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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.

For a fixed true parameter $0<p<1$, the usual asymptotic normality of maximum likelihood estimators givesEquivalently, for a large sample, $\hat{p}$ is approximately normally distributed with mean $p$ and variance $p^2(1-p)/(nr)$. Replacing the unknown variance by its consistent plug-in estimate gives the Wald confidence intervalHere $z_{1-\alpha /2}$ is a standard-normal quantile. Intersect with the physical parameter range $(0,1)$ if needed. This is an interior, large-sample approximation; the degenerate estimate $\hat{p}=1$ from $S=0$ does not justify a zero-width finite-sample interval.

Put $Q={\sum }_i{X}_i^2$. For $\theta >0$, the log-likelihood is $\ell (\theta )=\text{constant}-(n/2)\log \theta -Q/(2\theta )$. Its derivative is $-n/(2\theta )+Q/(2{\theta }^2)$, changing from positive to negative at $Q/n$. ThusThe known mean is zero, so no centring at the sample mean is needed. Taking the expected negative second derivative gives $I_n(\theta )=n/(2{\theta }^2)$. The large-sample variance is consequently $2{\theta }^2/n$. With $z_{1-\alpha /2}$ the indicated standard-normal quantile, a Wald confidence interval isThe interval may be restricted to the parameter space $\theta >0$. The parameter is the variance itself, so the standard error is proportional to $\hat{\theta }$, not to its square root.

Let $z={\Phi }^{-1}(1-\alpha /2)$ for the standard normal cumulative distribution function. Consistency of the maximum-likelihood estimator and Slutsky's theorem give the asymptotic standardizationIts probability of lying between $-z$ and $z$ tends to $1-\alpha$. Solving those inequalities for the unknown parameter yields the Wald confidence intervalThus the estimated standard error is $I_n(\hat{\theta }{)}^{-1/2}$. Its coverage is approximate and relies on the large-sample regularity assumptions.

Let $z=z_{1-\alpha /2}={\Phi }^{-1}(1-\alpha /2)$, where $0<\alpha <1$. By the preceding asymptotic normality and Slutsky's theorem, $\sqrt{n}(\hat{\theta }-\theta )/\hat{\theta }$ is approximately standard normal. Inverting $|\sqrt{n}(\hat{\theta }-\theta )/\hat{\theta }|\le z$ gives the Wald confidence intervalIts coverage approaches $1-\alpha$ as $n\to \infty$. The interpretation of the confidence interval is repeated-sample coverage of the fixed parameter; it does not assign a posterior probability to $\theta$.

For $S={\sum }_i{t}_i$, the likelihood is $L(\lambda )={\lambda }^n{e}^{-\lambda S}$, and its log derivative is $n/\lambda -S$. Thus, almost surely,The large-sample distribution is that of the estimator, rather than of the fixed parameter: $\sqrt{n}(\hat{\lambda }-\lambda )\stackrel{d}{\to }N(0,{\lambda }^2)$. A Wald confidence interval is consequentlyintersected with the positive parameter space if necessary. The unadorned parameter in the source's distribution request is interpreted as this estimator.

A single aphid survives to $T$ with probability $r=e^{-\lambda T}$, independently of the others. Hence $N_T\sim \mathrm{Bin}(n,r)$, andfor $0<n_T<n$. If $n_T=n$, the optimum is at the boundary $\lambda =0$; if $n_T=0$, the likelihood increases to its supremum as $\lambda \to \infty$, so there is no finite maximum likelihood estimate. Fisher information transforms from the binomial probability parameter asThe given strict inequality showsThresholding discards information about how long each survival time is. The second experiment therefore has a larger information bound on variance and larger asymptotic uncertainty for an efficient estimator. This is an expected-information comparison, rather than a claim that every realised finite-sample confidence interval must be wider.

For $S={\sum }_i{t}_i$, the likelihood is $L(\lambda )={\lambda }^n{e}^{-\lambda S}$, and its log derivative is $n/\lambda -S$. Thus, almost surely,The large-sample distribution is that of the estimator, rather than of the fixed parameter: $\sqrt{n}(\hat{\lambda }-\lambda )\stackrel{d}{\to }N(0,{\lambda }^2)$. A Wald confidence interval is consequentlyintersected with the positive parameter space if necessary. The unadorned parameter in the source's distribution request is interpreted as this estimator.

A single aphid survives to $T$ with probability $r=e^{-\lambda T}$, independently of the others. Hence $N_T\sim \mathrm{Bin}(n,r)$, andfor $0<n_T<n$. If $n_T=n$, the optimum is at the boundary $\lambda =0$; if $n_T=0$, the likelihood increases to its supremum as $\lambda \to \infty$, so there is no finite maximum likelihood estimate. Fisher information transforms from the binomial probability parameter asThe given strict inequality showsThresholding discards information about how long each survival time is. The second experiment therefore has a larger information bound on variance and larger asymptotic uncertainty for an efficient estimator. This is an expected-information comparison, rather than a claim that every realised finite-sample confidence interval must be wider.

