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Hi everyone,

I have a data set with two lists of hourly wages, one for each of two type of subjects. The subjects are divided in couples (1 type one + 1 type 2). I want to compute the average and median percentage difference between each couple of subjects, and its distribution. Can anyone help me do that? So far I only managed to load the data set and open it in a workfile. I only know how to estimate an equation and how to conduct a couple of test! :?

P.S.: I want to make this calculation because I think it could be representative of the average difference between the two types, but I could change it if anyone has a better idea.

Thank you :D ,

Fabio

Open the two series up as a group, then click on View->Tests of Equality

Thanks for the quick reply!

How can I do that programming?

FabioIM

edit: I have done it, but I am not sure how to interpret the results... :(

If I'm understanding your design, you have two series representing the coupled data, where the two values for a given observation are to be compared. If that's the case, then unfortunately the test Gareth gave you isn't going to do what you want since you want to compare the pairwise observations. The one that we provide compares the overall means and/or medians without accounting for the matched design.

I think that the easiest thing for you to do is to simply compute the difference by hand, and then test that. If you want the pct different between the two series, then just generate a new series. In fractional values, Then you can test whether the average or median of that is zero, as in

After performing it, I had the impression that it wasn't what I am looking for...
I didn't understand what you mean from "if you want the pct..." on. Could you be clearer? I am a beginner ;)
What does pct mean?

Thanks a lot!

FabioIM

Just saying that you can compute the difference between the two by taking the difference, or the pct. difference by taking the difference divided by a basis. You can obviously choose either as a basis. The examples I give are in fractional terms.

Got it,

Thanks!