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Let $s_i$ be speed and $j_i\in \{1,2,3,4\}$ the tool type. Each fit is a normal linear model with independent $N(0,{\sigma }^2)$ errors. The three mean specifications areThe constraints are treatment coding with type 1 as reference. The parameter counts are $2,5,8$, respectively. Model 2 gives parallel lines; model 3 allows interaction terms and distinct slopes.

In the analysis of variance interaction row, adding three independent slope differences costs three degrees of freedom. Its extra sum of squares is the reduction in residual sum of squares, $118.53-91.51=27.02$. Its mean square is $27.02/3\approx 9.007$, and its F-test statistic isThus the four missing entries are $3$, $27.02$, $9.01$, and $1.18$, to the precision allowed by the rounded output. This tests $H_0:{\delta }_2={\delta }_3={\delta }_4=0$ against the alternative that at least one slope difference is nonzero. Under $H_0$ and the normal linear model assumptions, the statistic has distribution $F_{3,12}$. Its p-value $0.3579213$ gives no reason to reject at 5%.

The simpler common-line model is inadequate compared with the parallel-line model. Testing $H_0:{\tau }_2={\tau }_3={\tau }_4=0$ against at least one nonzero type effect, while retaining speed, gives the partial F-testIts p-value is far below $0.001$, so reject the common-line restriction. This reduction differs from the sequential type sum of squares in the displayed table, because that table adds type before speed. Recommend model 2: different intercepts and a common decreasing slope.

The selected coefficient estimates giveThe intercept estimates the expected lifetime of type 1 at speed zero; if zero speed is outside the data range, it is only an extrapolated intercept. Its standard error is $4.039533$, and its Student t-test ratio is $8.885$, with two-sided p-value $2.31\times 10^{-7}$. At any fixed speed, type 2 differs from type 1 by $-0.833408$ hours, with standard error $1.904707$, $t=-0.438$ and p-value $0.668$, giving little evidence of a difference. Types 3 and 4 exceed type 1 by $12.607660$ and $16.539121$ hours, with standard errors $1.777937$ and $1.876116$; their test ratios $7.091$ and $8.816$ and p-values $3.68\times 10^{-6}$ and $2.55\times 10^{-7}$ support positive differences. Each of these individual coefficient tests has null value zero, alternative nonzero, and null distribution $t_{15}$.

The speed coefficient means a reduction of $0.024585$ hours per extra revolution per minute, or $2.4585$ hours per additional 100 rpm, for every type. Its standard error is $0.004682$; the ratio $-5.251$ has null distribution $t_{15}$ when the common slope is zero, and two-sided p-value $9.77\times 10^{-5}$.

The residual standard error $2.811$ estimates the common noise standard deviation in hours using 15 residual degrees of freedom, since $20-5=15$. The coefficient of determination $0.9247$ means that about $92.47\%$ of the corrected lifetime variation is explained by this fit. The overall F-test statistic $46.08$ tests the simultaneous restriction ${\tau }_2={\tau }_3={\tau }_4=\beta =0$ against at least one nonzero coefficient; its null distribution is $F_{4,15}$ and its p-value $2.974\times 10^{-8}$ rejects an intercept-only mean. These conclusions remain conditional on suitable regression diagnostics for independent, homoscedastic, approximately normal errors.

The required sketch plots the four fitted lines. Its speed interval is illustrative because the observed speeds are not supplied; the ordering and vertical gaps are determined by the fitted coefficients.

Both fits use the same Gamma distribution mean model and inverse link function. Their coefficient estimates, fitted means and hence their weight matrices coincide: the common dispersion parameter only multiplies the likelihood score by a scalar and therefore does not change its zero. The displayed calls likewise show the same family and formula; fixing dispersion in a summary changes the uncertainty calculation, not these coefficient estimates.

