For individual $i$, let $Y_i$ be the reported count, $g_i\in\{0,1\}$ the gender indicator, and $m_i\in\{0,1\}$ the minority indicator. The fitted <Poisson regression> is
$$
Y_i\mathrel{\perp\!\!\!\perp}Y_j,
\qquad
Y_i\sim\operatorname{Poisson}(\mu_i),
\qquad
\log\mu_i=\beta_0+\beta_1g_i+\beta_2m_i.
$$
Its <log-likelihood> is
$$
\ell(\beta)=\sum_{i=1}^{1308}
\{Y_i x_i^T\beta-e^{x_i^T\beta}-\log(Y_i!)\},
\qquad x_i=(1,g_i,m_i)^T.
$$
The <maximum-likelihood estimator> is
$$
(\widehat\beta_0,\widehat\beta_1,\widehat\beta_2)
=(-2.2959,-0.1916,1.7293).
$$
Holding minority status fixed, changing the gender indicator from zero to one multiplies the fitted <conditional expected value> by $e^{-0.1916}=0.826$. Thus the fitted mean count for men is about $17.4\%$ lower than that for women with the same minority status.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 218 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-218/1/a/solution