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A latent factor is an unobserved coordinate introduced to explain shared variation among measurements. In an orthogonal factor model, latent factors have variance one and are mutually uncorrelated, with coefficients specified by the factor loadings. They differ from a regression factor, which is a categorical predictor. A fitted covariance alone generally leaves the latent factor orientation undetermined; factor rotation gives equivalent representations.

The S-Plus input and the data table differ in one entry: the former uses $33$ for the first year in the second male age group, whereas the table has $22$. The displayed coefficients are reproduced by the software vector with $33$, so that is the input analysed here. Using $22$ instead would give a different fit, including an intercept of $36.175$ rather than $37.275$.

The factor grid has $5\times 4\times 2=40$ rows. Its first factor varies fastest, so the five years run within each age group, and the age groups run within each sex. All three predictors are regression factors: in particular the year coefficients are category contrasts, not a fitted linear slope in calendar time. Under reference-level coding, take the first year, youngest age group and men as the reference levels in a regression factor. Write $s=1$ for women and $0$ for men, and $I_{yk},I_{aj}$ for the other year and age indicators. The normal linear model isThe star in the formula includes both main effects and their statistical interaction. There are twelve coefficients: an intercept, four year contrasts, one sex contrast, three age contrasts and three sex-by-age contrasts. Year effects are assumed common across all sex-age combinations; there is no year interaction term. Independent errors with common variance are additional assumptions of the reported exact tests.

The least-squares estimator is $\hat{b}=(X^{T}X{)}^{-1}{X}^{T}P$, where $X$ has rank twelve. Under the normal linear model,and these quantities are independent. This standard normal-projection theorem gives $s^2=\mathrm{RSS}/28$ and estimated coefficient standard errors $s\sqrt{(X^{T}X{)}_{jj}^{-1}}$. A coefficient estimate divided by its standard error has a Student t-distribution with 28 degrees of freedom under its zero-coefficient null. These are the printed t statistics and two-sided p-values; a printed zero is rounding, not a probability of exactly zero. The option suppresses the coefficient-correlation display.

The fitted intercept $37.275$ is the expected percentage for the reference sex, age and year. Relative to that year, the year changes are $0.375,0.500,1.750,3.000$ percentage points. The last two are individually significant at the conventional five-percent level; the first two are not. The sex coefficient $-18.8$ is the female-minus-male difference in the youngest group. The male age changes are $-7,-12.8,-23$ points. Adding the three interaction term coefficients gives the female age changes $-5.2,-8.2,-13.8$. Thus the sex gap shrinks with age. An individual insignificant interaction term coefficient does not justify deleting the whole interaction: the joint partial F-test for the three interaction terms gives $F=18.6822$ on $(3,28)$ degrees of freedom, with $p\simeq 7.46\times 10^{-7}$.

The regression residuals are observed minus fitted percentages; their printed five-number summary describes their spread. Direct calculation gives $\mathrm{RSS}=60.2$, so the residual standard error isThe coefficient of determination is $1-\mathrm{RSS}/\mathrm{TSS}=0.985827$. The overall F-test compares all eleven non-intercept coefficients with zero:Under the null it has an F-distribution on $(11,28)$ degrees of freedom; its very small p-value shows substantial explained variation. This does not independently validate the Gaussian error assumptions. Survey percentages can have different sampling variances and correlations, and the survey denominators and design are not supplied. The output's exact inferential interpretation is conditional on the stated normal linear model.

The final command makes an interaction plot, not another regression fit. Its points are arithmetic averages over the five years for each sex-age combination. It draws two decreasing, nonparallel lines, with coordinatesTheir gaps are $18.8,17.0,14.2,9.6$ percentage points, illustrating the sex-by-age statistical interaction. The average uses the software vector's disputed entry, consistently with the displayed fit. These are means of percentages, not denominator-weighted pooled prevalence estimates.

The prose names a second year that differs from the table label. The calculations use the four supplied counts, interpreting the first and second rows as the two displayed periods; no numerical conclusion depends on resolving that label discrepancy.

The first generalized linear model treats the four counts as independent Poisson random variables, with a log link and both row and column regression factors. The row-by-column interaction makes the log-linear model saturated. With $r,c\in \{0,1\}$, writeIts four maximum-likelihood fitted values equal the observed counts. Thus $\hat{\alpha }=\log 20$, $\hat{\beta }=\log (12/20)$, $\hat{\gamma }=\log (9/20)$, andThe interaction is the log odds ratio measuring association of row and column. Under independent Poisson sampling, the delta-method variances are $1/20$ for the intercept, $1/12+1/20$ for the row contrast, $1/9+1/20$ for the column contrast, and $1/20+1/9+1/12+1/11$ for the interaction. Their square roots reproduce the printed standard errors. Although the software labels the standardized ratios as t values, these are asymptotic normal Wald statistics, with known Poisson dispersion one, not exact finite-df Student tests.