For $V(\mu )={\mu }^2$ and $g^{\prime }(\mu )=-{\mu }^{-2}$, the Fisher information weights areThus the common matrix $(X^{T}WX{)}^{-1}$ is multiplied by dispersion $1$ in the exponential case and by $\hat{\varphi }=0.3103711$ in the fitted gamma case. Taking square roots of the diagonal covariances givesFor instance $0.03607\times 0.5571\approx 0.02009$ for the intercept, and the same factor applies to all the coefficients. The standard errors are smaller because the fitted dispersion parameter is below one, not because a different mean function was fitted.

Let $W_{aj\ell }$ be recall for age level $a$, processing level $j$, and replicate $\ell =1,\dots ,10$. Write $u_a=1$ for Younger and $0$ for Older. With Older and A as the reference levels in a regression factor, the additive two-factor normal linear model under treatment coding iswhere the errors are independent $N(0,{\sigma }^2)$, with common positive variance. The design matrix is fixed; subjects supply independent observations; the specified conditional mean is correct; and the absence of an interaction term means the age difference is constant across processing levels. These are model assumptions, not consequences of random assignment. The treatment allocation supports independence between processing assignment and background characteristics, but age itself was not randomized.

An older subject in A has $u_a=0$ and ${\gamma }_A=0$, so the estimated mean is $11.35$ words. Its standard error is the intercept's $0.7632$. There are $100-6=94$ residual degrees of freedom, giving the 95% confidence interval for this mean:This is a mean-response confidence interval, not a prediction interval for a new individual's recall; the latter also includes the new observation's error variance.

The second two-factor normal linear model adds age-by-processing interaction terms:Its ten free mean parameters are equivalent to one mean for each of the ten cells. To compare the fits, the null hypothesis is ${\delta }_B={\delta }_C={\delta }_D={\delta }_E=0$. The full residual sum of squares is $722.30$ on $90$ degrees of freedom. Adding interaction reduces the additive model's residual sum of squares by $190.30$, so the reduced value is $912.60$. The nested-model F-test statistic isIts p-value is $0.0002793$. Reject additivity and retain the interaction model. The significant interaction means that a single age effect averaged over all processing methods does not adequately summarize the data.

The regression intercept $11.0$ is the older-A mean. The age coefficient $3.8$ is the younger-minus-older contrast specifically in A. The processing coefficients $(-4.0,2.4,1.0,-4.1)$ compare B, C, D and E with A specifically among older subjects. The interaction coefficients $(-4.3,0.4,3.5,-3.1)$ are differences between the age contrast in each of those methods and its value in A. Adding the relevant coefficients givesThus A, C and especially D favour younger subjects in their estimated means, whereas B and E have little estimated age difference. In the younger group, D has the largest estimated recall, followed by C and A; B and E are much lower. Among older subjects C is largest, D and A are intermediate, and B and E are lowest. Comparing these orders is a description of estimates, not a claim that every pairwise difference is significant.

Each coefficient's printed standard error estimates its uncertainty; its $t$ statistic divides the estimate by that error, and its two-sided p-value uses $t_{90}$ under the relevant zero-contrast null hypothesis. For example, the older B-versus-A and E-versus-A contrasts have $p=0.00217$ and $0.00170$. The C-versus-A and D-versus-A older contrasts have $p=0.06139$ and $0.43201$. The B interaction has $p=0.01846$, while the D interaction is borderline at $0.05387$ and the other two individual interaction tests are not significant at $5\%$. These individual tests do not override the joint interaction F-test, and multiple comparisons require care. A simple younger-versus-older contrast outside A combines coefficients and must use their estimated covariance, rather than adding their marginal errors.

The residual standard error $2.833$ estimates $\sigma$. The coefficient of determination $0.7293$ is the fraction of centered sample recall variation explained by the ten-cell model. The overall $F=26.93$ with null law $F_{9,90}$ tests all nine non-intercept coefficients jointly against zero, giving extremely strong evidence that the cell means are not all equal. The sequential analysis of variance separates age, process, and their interaction; the balanced design makes the main-effect sums orthogonal, whereas the coefficient table uses the specified reference-level contrasts.

