Fitting the Potato Yield Model with Superphosphate Fertiliser

[squenshl]

Homework Statement

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[Y_i] = $$\beta$$₁ + $$\beta$$₂x_i + $$\beta$$₃z_i where z_i = 3x_i² - 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$$\beta$$

Homework Equations

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(Y_i) 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

 

[squenshl]

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

 

[Tedjn]

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

 

[squenshl]

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

 

[statdad]

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

 

[squenshl]

Is that $$\beta$$ = (X^TX)^-1X^TY. I'm still not sure that my design matrix is right.

 

[Tedjn]

Yes, I think that's right.

 

[squenshl]

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

 

[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.

 

[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.

 

[squenshl]

I think I did my matrix multiplication wrong for beta hat. I get a 3x3 matrix for (X^TX)^-1 which is
(77.3577 -59.0788 4.7523
-59.0788 45.4737 -3.6883
4.7523 -3.6883 0.3036).
For my X^TY 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?

 

[Tedjn]

Now multiply (X^TX)^-1 by X^TY to find your beta hat vector. The dimensions should work out.

 

[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.

 

[squenshl]

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