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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2024

Paper 1

Let $X$ be the $173\times3$ design <matrix> and write $p=3$. The displayed estimate in row $j$ is
$$
\widehat\beta_j,\qquad
\widehat\beta=(X^TX)^{-1}X^TY.
$$
With
$$
s^2=\frac{\operatorname{RSS}}{n-p}
=\frac{\|Y-X\widehat\beta\|^2}{170},
$$
its standard error is
$$
\operatorname{se}(\widehat\beta_j)
=s\sqrt{[(X^TX)^{-1}]_{jj}}.
$$
The $t$-value and two-sided $p$-value are
$$
t_j=\frac{\widehat\beta_j}{\operatorname{se}(\widehat\beta_j)},
\qquad
2\mathbb P\{|T_{170}|\geq|t_j|\},
$$
respectively.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 13L / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 13L / b / Solution

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Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 13L / b / Solution

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