To assess pooling into three processing types, retain age-by-type interaction because the previous test rejects additivity. The reduced model $W\sim \mathrm{Age}\ast \mathrm{Type}$ has six cell means. Its null hypothesis says A and C have equal means within each age, and B and E have equal means within each age: four restrictions. Compare it with the ten-cell model usingIn fact the supplied fitted cell means and equal cell sizes determine the increase without raw observations. Pooling two ten-person cells with means $m_1,m_2$ adds $5(m_1-m_2{)}^2$ to the residual sum of squares. Here the four differences are $2.4,2.8,-0.1,1.1$, soTherefore $F_{\mathrm{pool}}=2.3083$ and $p\simeq 0.0640$. At $5\%$, the data do not reject the three-type reduction, though the result is borderline and is not proof that the pooled means are identical. This factor-level pooling test keeps the previously supported interaction structure; reducing the categories and removing all interaction simultaneously would test a different set of restrictions.

Let $a_i$ be age in years, $g_i$ the male indicator and $m_i$ the indicator for the Messiah group. The fitted Cox proportional-hazards model iswith an unspecified common baseline hazard. Its coefficients were estimated by Cox partial likelihood, using the Breslow approximation for tied event times. There are $100$ participants, $96$ recorded completions and four censored observations.

A higher hazard function means a greater instantaneous chance of completion among those not yet completing, so it describes faster completion rather than greater mortality. Holding gender and group fixed, an extra year of age multiplies the completion hazard by $1.0066$, about a $0.66\%$ increase. The 95% confidence interval is $(0.9865,1.0270)$ and the Wald test has $p=0.5241$, so there is little evidence of an age association.

Holding age and group fixed, male participants have a hazard ratio $0.5469$ relative to female participants, about a $45.3\%$ lower completion hazard. Its 95% confidence interval is $(0.3560,0.8401)$ and $p=0.00586$. Under the fitted model this corresponds to slower completion for males. It is not a ratio of mean or median completion times.

Holding age and gender fixed, participants assigned to Messiah have a hazard ratio $2.1283$ relative to The Kingdom, with 95% confidence interval $(1.3653,3.3176)$ and $p=0.000854$. This is strong evidence of a group difference, with faster completion in the Messiah group under the fitted model. Random assignment supports interpreting a group contrast as an effect of the assigned condition, subject to the assumptions about follow-up and censoring; age and gender contrasts remain observational associations.

Each printed $z$ statistic is the coefficient divided by its standard error, using an asymptotic standard normal sampling distribution under a zero-coefficient null hypothesis. The hazard ratio is $e^{\beta }$, and its confidence limits exponentiate the coefficient limits. Proportional hazards assumes these covariate-specific hazard multipliers stay constant over follow-up, together with the specified linear age effect and independent censoring conditional on covariates.

Multiple imputation by chained equations specifies one conditional distribution for each incomplete variable. In general, let $V_1,\dots ,V_q$ be the incomplete variables, $Z$ the fully observed predictors, and ${\varphi }_j$ the parameters of a model $p_j(V_j\mid V_{-j},Z;{\varphi }_j)$. The notation $V_{-j}$ means all variables except $V_j$. The models should respect each variable's support and contain the terms needed by the planned analysis.

First choose suitable models and initialise only the missing entries with admissible provisional values. One sweep updates $j=1,\dots ,q$ in turn. At update $j$, fit its model to records whose $V_j$ is observed, using the latest filled-in predictors. DrawThe first draw represents parameter uncertainty through a Bayesian posterior; the second represents residual uncertainty through a posterior predictive distribution. Replace the missing entries with these draws, leaving every genuinely observed entry fixed. Repeat whole sweeps until the chain has settled, assessing traces, mixing and plausibility. Obtain $M$ completed datasets from independently initialised chains, or sufficiently separated draws after convergence. Both parameter and missing-value uncertainty must be propagated; repeated deterministic prediction is not multiple imputation.

