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In multiple hypothesis testing, let $R$ count all rejections and $V$ count rejected true null hypotheses. The false discovery proportion is the random variable $V/(R\vee 1)$, defined to be zero when $R=0$. Its expectation is the false discovery rate. Distinguishing the realized proportion from its expectation matters: controlling the false discovery rate does not by itself give a high-probability bound on the realized proportion.

Conditioning on marital status produces five contingency tables. In each, compare the conditional row percentages. The following differences are the sample proportion with some bad experiences minus the sample proportion without them, in percentage points; negative very-safe differences mean fewer respondents feeling very safe in the former group.

The p-values test independence within the respective strata; the magnitudes and directions come from the conditional percentages and association measures. Differences in the table are calculated from the counts before rounding.

Among single respondents, the very-safe sample proportion changes from $33.95\%$ to $29.52\%$, while the unsafe sample proportion changes from $36.28\%$ to $38.10\%$. The direction resembles the weak pooled pattern, but $p=0.6114$ supplies little evidence against independence in this stratum. This is not evidence that the population association is exactly zero.

Among married respondents, the very-safe sample proportion changes from $41.18\%$ to $32.75\%$ and the fairly-safe sample proportion rises from $27.68\%$ to $35.39\%$. Unsafe remains almost unchanged, $31.14\%$ versus $31.87\%$. Thus the within-stratum association again chiefly shifts very safe toward fairly safe. Its $p=0.0257$ is below the $5\%$ significance level, while $V=0.092$ still indicates a small association.

For respondents living with a partner, very safe changes from $41.46\%$ to $28.32\%$, fairly safe from $30.08\%$ to $40.71\%$, and unsafe from $28.46\%$ to $30.97\%$. The very-safe difference is larger than in the married stratum. The printed p-value $0.04885$ would be just below the conventional $5\%$ significance level. However, this particular output is internally inconsistent: recalculating Pearson's chi-squared test from its six displayed counts gives $X^2=4.910389$, $p=0.08585$, and $V=0.14425$. The reported $V=0.144$ agrees with the counts, whereas the printed $X^2=6.900389$ would itself imply $p=0.03174$ for two statistical degrees of freedom, rather than the printed $0.04885$. Using the displayed counts, this stratum does not reject independence at $5\%$. The source PDF itself contains these incompatible figures; this is not resolved by rounding.

Among divorced respondents, very safe changes from $41.18\%$ to $37.17\%$, but unsafe also decreases, from $36.76\%$ to $30.97\%$, as fairly safe increases from $22.06\%$ to $31.86\%$. This is a redistribution toward the middle category, not a simple shift toward greater unsafety. The near-zero gamma, $-0.013$, reflects the lack of a consistent ordinal direction, while $V=0.106$ can still register distributional differences without a direction. The p-value $0.3583$ does not reject independence.

Among widowed respondents, very safe changes from $51.43\%$ to $42.47\%$, and unsafe from $14.29\%$ to $27.40\%$. The ordinal association is positive on the chosen ordering, with gamma $0.220$. Its lack of conventional statistical significance, $p=0.3158$, does not imply a smaller estimated association than in the married group: the widowed stratum has only $108$ complete observations, versus $857$ in the married stratum, and therefore less precision. All five tables have sufficiently large expected counts for the reported approximations; even the smallest expected count is about $8.10$.

The plot uses large-sample Wald confidence intervals for the difference of independent sample proportions within each stratum. They concern the individual category contrasts, whereas the tabulated Pearson's chi-squared tests compare the full three-category distributions. The intervals are unadjusted for multiple hypothesis testing; the pooled row reuses the stratum observations, so it is not a sixth independent sample.

In the elaboration model, the persistence of a generally weak positive ordinal pattern within four strata is closer to partial replication than to a complete explanation of the pooled association by marital status. The differences in estimated patterns suggest possible specification: the divorced group's shift toward the middle category differs from the widowed group's shift toward unsafe. They do not by themselves establish population effect modification. In particular, significant and nonsignificant results need not differ significantly; compare the effects using an interaction contrast or a fitted model with the appropriate interaction terms, rather than comparing their separate p-values.^[ref]

One additional omnibus check can be derived from the supplied tables without assuming a common effect. Recalculating the five statistics from the counts, including the corrected living-with-partner statistic, the joint conditional independence test for stratified contingency tables givesUnder independent-stratum sampling this tests independence within every stratum jointly; it does not test whether the association is homogeneous across strata. It does not reject at $5\%$, but this alone does not establish that stratification removed the population association: its ten-degree-of-freedom alternative is less focused than the pooled two-degree-of-freedom test, and nonrejection is not proof of equality.

