is 4,9. Backtransformation gives me 4,9^2-0,5 = 23,51
    LSD of transformed data is 0,6. Backtransformation gives -0,14

    a negative LSD value can of course not be interpreted, so I guess I'm missing something.

    I hope this helps you in helping me ^^. thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,826
    Thanks
    669

    Re: Fisher's LSD test on transformed data: valid backtransformation of the LSD valiue

    One thing about invariance principles of estimates (i.e. point estimates) for any statistic is that you don't always get the freedom of transforming the estimates by a function.

    One case where it is gauranteed is when you use a Maximum Likelihood Estimate or the MLE estimator as it's commonly known as.

    I'm not sure what technique the LSD is based on, but if it is based on a different scheme other than MLE then you can't just assume that transformations in general will preserve the nature of the estimate under that transformation.

    I took a look at the Wiki page and it references a 'q' (It might refer to a general quartile) but if you can point out how the LSD is calculated and what estimator is used then that would be appreciated.

    You said you used the GLM procedure so I'm guessing if you tried to estimate means with a linear least squares approach that you would be using the least squares estimator rather than the MLE.

    I think there is a way to do MLE estimation on least squares and this gives a good outline of the ideas behind certain classes of linear models:

    http://users.polisci.wisc.edu/frankl...ELec02p4up.pdf

    This is one reason why MLE estimators are useful and are preferred when you have crazy transformations.

    You might also want to look at non-parametric tests like those using medians including Moods test, Mann-Whitney, Friedman and others if you don't meet the assumptions of a standard vanilla ANOVA: just check whether your data meets the assumptions and also whether the context of your experiment and your goals also make sense relative to these assumptions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Test for normality - non-normal data ?
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: September 17th 2012, 11:34 PM
  2. Looking for some statistical data for Lilliefors test
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 17th 2011, 09:58 AM
  3. Replies: 2
    Last Post: December 14th 2010, 11:06 AM
  4. Fisher's Exact Test
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: February 9th 2010, 02:44 PM
  5. error bars in log-transformed data
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 5th 2009, 05:54 AM

Search Tags


/mathhelpforum @mathhelpforum