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Use the normal linear modelHere $Y$ is the response random vector, $X$ is the known design matrix, $\beta \in {\mathbb{R}}^p$ contains the unknown regression coefficients, and ${\sigma }^2>0$ is the common error variance. Conditional on $X$, the errors have normal distributions and are independent random variables. Require $\mathrm{rank}X=p\le n$ for identifiability of $\beta$ and invertibility of $X^{T}X$. Usually $n>p$ is needed to estimate the error variance from the regression residuals. An intercept, when included, is represented by a column of ones in $X$.

The ordinary least squares estimator is $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$. Consequently the fitted values and regression residuals areAn affine transformation of a multivariate normal distribution is again multivariate normal, possibly with a singular covariance matrix. For a random vector with covariance matrix $\Sigma$, its transformed covariance matrix is $A\Sigma A^T$. Since $PX=X$, $P^2=P=P^T$ and $(I-P{)}^2=I-P$, these results giveBoth multivariate normal distributions are supported on their respective projected subspaces. In particular, ${\hat{Y}}_i$ has variance ${\sigma }^2{P}_{ii}$ and $e_i$ has variance ${\sigma }^2(1-P_{ii})$; the regression residuals need not be mutually independent.

The simple linear regression assumes a straight conditional mean, $E(Y_i\mid x_i)={\beta }_0+{\beta }_1{x}_i$, so that errors are centred at zero throughout the predictor range. The displayed regression residuals are predominantly positive at both ends and negative in the middle. This is a residual curvature diagnostic: the fitted straight line misses a curved conditional mean. The main concern is the shape of the mean function. The plot alone does not establish failure of the normal distribution assumption or a particular error variance model.

Group means are independent random variables with normal distributionsThe resulting univariate sampling distributions areThe same variance calculation holds for every $i\ge 2$, since the difference involves two independent group means. Therefore the standard errors satisfyReplacing $\sigma$ by a common estimated error standard deviation preserves this ratio. Although the individual group means are independent, the contrasts ${\hat{\alpha }}_i$ share the baseline mean and have covariance ${\sigma }^2/J$ for distinct $i\ge 2$.

Index chocolate by $a\in \{A,B,C,D\}$, day by $d$ in the three observed categories, and replicate by $r\in \{1,2\}$. The additive two-factor normal linear model isHere $Y_{adr}$ is the board count, $\mu$ is the mean for the reference chocolate and reference day, and ${\alpha }_a,{\gamma }_d$ are chocolate and day fixed effects. With corner-point constraints, set ${\alpha }_A=0$ and ${\gamma }_{d_0}=0$, where $d_0$ is the first level in the day factor. The printed coefficient-free output does not determine that factor ordering; the model is unchanged by a different reference category. There is no chocolate–day interaction term in this fit. The six free mean statistical parameters consist of one baseline, three chocolate contrasts and two day contrasts; the common error variance supplies a further statistical parameter.

For $y=0,1,2,\dots$ and $\lambda >0$, the Poisson distribution has massMatch this to the exponential dispersion family withThen $\mu =b^{\prime }(\theta )=e^{\theta }=\lambda$ and $V(\mu )=b^{\prime \prime }(\theta )=\mu$. The canonical link function expresses the natural parameter in terms of the mean, so the Poisson canonical link is $g(\mu )=\log \mu$. Its conditional variance equals its conditional mean.

The Quasi-Poisson regression retains the same conditional mean but permitsThis is a mean–variance function specification through quasi-likelihood; it does not assign a full probability distribution to each count. For independent observations, the quasi-score equation is proportional to ${\sum }_i{x}_i(Y_i-{\mu }_i)=0$, so the mean statistical parameter estimates equal those from Poisson regression. The output estimates $\hat{\varphi }=3.515351$ through the Pearson dispersion estimator, and inflates the standard errors by approximately $\sqrt{\hat{\varphi }}=1.875$.

The large residual deviance relative to 728 residual statistical degrees of freedom also signals substantial overdispersion. Daily weather, traffic and other omitted conditions may produce greater count variation than a homogeneous Poisson distribution allows. The Quasi-Poisson regression accounts for that extra marginal variance. It still requires a correct conditional mean and an appropriate independence assumption; a common dispersion parameter alone does not repair serial correlation.

Under the more plausible Quasi-Poisson regression, the year Wald statistic is $-1.686$, with reported p-value $0.0922$. An approximate 95% confidence interval isThus the estimated fall is about 9.35%, but the data do not establish a reduction at the 5% level after allowing for overdispersion. The interval allows both a sizeable reduction and a small increase. The before–after comparison also lacks a contemporaneous randomized control, so causal inference about the campaign would require addressing other changes between years. The small Poisson regression p-value is not sufficient evidence of a campaign effect when its variance assumption is unsuitable.

