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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 $i\ge 2$.

The analysis of variance provides evidence of a chocolate effect both with day included ($F_{3,18}=3.9665$, p-value $0.02473$) and with day omitted ($F_{3,20}=4.039$, p-value $0.02133$). Because every chocolate–day cell has the same replication, balanced factorial orthogonality separates the two main effects; the chocolate row is meaningful despite being entered first.

The chocolate-only fitted group means are approximately $14.17,11.67,9.33,11.33$ boards for A, B, C, D respectively. Thus A has the largest fitted lecturing speed and C the smallest. Relative to A, the fitted differences are $-2.50,-4.83,-2.83$ boards. The printed individual Student t-tests give strong evidence for the A–C contrast (p-value $0.00245$); B and D versus A have p-values $0.08835$ and $0.05582$, respectively. Those latter contrasts are not significant at 5%, and the output does not test all other pairwise comparisons. Simultaneous claims would require accounting for multiple hypothesis testing.

There is little evidence of a day effect after adjusting for chocolate. A chocolate-only normal linear model is therefore a reasonable simpler summary, with residual standard deviation about $2.42$ boards and explained variation $R^2\simeq 0.377$. This is an association under the additive statistical model; a causal claim would additionally require an appropriate assignment of chocolate and checks of the regression residuals and possible interaction terms.

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.

At zero a centered normal distribution with standard deviation $\sigma$ has density $1/(\sqrt{2\pi }\sigma )$. The two prior standard deviations are $1/\sqrt{n_0}$ and $c/\sqrt{n_0}$, givingThe wider prior distribution has a lower density at its center.

Under the point hypothesis $\beta =0$, the test statistic $\sqrt{n}\overline{Y}$ has a standard normal distribution. The observed statistic is three, giving a two-sided p-value $2[1-\Phi (3)]\simeq 0.00270$. This is conventionally strong evidence against that point hypothesis. The narrow model $H_0$ is a continuous prior distribution around zero, rather than a point hypothesis.

For $n_0/n=100$ and $c=100$, Normal-normal conjugacy givesThe narrow-model posterior mean is strongly pulled toward zero and its standard deviation is about $0.0995/\sqrt{n}$. The wide-model posterior mean is about $2.9703/\sqrt{n}$, very close to the observation, with standard deviation about $0.9950/\sqrt{n}$. Each model produces a markedly different posterior, so choosing between them requires their predictive evidence.

Use a proper normal distribution prior centered at zero on the log odds ratio. For example,puts approximately 95 percent of its mass between $-2$ and $2$, giving odds ratios roughly between $e^{-2}$ and $e^2$, close to $1/8$ and $8$. Centering at zero treats reciprocal odds ratios symmetrically. If the desired central 95 percent interval is exactly $(1/8,8)$, use standard deviation $\log 8/{\Phi }^{-1}(0.975)\simeq 1.0610$. A soft prior is appropriate for implausibility, whereas a bounded uniform prior would declare effects outside the limits impossible.

Conditional on $\mu ,\psi >0$, let $\lambda \sim {\chi }_4^2$ and independently $Z\sim N(0,1)$. The normal distribution with the stated precision parameter can be generated asThereforeby the defining normal distribution and chi-squared distribution representation of Student's t-distribution. Its density isThus $\mu$ is the location and $\psi$ is the scale, not the standard deviation: $\mathrm{Var}({\beta }_j\mid \mu ,\psi )=2{\psi }^2$. Each study gets its own independent chi-squared draw in this Student t random-effect model.

Fit both the normal and Student t random-effect model with comparable proper prior distributions. Compare priors on the same spread measure: a normal distribution scale $\tau$ is a standard deviation, whereas the $t_4$ standard deviation is $\sqrt{2}\psi$.

Use a posterior predictive check: draw study effects and binomial counts from each fitted hierarchy and compare replicated dispersion and extreme study contrasts with the observations. For predicting a new study, generate a new effect from the hierarchy rather than reusing an existing fitted effect. A Leave-one-out cross-validation with entire studies held out can compare integrated predictive probabilities for both arms of each omitted trial, averaging over hyperparameters and its unobserved study effect. Leave-one-study-out influence analysis also reveals whether the difference is driven by a single trial.

Prefer the heavier-tailed hierarchy if it improves the relevant predictive checks and held-out study predictions robustly to reasonable prior choices. The deviance information criterion can supplement the comparison, but its effective parameter count can depend on the latent-variable representation, and six studies give limited information about tail shape. A small numerical criterion difference alone is insufficient evidence.