Here use a logistic regression for $C$ conditional on $P$, age and sex. For $P$, use an appropriate discrete count model conditional on $C$, age and sex, such as a suitably checked negative binomial regression, allowing enough flexibility for skewness and nonlinear effects. Include interactions or transformations needed by the analysis in the imputation models. If condom use is structurally undefined for a record, preserve that applicability rule rather than treating it as ordinary missing data.

Impute the eligible survey information before discarding records solely because the observed $P$ is absent. In each completed dataset select its own $P>1$ records and fit the requested logistic regression. This allows a record with missing $P$ to contribute to the target group in some imputations and not others. Models must respect any known eligibility restrictions; if eligibility was independently known, use that information.

For a scalar regression coefficient, let ${\hat{Q}}_m$ be its estimate and $U_m$ its estimated variance in completed dataset $m$. Rubin's rules giveThe standard error is $\sqrt{T}$. Use the corresponding multiple-imputation confidence interval, rather than treating one completed dataset as fully observed. For vector coefficients replace the squared differences by outer products and $U_m$ by the covariance matrix. This combines within-imputation and between-imputation uncertainty.

Three distinct benefits of a meta-analysis are greater precision, an explicit synthesis of the effect, and assessment of variation across studies.

Combining compatible independent estimates can reduce the standard error, improve statistical power and provide more informative confidence intervals than individual small trials. A prespecified pooled estimate summarises the direction and magnitude of the evidence quantitatively, instead of informally counting statistically significant results. Finally, between-study heterogeneity and subgroup comparisons can reveal whether effects vary with populations or interventions, helping assess generalisability. These benefits require comparable estimands and an appraisal of study validity; pooling cannot repair systematic within-study bias.

The event and nonevent cell counts are $(a,b)=(30,60)$ and $(c,d)=(20,40)$. Thus the estimated odds ratio and log odds ratio areFor independent binomial groups, the delta method givesUsing the permitted normal approximation quantile of 2, the approximate 95% interval isThe observed odds of death are equal between groups. The confidence interval includes zero and allows appreciable effects in either direction; it does not demonstrate equivalence. Exponentiating gives an odds ratio interval approximately $(0.493,2.028)$, so the data are compatible with roughly half to twice the death odds for the first treatment relative to the second. The tabulated standard error is rounded.

A fixed-effect meta-analysis assigns inverse-variance weights. Using the unrounded within-study variance from part (b), this trial has $w=8$, so its percentage weight isUsing the printed rounded standard error of $0.35$ instead gives approximately $2.04\%$; the raw cell calculation yields the more precise answer.

For a random-effects meta-analysis, the corresponding weight and percentage weight areWith ${\hat{\tau }}^2=0.05$, its unnormalised weight is $40/7\approx 5.714$. The supplied sums of fixed-effect weights do not determine the denominator of the random-effects percentage; the individual $v_i$ are needed. Positive heterogeneity makes the weight distribution more even. Percentage weight measures the trial's contribution to a pooled mean with fixed weights; total influence can also involve re-estimation of heterogeneity.

For an independent subgroup contrast in meta-analysis, the difference of disjoint independent subgroup estimates has estimated variance equal to the sum of the two variances. The subgroup contrast and standard error areThe Wald test of equal subgroup mean log odds ratios therefore givesUnder the null, $Z$ is approximately standard normal and $Z^2$ has a chi-squared distribution with one degree of freedom. A quantile-2 confidence interval for the difference is $(0.29-0.8,0.29+0.8)=(-0.51,1.09)$. There is no statistically persuasive evidence of a subgroup difference. The result does not establish equal effects. Nor does a between-trial subgroup comparison, by itself, identify a causal advantage of one technique: other study characteristics may differ.