The statistical saturated model has zero residual deviance and zero residual degrees of freedom; this is a perfect interpolation, not evidence of predictive adequacy. The null deviance $5.016056$ compares the four observed counts with an intercept-only model, whose four fitted means are $52/4=13$. Fisher scoring is the numerical score/information iteration used to fit the model; the iteration count is a convergence report, not a test statistic.

Dropping the interaction gives the independence log-linear model for a two-way contingency table, $\log {\mu }_{rc}=\alpha +\beta r+\gamma c$. Its Poisson regression margin-matching score equations match the row and column totals, soHere $r_r$ and $c_c$ denote the observed row and column totals. In particular the fitted row multiplier is $23/29$, the column multiplier is $20/32$, and the intercept is $\log (29\cdot 32/52)=2.881788$. These produce the second coefficient table. Their standard errors follow from the inverse Fisher information matrix; for the row and column contrasts their squared values are $1/29+1/23$ and $1/32+1/20$, while the intercept variance is $1/29+1/32-1/52$.

The Poisson deviance relative to the saturated fit isThe linear terms cancel because fitted and observed totals agree. The additive model has three coefficients, leaving one degree of freedom. Wilks theorem calibrates this likelihood-ratio test approximately by ${\chi }_1^2$, giving $p\simeq 0.2164$. The interaction Wald test instead gives $p\simeq 0.2192$; the two are only asymptotically equivalent.

Fisher's exact test conditions on both margins under independence. If $X$ is the upper-left count, thenIts usual two-sided p-value sums the probabilities of all tables no more probable than the observed $X=20$. That sum is $0.259724$, reproducing the output without an asymptotic approximation. It tests the same absence of row-column association, but conditioning and discreteness explain the numerical difference from the likelihood-ratio test.

In ordinary language, the male proportion among the cases is $20/29\simeq 69.0\%$ in the first row and $12/23\simeq 52.2\%$ in the second. The estimated case-sex odds ratio is $2.037$, but its approximate 95-percent confidence interval is wide, $(0.655,6.338)$. These small samples do not give convincing evidence that the sex composition of cases changed between the two periods. Failure to reject is not evidence that the compositions are exactly equal. Nor do these counts establish a male-to-female disease-incidence ratio or an incidence trend: population denominators and comparable ascertainment would be needed for those interpretations.

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.

Converting size to a regression factor allows two unrestricted contrasts rather than imposing a linear trend on numerical size codes. The command relevel with a numeric second argument selects the second existing factor level as its reference level. For sorted codes $0,1,2$ this is medium, coded $1$. However, the displayed coefficient labels size1 and size3 instead correspond to codes $1,2,3$ with code $2$ as reference. The printed coding and coefficient labels are inconsistent. The actual factor levels and labels must be checked. Under the intended medium-size reference, the two reported contrasts compare small and large businesses with medium ones; with literal $0,1,2$ coding their printed names would be size0 and size2.

Using B's delayed-entry survival object, the Cox proportional-hazards model isHere $h_0$ is the unspecified baseline hazard for a medium-size village business and $c$ is the Cambridge indicator. The fitted hazard ratios are conditional on the other covariates and constant over business age under the proportional hazards assumption. The function fits regression coefficients through partial likelihood; its survival object must actually be B's version, rather than A's earlier object with the same name.

The location coefficient $-0.803$, with standard error $0.344$, gives $z=-2.335$ and a two-sided Wald test p-value about $0.02$. Its reported hazard ratio is $0.448$, with 95% confidence interval $(0.228,0.879)$. At the same age and size, a Cambridge business therefore has an estimated closure hazard about 55.2% lower than a village business. The reciprocal ratio is $2.23$. A hazard ratio is not a survival-probability ratio or a proportional increase in lifetime.

The intended small-versus-medium contrast is $-0.322$, with standard error $0.482$, $z=-0.669$ and p-value $0.50$. Its hazard ratio is $0.724$, with 95% confidence interval $(0.282,1.863)$. The intended large-versus-medium contrast is $-0.224$, with standard error $0.375$, $z=-0.598$ and p-value $0.55$; its hazard ratio is $0.799$, with interval $(0.383,1.666)$. Neither size contrast provides convincing evidence of an effect. The wide intervals also show why these results do not establish equality of size-specific hazards.

The sample has $60$ businesses. The likelihood-ratio test, Wald test and score test jointly test all three coefficients being zero; their statistics are $5.45$, $5.76$ and $6.01$ on three degrees of freedom, with p-values $0.142$, $0.124$ and $0.111$. None rejects at 5%. This is compatible with the significant one-coefficient location test: testing a specified one-dimensional contrast and testing three coefficients together are different questions. The location result offers some adjusted evidence, but the joint output does not demonstrate a strong overall improvement, and interpretation should acknowledge the uncertainty and the number of comparisons.

