# Homework Help: Linear models

1. May 22, 2010

### squenshl

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] = $$\beta$$1 + $$\beta$$2xi + $$\beta$$3zi where zi = 3xi2 - 4.4875 for i = 1,...,4.
Suppose that the observations (Y1, Y2, Y3, Y4) are independent with common variance $$\sigma$$2
How do I find the design matrix X and hence write the model in the form E(Y) = X$$\beta$$
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$$\beta$$ where $$\beta$$ = ($$\beta$$i)T

Last edited: May 22, 2010
2. May 23, 2010

### squenshl

I tried it again and got the same but I'm not still not sure.

3. May 23, 2010

### Tedjn

Yes, this seems right if you are doing ordinary least squares.

4. May 24, 2010

### squenshl

But what happens to (Y1, Y2, Y3, Y4), do we use those to find the residuals, fitted values and leverages.

5. May 24, 2010

To find the residuals you need to find the estimates of $\beta$ first. Then

a) fitted values are $X \hat{\beta}$

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

6. May 24, 2010

### squenshl

Is that $$\beta$$ = (XTX)-1XTY. I'm still not sure that my design matrix is right.

Last edited: May 24, 2010
7. May 24, 2010

### Tedjn

Yes, I think that's right.

8. May 24, 2010

### squenshl

Yes it is. Because my Hat matrix H is idempotent (H2 = H) and the trace of H = p = 3.

9. May 24, 2010

### squenshl

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.

10. May 25, 2010

### Tedjn

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.

11. May 25, 2010

### squenshl

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?

12. May 25, 2010

### Tedjn

Now multiply (XTX)-1 by XTY to find your beta hat vector. The dimensions should work out.

13. May 25, 2010

### squenshl

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
14. May 26, 2010

### squenshl

Never mind. Got it. The fitted values are
(5.023
7.409
11.580
17.534)
and the residuals follow on from that.