OK, question edited in response to comment from @user158565 for those not in possession of the Jackman book. I tried to abstract the relevant information below.
I'm struggling with Simon Jackman's example 2.12 on p82 of his 2009 book Bayesian Analysis for the Social Sciences, specifically what happens with the value of $n$ (from proposition 2.4 on p81) in the calculation $\theta^*$ and $\sigma^{2*}$.
For example in the calculation of $\theta^*$ below there are two terms in the left parentheses $\frac{.491}{.000484}$ and $\frac{.548}{.000486}$ the second of which corresponds to $\bar{y}\frac{n}{\sigma^2}$ in Proposition 2.4 on p81. Since $\bar{y}=.548$ and $\sigma^2=.000486$ then $n=1$. However, Jackman gives $n=\alpha+\beta$ where $\alpha=279$ and $\beta=230$. If anyone could help me understand what's going on here I'd appreciate. Some more details:
The example is about voting intentions for one of two candidates with $\theta$, the probability of voting for one, given by $\theta \sim Beta(\alpha,\beta)$. The exercise is to approximate this Beta distribution with a Normal distribution $\theta \sim N(\tilde{\theta},\sigma^2)$ with
$$\sigma^2=V(\theta; \alpha, \beta)=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$$
known and
$$\tilde{\theta}=E(\theta; \alpha,\beta)=\frac{\alpha}{\alpha+\beta}$$
to be estimated given a prior $\theta\sim N(.491,.000484)$ and data $\alpha=279$ and $\beta=230$ giving $\tilde{\theta}=.548$ and $\sigma^2=.000486$
The text now reads "the poll result $\tilde{\theta}$ is observed data, generated by a sampling process governed by $\theta$ and the poll's sample size (n.b. $n=\alpha+\beta$). Specifically, using the normal approximation, $\tilde{\theta} \sim N(\theta,\sigma^2)$ or $.548 \sim N(\theta, .000486)$. We are now in a position to use the result in Proposition 2.4".
Proposition 2.4 gives:
$$\mu \vert\mathbf{y} \sim N \left(\frac{\mu_0\sigma_0^{-2}+\bar{y}\frac{n}{\sigma^2}}{\sigma_0^{-2}+\frac{n}{\sigma^2}},\left(\sigma_0^{-2}+\frac{n}{\sigma^2}\right)^{-1}\right)$$
where the subscript 0 indicates prior information. The example continues giving the posterior estimate for $\theta$:
$$\theta^*=\left(\frac{.491}{.000484}+\frac{.548}{.000486}\right)\left(\frac{1}{.000484}+\frac{1}{.000486}\right)^{-1}=.519$$