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Take the original synthetic model $T\sim N(0,0.6)$ and ${\epsilon }_1,{\epsilon }_2\sim N(0,0.4)$, mutually independent random variables, with the second argument denoting variance. The pair $(X,Y)$ has a multivariate normal distribution with mean zero, unit marginal variances, and Pearson correlation coefficient $0.6$. Its conditional multivariate normal distribution isFor a first reading $x=2$, the second reading has conditional expectation $1.2$ and conditional variance $0.64$, hence standard deviation $0.8$. There is still conditional probability $1-\Phi (1)\approx 0.1587$ of $Y>2$, a movement farther from zero, where $\Phi$ is the standard normal distribution function. The line $0.6x$ describes a conditional mean, not an individual trajectory.

A short derivation isolates $R=Y-0.6X$. Its covariance with $X$ is zero and its variance is $0.64$. Because $(X,R)$ has a multivariate normal distribution, zero covariance implies independence: $R\sim N(0,0.64)$ independently of $X$. This also explains why reversing the two measurements gives $\mathbb{E}[X\mid Y=y]=0.6y$. The statistical association does not identify a direction of causal effect.

read.table(...,header=T) reads four grouped rows with column names. The displayed outcome counts give the observed risks $0.1519,0.2145,0.2611,0.5111$ in the indicated factor order. rbind in the first test pools over predisposition: the two rows represent no cannabis use and some use, with event/non-event counts $(341,1775)$ and $(82,238)$. Thus the pooled risks are $341/2116=0.1612$ and $82/320=0.2563$. This test concerns the marginal statistical association between use and outcome, not an association adjusted for predisposition.

Under independence, expected counts in the two-by-two contingency table are $E_{ij}=O_{i+}{O}_{+j}/N$. The ordinary Pearson chi-squared statistic is $17.5183$. The printed $16.8618$ is reproduced by the default two-by-two Yates continuity correction:All absolute differences here exceed one half. Comparison with a chi-squared distribution on one statistical degree of freedom gives $p=4.02\times 10^{-5}$. There is strong evidence of marginal association. The simplified output omits the usual continuity-correction wording, but its numerical value shows which statistic was used.

attach makes the data columns available by name, and tot=with+without obtains group sizes $1936,275,180,45$. The grouped binomial regression fits $Y_g\sim \mathrm{Bin}(n_g,{\pi }_g)$, with response $Y_g/n_g$ and prior weights $n_g$. These weights tell R the binomial denominator; treating the four proportions as four equally precise individual observations would be wrong. The default logit link isLet $A$ indicate some use and $B$ indicate predisposition. The additive logistic regression specifies $\eta =a+bA+cB$. Its log-likelihood, including the binomial constants, isDifferentiation gives score equations $X^T(Y-n\pi )=0$ and Fisher information matrix $X^{T}WX$, where $W_{gg}=n_g{\pi }_g(1-{\pi }_g)$. Fisher scoring solves these equations, and the inverse fitted Fisher information matrix gives the reported approximate coefficient variances. The number of scoring iterations records numerical convergence, not degrees of freedom or statistical evidence.

The intercept corresponds to no use and no predisposition and gives baseline fitted risk ${\mathrm{logit}}^{-1}(-1.73881)=0.1495$. The additive coefficients imply common adjusted odds ratios:They multiply the odds, not the risks. The Wald tests divide the estimates by their standard errors and use a standard normal distribution; both predictors have small two-sided p-values. Approximate $95\%$ confidence intervals for these odds ratios are $(1.296,2.266)$ and $(1.685,3.110)$, obtained by exponentiating the log-scale intervals. The intercept test against zero asks whether the reference risk equals one half, and is not a test of either exposure association. The binomial dispersion is assumed to be one because $\mathrm{Var}(Y_g)=n_g{\pi }_g(1-{\pi }_g)$; the output does not demonstrate independence or rule out extra-binomial variation.

There are four grouped probabilities and three fitted coefficients, leaving one residual degree of freedom. The fitted risks are approximately $(0.1495,0.2314,0.2869,0.4080)$. The binomial deviance compares this fit with the four unrestricted observed proportions:The deviance goodness-of-fit test against ${\chi }_1^2$ has p-value $0.0796$. The common-odds-ratio fit is not rejected at $5\%$, although the doubly exposed group's observed risk exceeds its fitted risk appreciably and suggests possible interaction.