Use a common intercept and laboratory-specific slopes in the normal linear model. With $t_i$ denoting days and $I_{Bi},I_{Ci}$ the indicator variables for laboratories B and C, the design matrix has row $x_i^T=(1,t_i,t_i{I}_{Bi},t_i{I}_{Ci})$, andThe interaction terms change slopes, without introducing laboratory-specific intercepts. Thus the estimated daily growth rates arein millimetres per day, with fitted common intercept $0.6464$ mm. The residual degrees of freedom are $30-4=26$. The usual unbiased error variance estimate is the square of the residual standard error:The small difference from $5921.7/26\approx 227.758$ arises from rounding the standard error to two decimal places; the latter uses the more precise residual sum of squares supplied later. A maximum-likelihood estimate would instead divide that residual sum of squares by $30$.

For a new independent tumour grown for ten days in B, use $x_0=(1,10,10,0{)}^T$. Its fitted mean isWrite $h_0=x_0^T(X^{T}X{)}^{-1}{x}_0$. The prediction interval in a normal linear model iswhere $s^2=\mathrm{RSS}/26$ and $t_{26,1-\alpha /2}$ is the indicated Student's t-distribution quantile. The design information needed to evaluate $h_0$ is not supplied, so a fully numerical interval cannot be recovered from the coefficient standard errors alone: the coefficient covariances are also needed.

To prove coverage, write $Y_0=x_0^T\beta +{\epsilon }_0$, with ${\epsilon }_0\sim N(0,{\sigma }^2)$ independent of the original sample. The error $Y_0-x_0^T\hat{\beta }$ has a normal distribution with mean zero and variance ${\sigma }^2(1+h_0)$. By the joint distribution of least-squares and variance estimators, $26s^2/{\sigma }^2\sim {\chi }_{26}^2$ and is independent of $\hat{\beta }$; it is also independent of ${\epsilon }_0$. Hence the pivotal quantityConsequently $\mathrm{Pr}(-t_{26,1-\alpha /2}\le T\le t_{26,1-\alpha /2})=1-\alpha$, which is exactly the claimed prediction interval coverage. This is repeated-sampling coverage conditional on the design, assuming the model; it need not hold conditional on the particular observed training responses. The extra $1$ represents new-observation noise and must not be omitted as in a mean confidence interval.

Let $s_i$ be sunny days and $l_i$ latitude. The R linear-model formula expands to a normal linear model with both main effects and their interaction term:The fitted conditional mean isThe residual standard error is $0.2836$, giving the usual variance estimate $s^2=0.08042896$, and $n=96+4=100$.

The estimated interaction term is tiny relative to its standard error: $t=-0.103$ and the two-sided p-value is $0.91811$. A sensible initial simplification is to remove the interaction and refit an additive linear model, lm(energy ~ sunny + latitude). Both main effects are significant in the supplied fit, and should initially be retained. Do not simply copy their old coefficient estimates into the new fit: removing a design column generally changes all estimates. Centering predictors makes intercepts and main-effect interpretations more useful, but does not change the fitted surface. The Box–Cox transformation in part (d) suggests a further response-scale improvement; after changing the response scale, reassess the interaction and diagnostics.

The Pearson dispersion estimator is $\hat{\varphi }=3.099079$, well above the Poisson value one. A Quasi-Poisson regression retains the logarithmic mean model but usesThis is a quasi-likelihood specification; it does not by itself identify a count probability distribution. For a common $\varphi$, the quasi-score equation is the Poisson score divided by $\varphi$, so its roots and fitted means are unchanged. ThusCoefficient standard errors are multiplied by $\sqrt{\hat{\varphi }}$. The altitude standard error becomes $0.00447\sqrt{3.099079}\approx 0.0078691$, giving the requested approximate normal confidence intervalIt excludes zero, so retain altitude on this approximate 5% test. Its test statistic is about $2.412$ after allowing for overdispersion, rather than the Poisson value $4.245$. Rescaling uncertainty cannot repair a systematically misspecified mean; the diagnostics in part (e) motivate reconsidering that mean as well.