The elaboration model also distinguishes an antecedent confounder from an intervening causal mediator. These cross-sectional tables do not establish the relevant time ordering or identify either role for marital status. Five separate stratum tests further require caution about multiple hypothesis testing: for example a Bonferroni correction at familywise $5\%$ would use $0.01$ per test, below all five individual p-values. That caution changes the interpretation of selected significance labels, not the observed conditional percentages.

Both fits are the same additive two-way analysis of variance model: a common intercept, five country effects represented by four contrasts, and eight item effects represented by seven contrasts. The three missing cells are omitted, leaving $37$ observations and model rank $1+4+7=12$, hence $25$ residual degrees of freedom. The residual sum of squares is $659.44$ in either ordering, giving

These are sequential sums of squares, so each row measures the improvement when that factor is added after the preceding factors. The country-first sum $1115.56$ describes its unadjusted contribution before accounting for item; its displayed $F=10.573$ and $p=3.73\times 10^{-5}$ are not the country-adjusted-for-item test. When country is added after item, its extra sum is $1616.74$, givingThus the additive normal linear model gives strong evidence of country differences after accounting for the different products. Conversely, item added after country has extra sum $16910.20$, soThe printed zero for this probability is numerical formatting, not a probability literally equal to zero. Product differences are much larger than residual variation, as expected from their different scales.

The explanation for the order dependence is that missing cells can destroy factor orthogonality in two-way ANOVA. A complete equally replicated crossed design would make centered country and item indicator columns orthogonal. Here the missing observations change the item mix across countries and the country mix across items, so the two factor subspaces are not orthogonal. The two sequential partitions still have the same total explained sum: $1115.56+16910.20=16409.02+1616.74=18025.76$ to the displayed precision. For inference about either factor conditional on the other, compare the full model with the reduced model that omits that factor, rather than interpreting its first-entry sum as adjusted evidence.

The omnibus country test does not identify the particular country contrasts. As an extension, fitting the supplied price table gives the item-adjusted UK-minus-US contrast about $21.33$ pounds, with standard error $2.824$ and an ordinary Gaussian-model $95\%$ confidence interval approximately $[15.51,27.14]$ pounds. This is a derived model contrast, with Student t-distribution reference on $25$ degrees of freedom; simultaneous exploration of many country pairs requires control for multiple hypothesis testing.

Check residuals against fitted values, product and country, together with a normal Q-Q plot and influential observations. The homoscedasticity assumption on absolute prices may be doubtful because expensive products can have larger absolute variation. A model for logarithmic prices gives multiplicative country effects and can make the scale assumption more plausible. Country-by-item interaction terms would allow discounts to differ by product, but there is at most one observed value per cell: an unrestricted interaction saturates the $37$ observed means and leaves no pure-error degrees of freedom, so it cannot be tested against independent measurement noise from these data alone. Additional products or genuine cell replication would support that extension. The selected products and non-random missing entries also limit claims about a whole country's consumer prices; the conditional inference is for the stated additive model and observed price collection.

scan reads the numerical prices, including the three missing-value markers, into a vector. The data must be read row by row for the subsequent factor construction to match them. The sample size available for analysis is therefore $37$, rather than $40$. The numerical summary describes the observed prices: the sample median is £43.54 and the sample mean is £46.94. The upper tail, particularly the boots, raises the sample mean above the sample median. These summaries mix very different products and are not adjusted country comparisons.

The two character scans create labels. gl(8,5,length=40,labels=item) makes a regression factor with each of eight item labels repeated five times. gl(5,1,length=40,labels=country) cycles through the five country labels once for each item. For a regression factor and numerical response, the two plot commands produce grouped box plots. The country box plots suggest lower US prices, while the item box plots show particularly expensive boots and inexpensive cardigans. They are descriptive displays of this small basket, with different item compositions where prices are missing; they cannot separate country effects from item effects.

