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In article <5ah83k$sb6@boursy.news.erols.com>, cglijoi@erols.com says...
>
>I want to track results of a survey. Problem is that last year's survey 
>was on a five point scale and this year's is on a seven point scale. What 
>is the best way to convert from one form to another for comparison? I am 
>an SPSS user.
>
>Thanks for the interest,
>
>Claude Lijoi
>cglijoi@erols.com
>
I've wrestled with this one before.  With different distributions and variances
your choices are practically nil.  Even if your variances are equal
across years, if the shape of the distributions are not very normal (Gaussian),
I don't think you can standardize each year's data to some mean
and that variance.
I'm guessing it isn't feasible to do a one-time survey this year using a
five-point scale on a different, but similar sample.  This would give you
a starting place for converting things from this year to last year format,
but it is still iffy, and what about next year's survey if there is one?
If your data for this year is skewed toward one end of the scale or the other,
you could possibly chop off two points of the other tale, but you can't 
really state statistical confidence levels to anyone, and you'd have to
limit your numbers to percents rather than means.
I'd say chalk it up to experience and live without comparing this year's
results to last year's.
-- 
Kurt Salmela              .--------------------------.
DiffSimilar Analytics    | Marketing Data Analysis,   |
3399 Kent Street #310  
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1.      Mann-Whitney assumes distributions in 2 groups IDENTICAL except for
a shift in mean.
so, skew and variance should be SAME (or similar) in both groups.
2.      you vould try transforming to normality, try log
        then do the t-test.
3.      BUT you really want an odds ratio test since what you are
interested in is the probability for the cancer given some particular
cut-off value of the marker.
odds ratio = p(cancer|marker>crit value)/p(cancer|marker Tony Mak (tonymak@sco1.med.cuhk.edu.hk) wrote:
>
> : I need to compare a parameter in two independent groups(a new
> : 'cancer-marker' in patients with and without cancer).  I have 29 and 31
> : patients in each group.  The parameter is skewed to the right in both
> : groups.  I used Mann-Whitney U test, the p value is 0.010.   I have
> : submitted the paper to a scientific journal, one of the referree commented
> : that I should be able to use t-test because of the sample size.
>
> : My question is : I understand that the parametric test is more powerful,
> : but how can I tell if the distribution is suitable for the t-test or not?
>
>
>  -- Is the distribution suitable for a t-test?  If you should be
> comfortable in using the *means*  to describe the outcomes, and to
> compare them, then the t-test is proper and appropriate.  If the
> means do not give a good representation, then it is hard for mere
> sample size to compensate, even though the t-test is rather robust
> (especially with equal n).  If you considered the means useful to
> present, then there is implicit support for the use of a t-test.
>
> What might be ideal would be to perform the appropriate transformation,
> (if it is usual, for instance, that the biological parameter should be
> looked at as the log of what is measured, and that produces a symmetric
> distribution,)  and then carry out a t-test.
>
>
> : (I used ROC curve analysis to test the usefulness of the marker also)
>
>  -- You might try to explain to the referee that a rank-order test
> was used because (inexplicable?) outliers make these means unreliable,
> regardless of the sample size, and so the ranks were considered to be
> appropriate for the testing.  If what you cited was the totality of the
> reasoning of the referee, then you may hope that the editor realizes this
> possibility, that a referee is not always a good source on statistics.
>
>
>
> Rich Ulrich, biostatistician                wpilib+@pitt.edu
> http://www.pitt.edu/~wpilib/index.html   Univ. of Pittsburgh
>