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Analysis of variance decomposes variation into contributions associated with model terms and residual error. For nested normal linear models, differences of residual sums of squares give sequential or partial F-tests.

For one response per cell in an order-$q$ Latin square, fit an intercept and additive row, column and treatment effects with sum-to-zero constraints. Each factor contributes $q-1$ statistical degrees of freedom; the residual has $q^2-1-3(q-1)=(q-1)(q-2)$. The centered factor spaces are mutually orthogonal, so sequential sums of squares do not depend on fitting order. With an additional orthogonal symbol factor from a Graeco-Latin square, residual statistical degrees of freedom become $(q-1)(q-3)$. Exact F-tests require independent equal-variance normal errors; unmodeled interactions can invalidate the error term.

For nested normal linear models, the decrease in residual sum of squares obtained by adding coefficients is their extra sum of squares. Dividing by the number of added coefficients and by the full-model residual mean square gives the partial F-test. A sequential analysis of variance attributes this decrease according to the specified order of entry; it need not equal the partial comparison when other terms are retained.

An analysis of variance comparing a quantitative response variable across levels of one factor. The ordinary independent-groups F-test evaluates equality of group population means under the one-way normal linear model. Rejection says that not all means are equal, rather than identifying every differing pair.

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.

The ordinary least squares fit estimates the conditional mean of the continuous safety response as a linear function of the income coding:The regression coefficient for income is positive. Since larger response values mean greater safety, one additional coded income unit corresponds to an estimated increase of $0.2369$ safety-scale units in the fitted mean. Following the question's stated units, a unit is a £10,000 income step. The observed coding also uses income bands with open end categories, so this is a linear trend per coded band step; it is not evidence of a known constant effect for every extra pound, especially within the open top band. The output's abbreviated variable label does not override the stated units.

The intercept is the fitted mean at income code zero. It is estimated as $6.5933$, but zero is outside the observed income codes, so that value is an extrapolated model baseline, not an observed estimate for a particular income category. Its small p-value tests a zero intercept; it is not the main test of the income association.

For the income coefficient, the standard error is $0.0604$. The reported test statistic and two-sided test giveproviding strong evidence against the null hypothesis of zero linear slope under the stated model assumptions. A $95\%$ Student t confidence interval, using $1727$ residual statistical degrees of freedom, isThis is an interval for the population linear coefficient under the model, not an interval containing $95\%$ of individual safety scores. The large-sample estimate is precise enough to detect a positive association, but that does not establish substantial individual predictive power.

The coefficient of determination is $R^2=0.008828$: the fitted line accounts for about $0.883\%$ of the in-sample variation about the response sample mean. The adjusted coefficient of determination is about $0.825\%$. Thus income alone explains very little of the observed variation, despite the small slope p-value. The residual standard error of $3.997$ response units describes substantial scatter about the fitted line. This is an estimated residual standard deviation, rather than the standard error of the slope or a complete individual prediction interval.

The overall F-test has $F=15.38$ on $1,1727$ statistical degrees of freedom and $p=9.129\times 10^{-5}$. With one predictor and an intercept, it tests the same zero-slope hypothesis as the coefficient Student's t-test; $F=t^2$, up to rounding. It is not an additional independent confirmation. In this simple linear regression, the Pearson correlation coefficient has the slope's sign and magnitude $\sqrt{R^2}\simeq 0.0940$, another description of the weak linear association.

The regression uses $1729$ complete observations and estimates two coefficients, giving $1729-2=1727$ residual statistical degrees of freedom. The deletion of $88$ observations is complete-case analysis; extending its conclusions to the full target population requires suitable assumptions about the missing-data mechanism. Classical reported standard errors also rely on an adequate linear conditional mean, independent errors and homoscedasticity; a normal linear model gives the finite-sample Student t-distribution and F-test calibration, while large-sample approximations can justify similar tests under weaker distributional assumptions. The printed summary alone does not establish these assumptions. Regression diagnostics and, where appropriate, heteroskedasticity-consistent standard errors address different parts of that assessment.

