MultiLinear regression analysis - have answer just need someone to verify

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ajayhp
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Can you please help me with the following question.

I have obtained the following values of

\begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix}= \begin{bmatrix} 8.04 \\ 0.62 \end{bmatrix}

for b i)
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ghostwalker
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(Original post by ajayhp)
Can you please help me with the following question.

I have obtained the following values of

\begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix}= \begin{bmatrix} 8.04 \\ 0.62 \end{bmatrix}

for b i)
Just ran the data through LINEST on Excel, and I get 7.977777778 0.546153846 respectively for the two parameters.

Assuming it's the standard least squares regression, we appear to have a discrepancy.
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ajayhp
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(Original post by ghostwalker)
Just ran the data through LINEST on Excel, and I get 7.977777778 0.546153846 respectively for the two parameters.

Assuming it's the standard least squares regression, we appear to have a discrepancy.
Hi ghostwalker.

I have also run it in SAS and it appears to give your answer.

How would i go about answering the question then? Any Ideas?

Thanks for your reply btw.
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ghostwalker
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(Original post by ajayhp)
Hi ghostwalker.

I have also run it in SAS and it appears to give your answer.

How would i go about answering the question then? Any Ideas?

Thanks for your reply btw.
I'm not well up on the theory, and have no idea what an X matrix is in the context, nor how you got the values you posted, or what part a) of the question was.

If you can elaborate I may be able to help, but can't guarantee it.
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ajayhp
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For a generalised Linear model:

Y = X\hat{\beta} + e_{ij}

Where
\hat{\beta}=\begin{bmatrix} \hat{\beta_0} \\\hat{\beta_1} \end{bmatrix}

The X matrix is as folows:

 \begin{bmatrix} 1 & -2 \\ 1 & -1 \\ 1 &  0 \\ 1 & 1 \\ 1 &  2 \end{bmatrix}

We also have:

 \hat{\beta} = (X^{T}X)^{-1} X^{T}y

what would Y be in this case


(Original post by ghostwalker)
I'm not well up on the theory, and have no idea what an X matrix is in the context, nor how you got the values you posted, or what part a) of the question was.

If you can elaborate I may be able to help, but can't guarantee it.
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ajayhp
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Since Y has some repeated values maybe i can remake the whole of the matrix X

  X = \begin{bmatrix} 1 & -2 \\ 1 & -2 \\ 1 &  -2 \\ 1 & -1 \\ 1 &  -1 \\ 1 & -1 \\ 1 & 0 \\ 1 & 0 \\ 1 & 0 \\ 1 & 1 \\ 1 & 1 \\ 1 & 1 \\ 1 & -2 \\ 1 & -2 \\ 1 &  -2  \end{bmatrix}

so then Y would be:

 Y = \begin{bmatrix} 6.9 \\ 6.6 \\ 7.2 \\ 7.1 \\  0 \\  0 \\ 8.3 \\ 0 \\ 0 \\ 8.7 \\ 0 \\ 0 \\ 9.2 \\ 8.5 \\ 9.3 \end{bmatrix}
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ghostwalker
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(Original post by ajayhp)
...
I'm guessing there, and can't find anything in the literature available to me.

But since y is actually the Y matrix of values, I suspect you need to have all 9 values in the Y matrix, and add repeats to the X matrix as necessary.

As per your last post, but without the zero entries, so 9 columns in each.

But I am guessing!
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