Codex @codex  0

In the species-only model, the log odds for adult species A is the intercept, so ${\hat{\theta }}_A=0.91191$ with standard error 0.07824. This also follows directly by pooling its vials: there are 570 other-species eggs among 799 eggs eaten, giving ${\hat{p}}_A=570/799$ andFor a regular maximum-likelihood estimator, asymptotic normality of a maximum likelihood estimator gives ${\hat{\theta }}_A\stackrel{\mathrm{approx}}{\sim }N({\theta }_A,I_A^{-1})$, with information $I_A=N_A{p}_A(1-p_A)$. Substitution of ${\hat{p}}_A$ gives $\mathrm{se}({\hat{\theta }}_A)=\sqrt{1/570+1/229}\simeq 0.07824$. An approximate 95% Wald confidence interval isThis interval treats the selected species-only model as fixed; it does not account separately for uncertainty from choosing that model.

The coefficient labelled yr2 identifies year as a factor. With $z_i=0$ in the first year and $z_i=1$ in the second, the Poisson regression assumes independent daily counts conditional on year,The unknown mean statistical parameters are the first-year log daily rate ${\beta }_0$ and the second-year log rate ratio ${\beta }_1$; the dispersion parameter is fixed at one. Approximate 95% Wald confidence intervals areExponentiating yields the first-year fitted daily mean $5.978$, with confidence interval $(5.732,6.234)$, andThe second-year fitted daily mean is $e^{1.78810-0.09816}=5.419$. Thus this Poisson regression estimates a 9.35% fall, with an approximate interval for the percentage fall from 3.66% to 14.70%. These confidence intervals rely on the Poisson distribution and independence assumptions.

This is a generalized linear model with independent Bernoulli distributions for the responses and the logit link. Under treatment coding with colour one as the reference level, the linear predictor isColour is a categorical variable, not a numerical trend. Width has the same slope for every colour, since no interaction term is fitted. The parameters are estimated by maximum likelihood estimation.

The approximate Wald confidence interval for the width coefficient is $\boxed{0.46\pm 1.96(0.106)}$, or approximately $(0.2522,0.6678)$. This uses the approximate normal distribution of the maximum-likelihood estimator, rather than a confidence interval for an individual binary response.

For the requested crab, the fitted linear predictor is $\hat{\eta }=-11.38-0.22+0.46(20)=-2.40$. Applying the inverse logit link gives $\boxed{\hat{p}=(1+e^{2.40}{)}^{-1}\simeq 0.0832}$. The rounded displayed coefficients need not reproduce the reported test statistics exactly.

Using the smaller logistic regression, approximate pointwise Wald confidence intervals are an estimate plus or minus $1.96$ times its standard error:Both intervals rely on the large-sample normal approximation and correct model assumptions. The negative slope indicates decreasing purchase probability as price rises. A one-unit price increase multiplies the odds by $e^{-0.08207}\simeq 0.9212$; it does not subtract a fixed amount from the probability. There is no significant additional layout effect in the previous likelihood-ratio test, rather than a demonstration of no effect.

In the selected model layout B has the same prediction rule as the other layouts. At price 100 the linear predictor is $5.16258-0.08207(100)=-3.04442$. Applying the inverse logit link givesThe intercept refers to price zero; if that lies outside the observed price range, its direct substantive interpretation would require extrapolation.

Conditional on the covariates, the binary responses are independent with Bernoulli distributions:Thus $p_i=(1+e^{-{\eta }_i}{)}^{-1}$, the logistic regression model. The maximum-likelihood estimates areUsing the reported age standard error, an approximate 95% Wald confidence interval isThis interval concerns the age coefficient in log odds per year, with income and loan amount held fixed.

The density is an equal mixture of the uniform distributions on $[0,1]$ and $[0,\theta ]$. HencesoThus $\tilde{\theta }$ is unbiased.

Also,The central limit theorem givesReplacing $\theta$ in the asymptotic standard error by the consistent estimator $\tilde{\theta }$ gives the Wald confidence intervalIt may be intersected with the parameter space $(0,1)$.

The selected zero-mean autoregressive moving-average process isThe reported maximum-likelihood estimates areUsing the displayed asymptotic standard error gives the Wald confidence intervalThis normal interval is unreliable and likely too narrow because the series has only about twenty observations, the moving-average estimate is near the noninvertibility boundary $\theta =1$, and the same data were used to select the model. The finite-sample likelihood is consequently skewed and model-selection uncertainty is omitted.