Finally, this unadjusted observational study estimates an association. Other covariates, such as age or experience status, may be related to both income and reported safety. Without additional design or causal identification assumptions, the positive fitted slope does not establish that increasing income would cause an increase in safety.

TODO: the effect-specific $F$ values are unavailable without the supplementary output. Retrieve the rows for congruence, delay and congruence-by-delay interaction, if those terms were fitted. In an appropriate analysis of variance, each F-test compares a term's mean square in ANOVA with its designated error mean square in ANOVA; repeated observations can require different error terms. Neither the factor labels nor the recruited sample size determines those ratios.

TODO: an empirical Results section cannot be completed without the analysis and descriptive output. It must identify the actual factorial design, outcome scale and fitted model; summarize the four condition means with appropriately labeled standard deviations or standard errors; and report the congruence, delay and interaction tests with their actual F-test statistics, statistical degrees of freedom, p-values, effect sizes and relevant confidence intervals.

Interpret an interaction as evidence that the cue-congruence difference varies with delay, not merely as evidence that one separate test was significant and another was not. Any simple-effects analysis must identify its contrasts and multiple testing treatment. If the analysis used transformed reaction times, distinguish the model's scale from raw-second summaries. A nonsignificant result does not establish equality or absence of a theoretically meaningful effect. These are reporting requirements, not placeholders for invented results.

Use one-way analysis of variance to compare a quantitative response variable across levels of one categorical factor; for example, compare task accuracy across three independently assigned teaching methods. The ordinary independent-groups procedure tests the null hypothesis that all group population means are equal. Its F-test compares variation associated with the group factor against within-group error variation; a significant result indicates that not all population means are equal, without identifying which groups differ.^[ref]

The classical one-way normal linear model assumes independent observations and within-group errors approximately following a normal distribution with a common variance. Repeated observations on the same people require a different dependence structure; pronounced unequal variances may call for Welch's one-way analysis of variance. Planned contrasts or appropriately controlled follow-up comparisons identify scientifically relevant differences. Report the means, differences and confidence intervals, not just significance. Causal interpretation depends on the design of experiments, particularly randomization, rather than on the F-test alone.

The normality assumption concerns the model's error distribution, not necessarily the pooled raw observations. In an independent-groups analysis of variance, errors around each group mean are modeled as following a normal distribution; combining groups with genuinely different means can produce a nonnormal pooled distribution even when this assumption holds. In a paired t-test, the relevant distribution is that of within-pair differences. Linear regression conventionally assumes normal errors conditional on the predictors for exact small-sample Student t-tests and F-tests.^[ref]

Inspect suitable within-group or fitted-model regression residuals using a quantile-quantile plot and examine outliers and asymmetry. A formal normality test is not a decisive automatic gatekeeper: with small samples it may miss important departures, while large samples can detect minor departures. Approximate robustness depends on the sample size, balance, tail behavior and intended inference. The central limit theorem concerns sampling distributions and does not make every raw observation normal or repair dependence. Depending on the question, a justified transformation, a better-fitting probability model, robust inference or an appropriate bootstrap may be preferable. The normality assumption, independence and equal variance are distinct assumptions.

Write $1$ for the all-ones vector and setThe centering assumption makes $P_0{P}_X=0$. Thus the full normal linear model has fitted-value orthogonal projection $P_0+P_X$, and its residual sum of squares is $R=Y^{\mathsf{T}}(I-P_0-P_X)Y$. Assume $\nu >0$, as is needed to estimate the unknown error variance from residuals.

For the null that all slopes vanish, the reduced fitted value is $\bar{Y}1$, with residual sum of squares $R_0=Y^{\mathsf{T}}(I-P_0)Y$. The F-test isReject for an upper-tail value exceeding the $1-\alpha$ quantile of this F-distribution. The required theorem is Cochran's theorem: for an isotropic normal distribution, projections onto orthogonal subspaces are independent, and the squared norm on a rank-$r$ subspace divided by ${\sigma }^2$ has a chi-squared distribution with $r$ degrees of freedom. Here the null makes the projected mean in $\mathrm{col}(X)$ zero, so $(R_0-R)/{\sigma }^2\sim {\chi }_p^2$, independently of $R/{\sigma }^2\sim {\chi }_{\nu }^2$.

