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A random variable on a proposed causal pathway from a cause to an outcome. A causal mediator transmits part of a causal effect; statistical association with the cause and outcome alone does not establish that role.

Collider bias arises when analysis conditions on a common effect of two variables, thereby creating a statistical association between its causes.

The contingency coefficient transforms the Pearson chi-squared statistic as $C=\sqrt{X^2/(X^2+n)}$. It is an unsigned measure of statistical association, not a p-value or a direction of an ordinal relationship. Writing ${\varphi }^2=X^2/n$ gives $C=\varphi /\sqrt{1+{\varphi }^2}$. Unlike Cramér's V, its attainable maximum depends on table dimensions, so comparisons between differently sized tables need care.

For an $r\times c$ contingency table with $n$ observations and positive row and column totals, Cramér's V rescales the Pearson chi-squared statistic to a dimensionless measure of statistical association. It lies between zero and one and is unchanged by reordering category labels. It therefore describes the magnitude of dependence, without an ordinal direction. A small p-value can accompany a small $V$ in a large sample: evidence against independence and the magnitude of the departure answer different questions. The displayed quantity is the usual uncorrected sample measure.

The elaboration model compares a pooled statistical association with the associations conditional on a third variable. Replication describes a similar pattern within the strata. Specification describes a relationship whose size or direction varies between strata. Attenuation can suggest explanation by an antecedent confounder, or interpretation through an intervening causal mediator. These causal readings require justified temporal and causal assumptions; conditional tables alone do not identify the third variable's role.^[ref]

Assess the patterns through conditional probabilities or comparable effect sizes. Different p-values also reflect different sample sizes and standard errors, so significant and nonsignificant results need not differ significantly.

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.

Predict categorical response $Y$ by a most frequent category. Without knowing $X$, the number of mistakes in a contingency table is $E_0=n-{\max }_j{n}_{+j}$. Allowing a separate most frequent response for each predictor row gives $E_1=n-{\sum }_i{\max }_j{n}_{ij}$. Goodman and Kruskal's lambda is the fractional reduction in those mistakes when $E_0>0$. Its direction matters, and it can vanish despite statistical association: if every row has the same most frequent response category, knowledge of $X$ changes no modal predictions even when the row distributions differ.

These are row percentages, so the comparisons estimate the conditional probability of recorded vaccination within each NHS trust category. The percentages increase across the ordered trust categories asThe extreme-category difference is approximately 49.31 percentage points, rather than a 49.31% relative increase. The ordinal categorical variable records an order; the categories do not establish equal numerical trust increments.

Under the null hypothesis of independence, the expected count in a cell is its row total times its column total divided by 748. The Pearson chi-squared statistic compares observed and expected counts throughThere are $(5-1)(2-1)=4$ statistical degrees of freedom, giving $p\approx 6.49\times 10^{-21}$. The minimum expected count is approximately 5.015, so the usual expected-count concern is much less acute here than in the income strata. The table gives strong evidence of a statistical association. The likelihood-ratio test of independence in a contingency table agrees: its statistic is approximately 81.361 on four statistical degrees of freedom, with a very small p-value. The displayed 0.000 is rounding, not a zero probability.

For magnitude, Cramér's V isIt describes the size of the departure from independence without using the category order or giving a direction. Calling it “moderate” is only a rough contextual description; the actual percentage differences are easier to interpret.

The ordered coefficients describe a positive direction. Goodman and Kruskal's gamma uses only concordant pairs and discordant pairs:It excludes tied pairs, which are frequent when the response has only two categories. It is therefore not a claim that 54.7% of respondents were vaccinated because of their trust. The directional Somers' D for vaccination as response includes pairs tied on that response in its denominator:The smaller value reflects the retained ties; it does not contradict the positive statistical association.

The small column-dependent Goodman and Kruskal's lambda has a useful prediction interpretation. Without trust information, always predicting the more frequent vaccinated category gives 121 mistakes. Predicting the majority category separately in each trust row gives $13+15+26+34+28=116$ mistakes. HenceTrust reduces this particular modal-prediction error by only about 4.13%, despite the strong evidence against independence. Most rows retain the same majority category. Evidence of association, its ordinal direction and improvement in a chosen prediction rule are different questions. None establishes that changing trust would change an individual's vaccination status.

The fitted simple linear regression isIts regression slope is a predicted decrease of approximately 0.006314 trust-score units per extra year, or 0.06314 units per decade. This is an estimated statistical association between age and the score. A cross-sectional study does not establish that the same person's trust decreases at this rate as they age.

The estimated standard error of the slope is 0.005427. Testing a zero regression slope gives $t\approx -1.163$ on 746 residual degrees of freedom and $p\approx 0.245$. An approximate 95% confidence interval isThere is insufficient evidence of a nonzero linear age association at the 5% level. The interval includes both small negative and small positive slopes; failure to reject does not establish an exactly zero relationship or rule out a nonlinear one.

The regression intercept 6.388920 predicts the score at age zero. That age is outside the observed 19–89 range, so interpreting it substantively would be extrapolation. Its very small reported p-value tests a zero intercept, rather than the age hypothesis. The 746 residual degrees of freedom correspond to 748 observations minus the intercept and slope.

