1. Feb 11, 2013

### mintsharpie

Hello, I just have a quick question understanding multiple comparisons and I'd appreciate any help because I'm on the verge of failing :uhh: I'm reviewing for a test, and reading over questions and their corresponding answers in the back of the textbook. The question I'm stuck on deals with a psychologist testing the claim made by a drug company that a drug would help patients.

To do this, they selected 20 patients from their hospital, and randomly assigned them to one of four groups - group 1 receiving the new drug, group 2 receiving a different drug, group 3 receiving a different drug, and group 4 as the control group. Here is the answer given in the textbook:

I understand the first part, and how SSE is calculated and everything, but the second table with the contrasts totally baffles me. I have no idea how that table was filled in, or how I would be able to fill it in on a test if I had a different example. How were those 1s, -1s, and 2s determined? What do they mean and how were they calculated? In addition, once I go on to the appropriate post hoc test - in this case, Dunn - how do I utilise the table in terms of critical values?

2. Feb 11, 2013

### Stephen Tashi

I'm not an expert on this, but I was curious enough about the table to look up "Dunn Test" and found: http://www.google.com/url?sa=t&rct=...sg=AFQjCNENKSsWfC1UN4CdwHObbCR7NAkXpw&cad=rja

In the slide "What is a contrast anyway", it is explained that a contrast is a linear combination of means. For example, if the hypothesis $\mu_A = \mu_B$ is true then the contrast $1 \mu_A + (-1) \mu_B = 0$. So I suspect the table gives you the coefficients (such as 1 and -1) that are involved in the contrast being tested. I don't know anything else about the subject.

3. Feb 11, 2013

### mintsharpie

I'm so confused, and I don't understand this at all

4. Feb 13, 2013

### ssd

This is an "analysis of one way classified data".
In design of experiments you see it as CRD (completely randomized design).
The analysis is very straight forward.

Xi = ith observation
Ti= i th row total, G = sum of Ti. Then SS(T)= Sum(Ti*Ti/Ni)-cf, has df=4-1=3
where, cf= (G*G/N), Ni= values in i th row, N = sum of Ni.
SST= Sum(Xi*Xi)-cf, has df= 20-1=19.
SSE= SST-SS(T), has df=19-3=16.
F= MS(T)/MSE ~ F (3,16) (=> F distribution with 3,16 df)
MS(T)=SS(T)/3, MSE=SSE/16. Critical region:F> F(a,3,16). a=0.05 or 0.01 as you choose. Find F(a,3,16)
from Biometrica tables.