= Solution
The command fits independent grouped <binomial distributions> $Y_j\sim\operatorname{Bin}(N_j,p_j)$, using the supplied totals as trial counts. The <logistic regression> has the additive <linear predictor>
$$
\widehat\eta=-3.06073-1.27079\,\mathbf1_{\{\mathrm{Kid}\}}-0.37211\,\mathbf1_{\{\mathrm M\}},\qquad
\boxed{\widehat p=\frac{e^{\widehat\eta}}{1+e^{\widehat\eta}}.}
$$
The reference category is adult females. The age and gender coefficients are conditional <log odds ratios>: the fitted odds for a child are multiplied by $e^{-1.27079}\approx0.281$, and those for a male by $e^{-0.37211}\approx0.689$.
The coefficients are <maximum-likelihood estimates>, found by <Newton's method> or <iteratively reweighted least squares>. With <design matrix> $X$ and $W_{jj}=N_j\widehat p_j(1-\widehat p_j)$, the estimated <covariance matrix> is $(X^TWX)^{-1}$; the printed <standard errors> are square roots of its diagonal entries. Each <Wald test> compares a coefficient with zero using its estimate divided by its <standard error>, approximately a standard <normal distribution> under the null. The age and gender effects are strongly significant. The intercept test instead tests whether the reference probability is $1/2$.
The fitted probabilities are about $0.04476$ for adult females, $0.03128$ for adult males, $0.01298$ for female children, and $0.008981$ for male children. \b[Adult females have the largest fitted probability], about \b[4.48%]. The residual <deviance> is $0.06514$ on $4-3=1$ <degree of freedom>; comparison with a <chi-squared distribution> gives $p\approx0.799$. There is no evidence here that an age–gender <interaction> is needed, under the assumed independent <binomial distribution> model.
Back to article page
