How to complete a Scheffe's test by hand?

Disclaimer: I’ve never used Scheffe’s test for my post-hocs, so this is the first time I’m looking at it.

You might find this document useful. It gives an ANOVA output (from a test done in SPSS), and uses those values to do various post hocs, including Scheffe’s. Perhaps it will help you figure out which bits of your output you need to use.

@glacial Hmmm I saw that sheet and I was still left at a loss. Maybe a better question is how to find Sum of Squares contrast, and what the variables in that formula stand for, and how they can be found.

Well, in that example, they’re using MSerror, which is just SSerror/df. There’s nothing in the original ANOVA table that’s specifically linked to contrasts, and yet they’re able to calculate the f value. I’ll dig a little more and see what I can find.

http://books.google.com/books?id=xWENVdl6D0YC&lpg=PA405&ots=pYVXiYPRYC&dq=explaining%20psychological%20statistics%20sscontrast&pg=PA405#v=onepage&q=explaining%20psychological%20statistics%20sscontrast&f=false It’s formula 13.14 that I need to use that I’m having trouble with

Ok, from looking at your textbook, are you having difficulty with the coefficients in the formula for L? Because, I don’t think it’s very well explained in this section (and I can’t see large portions of the rest of the text)... so I can see how it might cause confusion. When I studied contrasts (in a different context), the coefficients came from decisions about which means to compare with which other means.

So, as an example, if your model includes mass of juvenile, yearling, and adult deer (so three different means), and your hypothesis is that juveniles and adults do not differ from each other, then the yearling mass must be given a coefficient of 0, and the juvenile and adult must be given 1 and -1 respectively, so that they are in contrast with each other:

phi = (1) mu(juveniles) + (0) mu(yearlings) + (-1) mu(adults)

The sum is zero, and juveniles and adults will be compared to each other.

If, on the other hand, your hypothesis is that adults and yearlings do not differ from juveniles, then the coefficient for juveniles must contrast those of adults + yearlings, so you end up with:

phi = (1) mu(juveniles) + (-0.5) mu(yearlings) + (-0.5) mu(adults)

This is very like the example given in your text.

These coefficients are your ci values. I think the rest should be easier to figure out once you have this down. From here, you can calculate SScontrast, then divide by MSw to get your F value.

Thank you this is exactly what I’ve been struggling with. One last question, if you don’t mind, @glacial . For my homework, I’m doing a problem with four groups, but I’m comparing one group to the other three. Would I weight the other three using fractions/decimals then, to get it to equal zero? (the one group is smaller than the other three, which are approximately the same)

@flash74686 That would be my approach. The one group would get (1) and each of the others would be (-⅓). Or the opposite signs. Now let’s hope I’m right. :P

It looks right to me, based on other stuff I’ve seen! Thank you so much, you’re an absolute life-saver!

You’re welcome. :)