= Solution
For an <exponential dispersion family>, differentiating the normalization integral with respect to $\theta$ gives an expected score of zero. The score is $(Y-b'(\theta))/\phi$, hence $E(Y)=b'(\theta)$. Differentiating again, the expected squared score is $b''(\theta)/\phi$; it is also $\operatorname{Var}(Y)/\phi^2$. Thus
$$
\boxed{\mu_i=b'(\theta_i),\qquad\operatorname{Var}(Y_i)=\phi b''(\theta_i).}
$$
The <canonical link function> sets the <linear predictor> equal to the natural parameter: $g(\mu)=\theta=(b')^{-1}(\mu)$. The <variance function> is $V(\mu)=b''((b')^{-1}(\mu))$, so $\operatorname{Var}(Y_i)=\phi V(\mu_i)$; the <dispersion parameter> $\phi$ is separated from the mean-dependent factor.
The <saturated statistical model> has a separate mean for each response and fits $\widetilde\mu_i=y_i$, with boundary parameters interpreted by limits. At the same fixed dispersion, its log <likelihood> and the fitted model's <log-likelihood> define the <scaled deviance>
$$
D^*=2\{\ell(\widetilde\theta;\phi)-\ell(\widehat\theta;\phi)\}.
$$
The unscaled <deviance> is
$$
D=\phi D^*=2\sum_i\{y_i(\widetilde\theta_i-\widehat\theta_i)-b(\widetilde\theta_i)+b(\widehat\theta_i)\}.
$$
Keeping dispersion fixed in this definition is essential; it is not a comparison with a separately estimated zero <variance> in a Gaussian saturated fit.
For the <normal distribution>, expanding its density exponent gives
$$
-\frac{(y-\mu)^2}{2\sigma^2}-\frac12\log(2\pi\sigma^2)
=\frac{y\mu-\mu^2/2}{\sigma^2}-\frac{y^2}{2\sigma^2}-\frac12\log(2\pi\sigma^2).
$$
Consequently
$$
\boxed{\theta_i=\mu_i,\quad b(\theta)=\theta^2/2,\quad\phi=\sigma^2,\quad c(y,\phi)=-y^2/(2\phi)-\tfrac12\log(2\pi\phi).}
$$
Here $V(\mu)=1$ and the canonical link is the identity, so $\mu_i=\beta^Tx_i$. The saturated natural parameter is $\widetilde\theta_i=y_i$. Substitution into the <deviance> formula yields
$$
D=\sum_i(y_i-\widehat\mu_i)^2,\qquad
\boxed{D=\operatorname{RSS},\qquad D^*=\operatorname{RSS}/\sigma^2.}
$$
For the cholesterol fits, all formulae include an intercept. Model 1 has age and BMI with three coefficients; models 2 and 3 have one of these <covariates> with two coefficients; model 4 has only the intercept. Their residual <statistical degrees of freedom> are respectively $27,28,28,29$. Their reported deviances are <residual sums of squares>, because these are Gaussian linear models.
The relevant hypothesis is $H_0:\beta_{\mathrm{BMI}}=0$ in the full model, against $H_1:\beta_{\mathrm{BMI}}\ne0$, with age already present. Thus compare model 2 with model 1 using the <partial F-test>. Its extra sum of squares is $31.64-27.10=4.54$, and the common <variance> is estimated from model 1 by $\widehat\sigma^2=27.10/27$. Therefore
$$
F=\frac{31.64-27.10}{27.10/27}=4.52325\sim F_{1,27}\quad\text{under }H_0.
$$
Since $4.52325>4.210$, reject at $5\%$; the two-sided coefficient-test equivalent has $p\simeq0.0427$. \b[BMI is associated with cholesterol after adjustment for age], under the linear Gaussian model. The <RSS> values alone do not identify the sign of its coefficient or the size of the association.
For comparison, the marginal BMI test would compare model 4 with model 3, giving $F=13.88/(35.82/28)=10.8498$ on $(1,28)$ <statistical degrees of freedom>, with $p\simeq0.00268$. This does not answer the adjusted question and makes the BMI evidence appear stronger than the conditional test. Age also contributes after BMI: comparing model 3 with model 1 gives $F=8.72/(27.10/27)=8.6878$ on $(1,27)$ <statistical degrees of freedom>, with $p\simeq0.00653$. Thus retaining both predictors is supported at the stated level.
A raw <deviance> difference must not simply be compared with the supplied $\chi_1^2$ cutoff: Gaussian dispersion is unknown and the differences here are unscaled. If $\sigma^2$ were known, the scaled difference $4.54/\sigma^2$ would have a $\chi_1^2$ law. Estimating it from the full fit gives the exact finite-sample $F$ comparison above, with the appropriate denominator <statistical degrees of freedom>.
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