The reported $R^2=0.087$ is a Cox–Snell likelihood pseudo-R-squared, here $1-e^{-5.45/60}\approx 0.087$. It measures likelihood improvement rather than a literal fraction of variance in lifetimes explained. Its attainable maximum is reported as $0.979$. It does not validate the proportional hazards assumption or establish useful predictive performance. Adjusted survival predictions would require estimating the baseline survival function as well: $S(t\mid z)=S_0(t{)}^{e^{{\beta }^{T}z}}$. The coefficient summary alone does not supply adjusted five- or ten-year survival probabilities.

Before reporting results, check the size coding discrepancy, the precise sampling dates and the survival records. The stated January 2000–December 2005 span is six calendar years, despite being called five years. Check nonnegative ages, entry no later than exit, correctly recorded closure indicators and the same opening-based clock throughout. Establish whether inclusion and censoring are plausibly independent of lifetime given the covariates, and whether different opening cohorts or calendar environments can reasonably share the fitted lifetime model. Simply including delayed entry does not cure informative selection, omitted confounders or dependence between businesses.

Assess proportional hazards using Schoenfeld residuals against time and a proportional hazards assumption test, supplemented by groupwise log-minus-log survival plots. Investigate influential observations with coefficient-deletion diagnostics and deviance residuals, and inspect overall fit using Cox–Snell residuals. Consider a city-by-size interaction or time-varying effects if supported by the available events, rather than relying on the displayed three main effects. Finally, report the uncertainty and declining late risk sets—only six remain at the final listed event—and distinguish a conditional association from a causal explanation or a prediction for all future businesses.

A regression factor is a categorical predictor represented by level indicators and contrasts. Treating year as a regression factor permits arbitrary differences between the three yearly means, reflecting conditions such as weather, rather than imposing a linear time trend.

Let $Y_{ijk}$ be the yield for seed variety $i\in \{1,2\}$, fertiliser level $j\in \{L,H\}$, and year $k\in \{2004,2005,2006\}$. A treatment-coded two-factor normal linear model with additive year effects iswith independent $N(0,{\sigma }^2)$ errors. The unrestricted coefficients are ${\beta }_0,a,b,c_{2005},c_{2006},d$: six parameters for twelve observations. Equivalently use baseline-zero variety, fertiliser, and interaction arrays. There are no year-treatment interactions in this fit.

Dividing each sum of squares by its statistical degrees of freedom gives the completed analysis of variance:In particular, the missing interaction statistic is $16.0083/(2.8414/6)\simeq 33.803$. Its null hypothesis is $H_0:d=0$, against $H_1:d\ne 0$: the change in mean yield from high fertiliser is the same for both varieties. Under the null hypothesis and the independent homoscedastic normal linear model, this nested-model F-test has distribution $F_{1,6}$. The printed $p$-value 0.001137 strongly rejects no interaction.

Do not drop fertiliser just because its main-effect ANOVA row has $p=0.7011$. In this balanced factorial design that row measures a fertiliser effect averaged over varieties; large opposite effects can cancel. The strongly supported interaction requires retaining its associated main effects by regression model hierarchy. Seed and year also have evidence of effects in their ANOVA rows, so this table alone gives no compelling simplification of the fitted terms.

The interaction contrast is $-4.6200$, with standard error 0.7946. The fitted high-minus-low fertiliser effect is $+2.4700$ for variety 1 but $2.4700-4.6200=-2.1500$ for variety 2. The fitted variety-2-minus-variety-1 contrast is $+3.6500$ under low fertiliser and $3.6500-4.6200=-0.9700$ under high fertiliser. Thus variety 2 performs best under low fertiliser, while high fertiliser benefits variety 1 and reduces variety 2's fitted yield. The marginal fertiliser effect is only $(2.4700-2.1500)/2=0.1600$.

The four fitted means in the baseline year are $5.5050,7.9750,9.1550,7.0050$ for $(1,L),(1,H),(2,L),(2,H)$ respectively. Add $1.5775$ in 2005 and $0.0875$ in 2006 to every fitted mean. The 2005 coefficient has $p=0.017646$ under a $t_6$ test of a zero contrast with 2004; the analogous 2006 contrast has $p=0.863216$. The latter is lack of evidence for a difference, not proof that the yearly means coincide. Standard errors for the combined contrasts require the coefficient covariance matrix. The experiment is small: residual standard error is 0.6882 on only six statistical degrees of freedom. If the same fields were reused, possible within-field dependence and field allocation would need checking before treating the fitted error independence as established.

A regression factor is a categorical variable encoded by indicator columns or treatment contrasts in a statistical model. A factor with $k$ observed levels normally contributes $k-1$ statistical degrees of freedom when an intercept is present. The reference level in a regression factor has coefficient zero under treatment contrasts. Coding a year as a factor permits arbitrary year effects rather than imposing a linear trend.