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Randomized Complete Block Design - Scheffe Multiple Comparison

  1. Apr 12, 2015 #1
    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:

    [tex]y_{hij} = \mu + \theta_h + \alpha_i + \beta_j + (\alpha\beta)_{ij} + \epsilon_{hij}[/tex]
    where [itex]\theta_h[/itex] refers to the [itex]h^{th}[/itex] block effect and [itex]\alpha_i[/itex] refers to the effect of factor A, [itex]\beta_j[/itex] refers to the effect of factor B and [itex](\alpha\beta)_{ij}[/itex] 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 [itex]2\sqrt{(3−1)F_{3−1,8(b−1),0.05}} \sqrt{MSE}[/itex]

    Where [itex]8(b−1)[/itex] 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 [itex]F_{2,8(b−1),0.05} \leq \frac{25}{670∗2}=0.0186567[/itex]

    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. jcsd
  3. Apr 13, 2015 #2

    RUber

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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?
     
    Last edited: Apr 13, 2015
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