The first lm fits an additive two-way analysis of variance:The constraints are R's usual treatment coding. The normal linear model used for the printed tests assumes independent errors with common variance and approximately a normal distribution, and no item-by-country interaction in the mean. Missing prices are omitted, leaving $37$ responses and $1+4+7=12$ independent mean coefficients. Thus the residual degrees of freedom are $37-12=25$. This deletion is defensible for the conditional price model when missingness does not depend on the unobserved error after conditioning on the two factors; the output itself does not establish such a missing-data mechanism.

The anova tables report sequential sums of squares, rather than order-independent adjusted effects. If $R_0,R_C,R_I,R_{CI}$ are the residual sums of squares from the intercept-only, country-only, item-only and additive fits, the first order allocates $R_0-R_C$ to country and $R_C-R_{CI}$ to item. The second order allocates $R_0-R_I$ to item and $R_I-R_{CI}$ to country. Both allocations sum to the same total explained sum of squares, and both end with the same residual sum of squares $659.4$. The three missing combinations destroy balanced factorial orthogonality, so the allocation changes with order. To test country after allowing for items, use the country row in the second table; to test item after allowing for countries, use the item row in the first table. The first-entered rows do not supply those adjusted tests.

Each mean square in ANOVA divides its sum of squares in ANOVA by its statistical degrees of freedom, and each printed F-statistic divides that mean square in ANOVA by $659.4/25=26.38$. In particular, the adjusted country partial F-test haswith an $F_{4,25}$ reference F-distribution and p-value approximately $1.86\times 10^{-6}$. The adjusted item partial F-test gives $F=91.58$ on $7,25$ statistical degrees of freedom. Both factors contribute strongly under the additive normal linear model. The larger item sum of squares in ANOVA measures the much greater differences between these products, but is not evidence that every pair of items differs significantly.

The intercept $50.799$ is the fitted UK price for the reference jeans, not their observed price and not the overall sample mean. Each country regression coefficient is its adjusted price difference from the UK, assumed common across items: Sweden is £5.224 lower, France £10.708 lower, Germany £7.676 lower and the US £21.328 lower. Each item regression coefficient is its adjusted difference from the reference jeans, assumed common across countries. Thus the fitted UK boots price is $50.799+52.854=103.653$, and the fitted US boots price is $103.653-21.328=82.325$.

The coefficient standard errors come from $s^2(X^{T}X{)}^{-1}$, where $s^2=659.4/25$ and $X$ is the design matrix. Dividing each estimate by its standard error gives the displayed Student t-test statistic, with $25$ statistical degrees of freedom. The stars summarize the corresponding two-sided p-values. The individual tests compare each indicated country or item with its reference level, conditional on all other additive terms; they are neither tests of every pairwise comparison nor a simultaneous confidence interval procedure. In particular, Sweden's p-value $0.0626$ provides weaker evidence than the other country contrasts, and does not establish equality with the UK. The intercept test has little substantive relevance, and interpreting many individual stars also raises multiple hypothesis testing issues.

The residual standard error $s=5.136$ measures unexplained prices in pounds. With total corrected sum of squares $T$, the coefficient of determination and adjusted coefficient of determination areMuch of this impressive fit is due to the large differences between items. The overall F-test compares the additive fit with a constant mean: $F=((T-659.4)/11)/(659.4/25)=62.12$ on $11,25$ statistical degrees of freedom. It tests that all eleven non-intercept coefficients vanish, not that the additive model is adequate. Regression diagnostics should examine regression residuals, changing price variability across items and possible interaction. Percentage differences may be more plausible than common pound differences, motivating a natural logarithm of price. No random sample of countries or products, and no independent replication of a price within a cell, is supplied; broad claims about all designer goods or the causes of country differences require more than these conditional tests.

Finally, Item*Country expands to both main effects and their full interaction. A cell-means parameterization has one separate fitted mean for every observed cell. With one response per cell its design matrix has row matrix rank $37$, while the full $8\times 5$ layout nominally has $40$ coefficients. Three coefficient combinations are aliased because their cells are missing. This is saturation of an unreplicated two-factor regression: every observed price is fitted exactly and the residual degrees of freedom are zero. The final fit has zero residuals, $R^2=1$ and no residual-based estimate of the error variance or ordinary interaction significance test. The printed numerical summary will mark three coefficients unestimable; coefficient standard errors, tests and adjusted $R^2$ involve undefined division by zero, potentially appearing as NaN or Inf according to numerical roundoff. Perfect interpolation here is a consequence of a saturated statistical model, not evidence for a superior predictive model.

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.