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Linear models

  1. May 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose (Y1, Y2, Y3, Y4) = (5.2, 6.8, 11.9, 17.0) are the average yields (in tonne/ha) of potato grown in soil with 4 different levels of superphosphate fertiliser, x1 = 1.20, x2 = 1.75, x3 = 2.30, x4 = 2.85. We want to fit the model E[Yi] = [tex]\beta[/tex]1 + [tex]\beta[/tex]2xi + [tex]\beta[/tex]3zi where zi = 3xi2 - 4.4875 for i = 1,...,4.
    Suppose that the observations (Y1, Y2, Y3, Y4) are independent with common variance [tex]\sigma[/tex]2
    How do I find the design matrix X and hence write the model in the form E(Y) = X[tex]\beta[/tex]
    2. Relevant equations



    3. The attempt at a solution
    I found z1, z2, z3, z4 using x1, x2, x3, x4 to get z1 = -0.1675, z2 = 4.70, z3 = 11.3825, z4 = 19.88 so from E(Yi) do I get
    X =
    (1 1.20 -0.1675
    1 1.75 4.70
    1 2.30 11.3825
    1 2.85 19.88)
    hence E(Y) = X[tex]\beta[/tex] where [tex]\beta[/tex] = ([tex]\beta[/tex]i)T
     
    Last edited: May 22, 2010
  2. jcsd
  3. May 23, 2010 #2
    I tried it again and got the same but I'm not still not sure.
     
  4. May 23, 2010 #3
    Yes, this seems right if you are doing ordinary least squares.
     
  5. May 24, 2010 #4
    But what happens to (Y1, Y2, Y3, Y4), do we use those to find the residuals, fitted values and leverages.
     
  6. May 24, 2010 #5

    statdad

    User Avatar
    Homework Helper

    To find the residuals you need to find the estimates of [itex] \beta [/itex] first. Then

    a) fitted values are [itex] X \hat{\beta} [/itex]

    b) residuals are original y - fitted values

    c) your text (or your notes) will explain how to get the leverage values (if you use software (such as R or S+, for two examples) you can get these - everything you need, actually - from there
     
  7. May 24, 2010 #6
    Is that [tex]\beta[/tex] = (XTX)-1XTY. I'm still not sure that my design matrix is right.
     
    Last edited: May 24, 2010
  8. May 24, 2010 #7
    Yes, I think that's right.
     
  9. May 24, 2010 #8
    Yes it is. Because my Hat matrix H is idempotent (H2 = H) and the trace of H = p = 3.
     
  10. May 24, 2010 #9
    I have found my leverages and fitted values (my fitted values are a 4x4 matrix) but when it comes to finding the residuals r = Y - Y(hat) i get a 4x1 matrix - a 4x4 matrix and that is impossible.
     
  11. May 25, 2010 #10
    How did you get your fitted values to be 4x4? Your beta hat is a column vector, since Y is a column vector, and then X*beta hat is a column vector.
     
  12. May 25, 2010 #11
    I think I did my matrix multiplication wrong for beta hat. I get a 3x3 matrix for (XTX)-1 which is
    (77.3577 -59.0788 4.7523
    -59.0788 45.4737 -3.6883
    4.7523 -3.6883 0.3036).
    For my XTY i get the matrix
    (Y1 + Y2 + Y3 + Y4
    1.2Y1 + 1.75Y2 + 2.3Y3 + 2.85Y4
    -0.1675Y1 + 4.7Y2 + 11.3825Y3 + 19.88Y4).
    How do I get a column matrix from this?
     
  13. May 25, 2010 #12
    Now multiply (XTX)-1 by XTY to find your beta hat vector. The dimensions should work out.
     
  14. May 25, 2010 #13
    How do I multiply the 2 matrices. I've never seen that type of matrix multiplication before.
    Is it just (77.3577 - 59.0788 + 4.7523)/(Y1 + Y2 + Y3 + Y4) etc.
     
    Last edited: May 26, 2010
  15. May 26, 2010 #14
    Never mind. Got it. The fitted values are
    (5.023
    7.409
    11.580
    17.534)
    and the residuals follow on from that.
     
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