Randomized Complete Block Design - Scheffe Multiple Comparison

MattMurdock
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Homework Statement



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?

Homework Equations

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
 
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[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?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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