For an interior true parameter and large $n$, standard asymptotic normality of a maximum likelihood estimator givesEstimate its standard error by $s_e=\hat{p}\sqrt{(1-\hat{p})/(nr)}$. A two-sided approximate Wald confidence interval is $(L,U)=(\hat{p}-z_{1-\alpha /2}{s}_e,\hat{p}+z_{1-\alpha /2}{s}_e)$, intersected with the parameter range $(0,1)$. The normal quantile is chosen so a standard normal lies between its positive and negative values with probability $1-\alpha$.

Put $Q={\sum }_i{X}_i^2$. For $\theta >0$, the log-likelihood is $\ell (\theta )=\text{constant}-(n/2)\log \theta -Q/(2\theta )$. Its derivative is $-n/(2\theta )+Q/(2{\theta }^2)$, changing from positive to negative at $Q/n$. ThusThe known mean is zero, so no centring at the sample mean is needed. Taking the expected negative second derivative gives $I_n(\theta )=n/(2{\theta }^2)$. The large-sample variance is consequently $2{\theta }^2/n$. With $z_{1-\alpha /2}$ the indicated standard-normal quantile, a Wald confidence interval isThe interval may be restricted to the parameter space $\theta >0$. The parameter is the variance itself, so the standard error is proportional to $\hat{\theta }$, not to its square root.

Let $z={\Phi }^{-1}(1-\alpha /2)$, where $\Phi$ is the standard normal distribution's cumulative distribution function. Statistical consistency and Slutsky's theorem allow replacement of $I_1(\theta )$ by $I_1(\hat{\theta })$:Thus the probability that this standardized estimation error lies between $-z$ and $z$ tends to $1-\alpha$. Rearranging gives the Wald confidence intervalThe coverage probability is approximate, with limiting value $1-\alpha$; it is a repeated-sampling statement about the random confidence interval.

Let $z=z_{1-\alpha /2}$ be the $(1-\alpha /2)$ standard normal quantile. Since $\hat{\mu }$ is a consistent estimator, Slutsky's theorem givesIts estimated standard error is $\hat{\mu }/\sqrt{3n}$, giving the approximate $(1-\alpha )$ Wald confidence intervalFor each fixed $\mu >0$, this procedure's coverage probability tends to $1-\alpha$. Intersecting with the positive parameter space preserves coverage of the true positive value.

The standard error suggested by part (iii) is $\theta /\sqrt{n}$. Substituting the consistent estimator $T_2$ gives $T_2/\sqrt{n}$; Slutsky's theorem givesUsing the central standard normal quantiles $\pm 1.96$, an approximate 95% Wald confidence interval isIts coverage probability approaches 0.95 for each fixed $\theta >0$ as $n\to \infty$. The intersection enforces positivity and does not change whether the true positive statistical parameter is covered. The value 1.64 would give the wrong central two-sided coverage probability.

The estimated standard error is ${\hat{\theta }}^{3/2}/\sqrt{n}$. Using the central 95% standard normal quantiles $\pm 1.96$, a Wald confidence interval isConsistency of $\hat{\theta }$ and Slutsky's theorem justify replacing the unknown scale ${\theta }^{3/2}$ by its estimate. The coverage probability is asymptotic, not exact at finite $n$. Intersecting with the positive parameter space removes impossible values without changing whether the true positive parameter is covered.

In a binary logistic regression, write the signed directional margin of observation $i$ as $(2y_i-1)x_i^{T}d$. Quasi-complete separation occurs when a nonzero direction $d$ makes all these margins nonnegative, with some zero and some strictly positive. Along coefficients $\beta +td$, $t\to \infty$, each observation's log-likelihood contribution is nondecreasing, and those with positive margins strictly increase toward their limiting value. Hence the unpenalized fit cannot have a finite maximum in that direction. A group containing only outcome zeros is a simple example: sending its indicator coefficient to $-\infty$ improves that group's likelihood while leaving all other observations unchanged. Zero fitted risk is then a boundary limit, and ordinary finite-coefficient Wald confidence intervals are unavailable.

A rate ratio compares event counts per unit exposure between two conditions. In a Poisson regression with a log link and binary indicator coefficient $\beta$, it equals $e^{\beta }$. Exponentiating an approximate Wald confidence interval for $\beta$ gives an interval for the rate ratio. A rate ratio is distinct from a relative risk or an odds ratio.

For $K\sim \mathrm{Bin}(n,p)$, $n>0$, put $\hat{p}=K/n$ and let $z$ be the $1-\alpha /2$ quantile of the standard normal distribution. Inverting the approximate score test retains precisely the parameters satisfyingMultiplication and collection of terms gives the quadratic inequalityIts two roots are the displayed endpoints. The Wilson score interval stays within $[0,1]$ and remains nontrivial when $K=0$ or $K=n$, unlike the unmodified Wald confidence interval. Its coverage probability is approximate and varies with $p$; it is not a uniformly exact 95% guarantee. Separate pointwise intervals for several groups are not a simultaneous confidence statement. The method is named after Edwin Bidwell Wilson.^[ref]