The coefficient of determination is $R^2=0.001811$, so the fitted line explains approximately 0.1811% of the observed centered variation, rather than 18.11%. The adjusted coefficient of determination is approximately 0.0004732. The residual standard error is 2.602 score units, close to the reported response standard deviation; most variation remains unexplained by the fitted line. With one predictor and an intercept, the slope's F-test is the square of its Student's t-test statistic, up to printed rounding, so $F\approx 1.354$ and $p\approx 0.245$ agree with the slope test.

Output 6 is a residual-versus-fitted plot: its vertical coordinates should be regression residuals $e_i=y_i-{\hat{y}}_i$. The fitted values range over only approximately 5.83–6.27, as the weak slope predicts over ages 19–89. The printed scatter shows no obvious strong mean curvature or clear widening funnel, but it has marked horizontal concentrations near its extreme residual levels. An absence of obvious curvature does not prove a correct linear conditional mean, and the plot alone cannot establish homoscedasticity, independent errors or a normal distribution. A normal probability plot and residuals against observation order address different aspects. Large residuals also do not by themselves establish influential observations; regression leverage and Cook's distance matter.

There is an additional inconsistency in the original figure. The descriptive output reports observed responses between zero and ten, so residual constraints from a bounded response requireYet the printed graphic contains points around $(\hat{y},e)=(6.25,4.2)$, which would imply $y\approx 10.45$, above the stated observed maximum. Some lower-edge points likewise imply negative responses. This is present in the original PDF, rather than introduced by the transcription. The graphic therefore cannot be treated as a fully consistent diagnostic of the reported dataset. The observations about its visible shape above describe the supplied picture; reliable model checking would require the actual fitted values and residuals.

The printed code has separate typographical problems. The displayed model call supports the intended formula with ~, and the plot needs one closing parenthesis:

model1 <- lm(t_guard ~ age, data = exam)
plot(fitted(model1), residuals(model1),
     xlab = "Fitted values", ylab = "Residuals")
abline(h = 0)

These calls correct the syntax; they cannot be rerun on the original observations because the individual-level dataset is not supplied. The regression interpretation above uses the available numerical summaries and states the diagnostic limitation explicitly.

Each row of the contingency table is a conditional distribution of the safety category among respondents with that experience status. Thus the relevant comparison uses row percentages, rather than interpreting the larger number of respondents with some bad experiences as itself an effect on safety.

The sample proportions reporting feeling very safe are $39.59\%$ without bad experiences and $32.78\%$ with some bad experiences: the latter is lower by about $6.81$ percentage points. Feeling fairly safe rises from $28.49\%$ to $34.63\%$, a difference of about $6.14$ percentage points. Feeling unsafe changes only from $31.92\%$ to $32.59\%$, about $0.67$ percentage points. The main descriptive shift is therefore from very safe toward fairly safe, rather than a large increase in the unsafe category.

Pearson's chi-squared test tests the null hypothesis that experience status and safety category are independent. If that were true, the expected cell count would be its row total times column total divided by $n$. The Pearson chi-squared statistic sums squared observed-minus-expected count differences divided by their expected counts. HereThe expected counts are comfortably large, with minimum approximately $234.7$, so the ordinary large-sample chi-squared distribution approximation is appropriate under the sampling assumptions. There is evidence against independence even at the $1\%$ significance level. The reported likelihood-ratio result gives the same conclusion.

Magnitude is a separate question. Cramér's V giveswhich describes a weak statistical association. It does not mean that $7.7\%$ of safety is explained. Nor does the p-value measure the size of that association.

For ordinal direction, order the experience categories as none, some, and the safety categories from very safe toward unsafe. Goodman and Kruskal's gamma compares concordant pairs with discordant pairs after excluding tied pairs:Its small positive sign then corresponds to a weak tendency toward less safety among respondents with bad experiences. The sign would reverse if exactly one category ordering were reversed. Because this contingency table has many ties, the response-direction Somers' D, which retains response ties among pairs with different predictor values, is smaller:These coefficients describe ordered association; the omnibus Pearson's chi-squared test tests any departure from independence and does not supply a separate p-value for gamma.

The column-dependent Goodman and Kruskal's lambda makes the weak predictive improvement tangible. Predicting the overall most frequent safety category makes $1165$ errors in this table. Using a separate most frequent category within each experience row makes $1145$, giving $\lambda =20/1165\simeq 0.0172$, only about a $1.7\%$ reduction in those classification errors.

The sample supports a statistically detectable but weak association, chiefly a reduced sample proportion feeling very safe. This observational study does not establish that bad experiences cause the change: reverse direction, reporting differences or confounding could also contribute. The numerical p-value also presumes the sampling and complete-case assumptions appropriate to the observations used.

Use the account of statistical association between two variables in part b of the Experimental Psychology paper. It distinguishes the measured association from a causal conclusion.

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.

In a simple linear regression with conditional mean $\mathbb{E}[Y\mid X=x]=\alpha +\beta x$, the slope $\beta$ is the change in that mean per unit increase in the predictor. The ordinary least squares estimate is $\hat{\beta }=S_{xy}/S_{xx}$ when $S_{xx}>0$. It describes a statistical association; a causal interpretation requires further assumptions.

A statistical association is a systematic dependence between variables. It need not represent a causal effect, because it may arise from confounding, selection, or reverse causation.