For subject $i=1,\dots ,100$ and $t_j\in \{0,2,4,6,8,10\}$, the random-intercept linear mixed model isAll subject random effects and measurement errors are mutually independent. The fixed effects ${\beta }_0,{\beta }_1$ describe the population mean baseline and weekly gain. Conditional on $b_i$, a person's observations have independent normal distributions; after integrating out $b_i$, their covariance is ${\tau }^2$ at distinct weeks and their common variance is ${\tau }^2+{\sigma }^2$. Therefore their correlation is ${\tau }^2/({\tau }^2+{\sigma }^2)$.

This Gaussian linear mixed model permits correlated repeated readings while keeping different subjects independent. The fitted standard deviations are $\hat{\tau }=21.67593$ kg and $\hat{\sigma }=14.23496$ kg, giving within-person correlation about $0.699$. All persons still have the same latent weekly slope in this model.

Allow each subject to have a separate baseline and slope through a correlated random-intercept and random-slope model:The subject random effects are independent of all measurement errors, and subjects are independent. The within-person covariance between times $s,t$ is ${\tau }_0^2+(s+t)\rho {\tau }_0{\tau }_1+st{\tau }_1^2$, with an additional ${\sigma }^2$ when the same reading is used twice.

Prefer the model with a random slope. The maximum likelihood estimation refits improve twice the log-likelihood by $227.9604$ while adding two covariance statistical parameters. Both the Akaike information criterion ($5165.779$ to $4941.819$) and Bayesian information criterion ($5183.367$ to $4968.200$) strongly favour it. The printed nominal likelihood-ratio test is overwhelming. The usual reference chi-squared distribution with two statistical degrees of freedom is not an exact regular calibration, since zero slope variance is a boundary and the intercept–slope correlation is then unidentified; a design-specific parametric bootstrap could calibrate it. This qualification does not undermine the substantial descriptive improvement shown by both information criteria.

The preferred fit estimates population mean baseline strength ${\hat{\beta }}_0=61.09286$ kg and weekly gain ${\hat{\beta }}_1=2.83143$ kg/week. The between-person baseline standard deviation is ${\hat{\tau }}_0=15.785582$ kg, and the between-person slope standard deviation is ${\hat{\tau }}_1=2.738666$ kg/week. Their estimated correlation $\hat{\rho }=0.117$ is weakly positive, giving random-effect covariance about $5.058$ kg${}^2$/week; its uncertainty is not supplied. The measurement-error standard deviation is $\hat{\sigma }=9.923280$ kg. The printed $-0.038$ instead describes the correlation between the estimated fixed effects, not between the subject random effects. These statistical parameter estimates come from restricted maximum likelihood; the model comparison uses the separate maximum likelihood estimation fits.

The population mean fitted trajectory is $61.09286+2.83143t$ kg. HenceUsing the reported slope standard error, an approximate 95% confidence interval for the population mean gain is $28.3143\pm 1.96(2.984464)=(22.46,34.16)$ kg; a Student t confidence interval with 499 statistical degrees of freedom is almost identical.

Successful individuals can improve far more than the average. Their latent ten-week gains have fitted normal distributionA clear estimate for an upper-performing group uses a between-person slope quantile: the 95th percentile is $28.3143+1.645(27.38666)\simeq 73.4$ kg, and the 97.5th percentile is about $82.0$ kg. The upper 5% of fitted underlying gains begin around 73 kg. These are person-to-person performance quantiles, not confidence intervals for the population mean. Identifying the best observed trainee would require that person's data or fitted subject random effects; the aggregate output cannot identify a literal maximum. If performance means the observed difference of endpoint readings, add the measurement-error variance $2{\sigma }^2$ to the latent gain variance.

Let $y_i$ be each trial's estimated log risk ratio, with estimated sampling variance $s_i^2$. A fixed-effect meta-analysis models the independent estimates as $y_i\simeq N(\mu ,s_i^2)$, with one common true log risk ratio $\mu$. Differentiating the Gaussian log-likelihood, or minimizing ${\sum }_i(y_i-\mu {)}^2/s_i^2$, gives inverse-variance weights $w_i=1/s_i^2$ andThe variance follows directly by adding independent variances of the weighted estimates: ${\sum }_i{w}_i^2{s}_i^2/({\sum }_i{w}_i{)}^2=1/{\sum }_i{w}_i$.

Using the printed rounded estimates and standard errors, ${\sum }_i{w}_i=50.2923$, soThe pooled risk ratio is about $0.902$. An interval using the same quantile two is $(-0.3848,0.1792)$ on the log scale, so it includes no effect. Recomputing all trial estimates from the event counts would change the last digits because the printed log estimates and errors are rounded.

The key assumptions are independent trials, a common true treatment effect on this chosen log scale, approximately unbiased and approximately normal trial estimates, and suitable sampling-variance estimates. The common-effect assumption is stronger than merely studying the same named treatment: systematic population, treatment or design differences may produce genuine heterogeneity. For an unbiased interpretation of the pooled evidence, inclusion of studies must also not depend selectively on their results. Treating the plug-in variances as fixed is the usual approximation in the displayed variance formula.