# Randomized Complete Block Design - Scheffe Multiple Comparison

1. Apr 12, 2015

### MattMurdock

1. The problem statement, all variables and given/known data

I'm working on a problem in Design and Analysis of Experiments by Dean and Voss. It's Chapter 10 question 11 part c.

We have an experiment with 2 treatment factors (each with three levels) and 1 blocking factor (with four levels. It's a randomized complete block design so only one experimental unit per treatment/block combination:

$$y_{hij} = \mu + \theta_h + \alpha_i + \beta_j + (\alpha\beta)_{ij} + \epsilon_{hij}$$
where $\theta_h$ refers to the $h^{th}$ block effect and $\alpha_i$ refers to the effect of factor A, $\beta_j$ refers to the effect of factor B and $(\alpha\beta)_{ij}$ refers to the interaction between the 2.

The experimenters want Scheffe 95% confidence intervals for normalized contrasts in the main effects of each factor to be no wider than 10. A pilot experiment was run to give an estimate for MSE equal to 670. How many subjects are needed?

2. Relevant equations

3. The attempt at a solution
The width of a Scheffe interval is defined to be $2\sqrt{(3−1)F_{3−1,8(b−1),0.05}} \sqrt{MSE}$

Where $8(b−1)$ is the degrees of freedom of SSE and MSE is the variance of the contrasts since they are normalized.

If this is to be less than or equal to 10 then $F_{2,8(b−1),0.05} \leq \frac{25}{670∗2}=0.0186567$

My problem is that I don't think I can possibly make this inequality work because the F critical values never go below 3 for a numerator degree of freedom equal to 2. I was wondering if anyone could shed some light on this for me. Am I making a mistake or is there something wrong with the question? Thank you

Last edited: Apr 12, 2015
2. Apr 13, 2015

### RUber

[EDIT]
I was wrong with my first post...
Your algebra was right...it is the MSE that will change.
You were given that the width is
$2\sqrt{(3−1)F _{3−1,8(b−1),0.05}} \sqrt{ MSE} < 10$ .
Rearranging and squaring gives: $(3−1)F _{3−1,8(b−1),0.05}< \frac{25}{MSE}$.
So you want to find: $F _{3−1,8(b−1),0.05}< \frac{25}{2 MSE}$.
Remember that your MSE for these purposes will be affected by the sample size...where does that fit in?

Last edited: Apr 13, 2015