= Solution
Write $Y_i$ for the colony count and $d_i$ for the dose on plate $i$. The first fit is the <Poisson regression>
$$
Y_i\mathrel{\perp\!\!\!\perp}Y_j\quad(i\ne j),
\qquad
Y_i\sim\operatorname{Pois}(\mu_i),
\qquad
\log\mu_i=\beta_0+\beta_1d_i.
$$
Thus its <log-likelihood> is
$$
\ell(\beta_0,\beta_1)
=\sum_{i=1}^{18}\left[y_i(\beta_0+\beta_1d_i)
-e^{\beta_0+\beta_1d_i}-\log(y_i!)\right].
$$
The <Poisson deviance> relative to the saturated model is
$$
D=2\sum_{i=1}^{18}
\left\{y_i\log\frac{y_i}{\widehat\mu_i}
-(y_i-\widehat\mu_i)\right\},
$$
where a summand with $y_i=0$ uses $0\log0=0$.
At dose zero the fitted <expected value> is $e^{\widehat\beta_0}=e^{3.321995}\simeq27.72$ eradicated colonies. Increasing dose by one unit multiplies the fitted mean by $e^{\widehat\beta_1}=e^{0.0001901}\simeq1.000190$; for example, an increase of $100$ units multiplies it by about $1.0192$. The positive fitted effect is small and its displayed two-sided $p$-value, $0.105$, gives little evidence against a zero dose coefficient.
Back to article page