Use the usual clinical trial sampling assumption that all patients' responses are independent, with fixed positive arm sizes and positive known variances. Marginal normal distributions alone would not specify the variance of a treatment contrast without this assumption. Put ${\overline{Y}}_i=n_i^{-1}{\sum }_j{Y}_{ij}$, $v_i={\sigma }_i^2/n_i$, and $s_i=\sqrt{v_i+v_0}$. The sample mean contrast estimates ${\delta }_i$ and has standard error $s_i$. Thus the one-sided Wald statistics areLarge positive values of these Wald test statistics provide evidence against the respective null hypotheses.

First, the absolute values in the printed limit are an error: the correct long-run variance of a stationary process is the signed sum of autocovariances. Directly,soEach coefficient tends to $1$ and has absolute value at most $1$. Absolute summability and the dominated convergence theorem therefore giveThe sum of absolute values is an upper bound, not the general limit. For a concrete counterexample to the printed assertion, take $X_t={\eta }_t-{\eta }_{t-1}$ with unit-variance iid noise. Then $\gamma (0)=2$, $\gamma (\pm 1)=-1$, and other covariances vanish. The absolute sum is $4$, but ${\overline{X}}_T=({\eta }_T-{\eta }_0)/T$, so $T\mathrm{Var}({\overline{X}}_T)=2/T\to 0$.

For correlated data the central limit theorem can still hold under suitable strong mixing of a stationary process and moment conditions, but the limiting variance is the long-run variance of a stationary process, rather than the one-observation variance. When it is positive,Positive serial dependence usually increases the standard error, while negative dependence can reduce it. The approximate effective sample size of a stationary sample is $T{\gamma }_X(0)/{\sum }_h{\gamma }_X(h)$ when that denominator is positive. A long-memory time series can require a different normalization or a different limit law. Absolute covariance summability alone does not prove a CLT: if $X_t=V{\eta }_t$ with a common independent random scale $V$ taking values $1$ and $2$ with equal probabilities and iid ${\eta }_t$ with the standard normal distribution, the off-diagonal covariances are zero, yet $\sqrt{T}{\overline{X}}_T$ has the nonnormal scale-mixture law $VN(0,1)$. The example is not an ergodic stationary process. If the long-run variance of a stationary process is zero, the usual nondegenerate square-root-$T$ CLT is unavailable.

For the causal AR(1), $|\varphi |<1$ andThe conditional distribution follows because the current innovation is independent of the past.

For a fixed known $\varphi$, condition on the observed $X_1$ and use the $n=T-1$ transitions $t=2,\dots ,T$. Their conditional maximum likelihood criterion is, up to constants,Differentiating in $\mu$ yieldsSince $X_t-\varphi X_{t-1}=(1-\varphi )\mu +{\epsilon }_t$,Thus it is an unbiased estimator, even conditionally on $X_1$, andIt has statistical consistency with mean-square convergence, and the iid-noise strong law of large numbers also gives almost-sure statistical consistency. Its conditional distribution is exactly normal with the displayed mean and variance. If $X_0$ is also observed and all $T$ transitions are used, replace $T-1$ by $T$.

The fixed-$\varphi$ qualification is necessary for the exact finite-sample claims. If $\varphi$ is jointly estimated, conditional likelihood function is linear regression with an intercept: writing ${\bar{X}}_{-}=n^{-1}{\sum }_{t=2}^T{X}_{t-1}$ and ${\bar{X}}_{+}=n^{-1}{\sum }_{t=2}^T{X}_t$, the unconstrained estimators areThis ratio is not generally an unbiased estimator and does not have the preceding finite-sample variance. Under the usual stationary regression conditions it has statistical consistency; its asymptotic variance is ${\sigma }_{\epsilon }^2/(1-\varphi {)}^2$. Indeed, it differs from ${\bar{X}}_{+}$ by $\hat{\varphi }(X_T-X_1)/(n(1-\hat{\varphi }))$, an asymptotically negligible endpoint term. Profiling an unknown innovation variance does not change the fixed-$\varphi$ estimate of $\mu$.