The second formula adds $\delta AB$. Its four coefficients fit all four probabilities, making a saturated statistical model with zero grouped residual degrees of freedom. The near-zero deviance is roundoff from exact interpolation; it supplies no remaining grouped lack-of-fit test. Unlike Q1's unreplicated Gaussian fit, however, these grouped responses represent many independent Bernoulli trials, and their binomial variances are specified by their sizes and probabilities. Coefficient standard errors remain available from binomial information without estimating dispersion from the zero residual degrees of freedom.

Indeed, let ${\eta }_{ab}=\log [Y_{ab}/(n_{ab}-Y_{ab})]$ be the fitted cell logit link value. The saturated coefficients areThese calculations yield the printed estimates. The delta method gives $\mathrm{Var}({\hat{\eta }}_{ab})\approx 1/Y_{ab}+1/(n_{ab}-Y_{ab})$, and independent groups make the variance of $\hat{\delta }$ the sum of all eight reciprocal counts. Its square root is $0.37857$, reproducing the printed standard error.

The saturated logistic regression describes the conditional odds ratios as follows:Thus the main coefficients now concern effects at the other factor's reference level. Logistic interaction as a ratio of odds ratios gives $e^{\delta }=1.939$, with approximate $95\%$ confidence interval $(0.923,4.073)$. The interaction Wald test has $p=0.0802$, while the nested likelihood-ratio test uses deviance difference $3.0733$ on one statistical degree of freedom and has $p=0.0796$. The data suggest a larger odds association in predisposed subjects, but do not establish this interaction at the conventional $5\%$ level. This conclusion is specific to the odds scale; absolute-risk differences can vary even with no logit-scale interaction.

The Akaike information criterion is $-2\hat{\ell }+2k$. The additive fit has $k=3$ and AIC $31.766$; the interaction fit has $k=4$ and AIC $30.692$. The extra coefficient improves twice the log-likelihood by $3.0733$ but incurs penalty $2$, so AIC falls by only $1.0733$. It weakly favors the larger fit rather than decisively establishing an interaction, and AIC is not a p-value.

The marginal odds ratio $1.793$ and the fitted conditional odds ratios are different quantities. Predisposition is associated with both use frequency and outcome, so confounding is a substantive possibility; noncollapsibility of the odds ratio also means marginal and conditional odds ratios need not agree. This is an observational association analysis with only one adjustment factor. It does not by itself establish a causal effect, exclude other confounding, or validate the assumed independent binomial sampling.

This is a sequential analysis of deviance for nested generalized linear models. Adding age to the intercept-only logistic regression reduces deviance by $6.5187$, with approximate ${\chi }_1^2$ p-value $0.01067$. Adding frailty to the age model reduces it by another $16.6474$, with p-value $4.501\times 10^{-5}$. Thus both variables have evidence of association with surgical success, under the model assumptions, and neither should be dropped solely on these results.

The second test assesses frailty adjusted for age, whereas the first assesses age before frailty is included. Sequential tests depend on term order. Evidence for age adjusted for frailty is additionally provided by its full-model Wald test, $p=0.01390$. Both fitted coefficients are negative: age and frailty are associated with lower fitted success probabilities. On the odds scale, one extra year at fixed frailty multiplies the odds by $e^{-0.08942}\approx 0.9145$, and frailty at fixed age multiplies them by $e^{-3.15336}\approx 0.04271$. These are odds ratios, not risk ratios, and an observational statistical association does not by itself establish a causal effect.

A Type I error is rejecting a true null hypothesis. Here it means declaring a statistical association between the biomarker and bowel cancer when, in the population and under the association model being tested, there is no statistical association. It is a false positive association finding, not merely a large estimation error. A test with level $0.05$ limits this probability to $0.05$ under the null hypothesis, subject to valid calibration; the p-value is not the probability that the null hypothesis is true.

For the electronic-record observational study, three strengths are:

Three weaknesses are:

For two-sample Mendelian randomization, three strengths are:

Three weaknesses are:

Thus the agreement supports the hypothesis, but does not establish the effect of a dietary intervention.

Education is a causal risk factor if an intervention that changes educational exposure changes the distribution of the corresponding potential outcome for myopia. An observed statistical association need not have that interpretation. Shared causes such as socioeconomic circumstances, parental behavior, genetic traits, or preference for indoor near work can confound education and myopia. Reverse causality is also possible: children with impaired distance vision may select more reading and indoor study, which can increase educational attainment without education causing the impairment.