Let $p_1,p_2$ be the numbers of columns in the two predictor blocks. To test the first block while retaining the second as nuisance, formThe partial F-test for nested linear models isAgain reject in the upper tail. The difference of the nested fitted-value projections has rank $p_1$, is orthogonal to the full residual projection, and annihilates the null mean, which gives the same independent chi-squared argument. If a block has no columns, the corresponding test is vacuous.

The phrase orthogonal statistical parameters refers to zero cross-block Fisher information, not to the numerical parameter vectors being perpendicular. In this model the cross-block information is $X_1^{\mathsf{T}}{X}_2/{\sigma }^2$, so the orthogonal coefficient blocks in a centered normal linear model condition isThen the two fitted subspaces are orthogonal, the block least-squares estimators have zero covariance and are independent because they are jointly Gaussian, and each block's extra sum of squares is unchanged by including the other block. Centering also makes both blocks orthogonal to the intercept.

A randomized complete block design groups similar experimental units into experimental blocks and places every treatment once in each experimental block, independently randomizing treatment allocation within experimental blocks. Blocking removes background between-block variation from within-block treatment comparisons. A balanced incomplete block design has $v$ treatments and $b$ experimental blocks of size $k<v$, with each treatment occurring in $r$ experimental blocks and every distinct treatment pair together in $\lambda$ experimental blocks. Counting treatment occurrences and co-occurrences gives $vr=bk$ and $r(k-1)=\lambda (v-1)$.

A Latin square of order $q$ places each of $q$ treatment symbols once in every row and every column of a $q\times q$ array. This controls two nuisance directions. A Graeco-Latin square superposes two mutually orthogonal Latin squares: each square is Latin, and each ordered pair of symbols from the two squares occurs exactly once. The two symbol factors can then be fitted together with row and column effects. Random permutations of row labels, column labels and treatment symbols provide a suitable randomized allocation without destroying these balances.

For the operator experiment, fit the additive normal linear modelHere $r_i,c_j$ account for the two fabric directions, and ${\tau }_o$ is the operator effect. Additivity, independent equal-variance errors and the normal assumption supply the exact F-test; there is only one observation per cell, so an unrestricted interaction model is not available.

In the analysis of variance for a Latin square, the three centered factor spaces each have dimension $4-1=3$. Together with the intercept the model has rank ten, leaving $16-10=6$ residual statistical degrees of freedom. The corrected total has $16-1=15$. The completed calculation isUnder the null hypothesis ${\tau }_1=\cdots ={\tau }_4=0$, the operator F-statistic isThe upper five-percent critical value for $F_{3,6}$ is $4.76$, so the operator effect is not significant at five percent in this model.

The decision is independent of the order of fitting rows, columns and operators. To prove the required orthogonality, let $u_i$ be any centered row coefficients and $w_o$ any centered operator coefficients. Their observation-space inner product is ${\sum }_{i,j}{u}_i{w}_{o(i,j)}={\sum }_i{u}_i{\sum }_o{w}_o=0$, since every operator occurs once in each row. The same calculation applies to columns and operators, and all row-column pairs occur once as well. Thus the three centered spaces are mutually orthogonal; their projections commute and sequential sums of squares equal adjusted sums of squares in every fitting order.

With machines included, the layout is a Graeco-Latin square: the machine symbols are Latin, and all sixteen operator-machine pairs occur exactly once. Fitin addition to the earlier constraints. The machine space contributes three new statistical degrees of freedom and is orthogonal to the previous factor spaces, so the operator sum of squares remains $18.25$. The residual sum of squares and statistical degrees of freedom becomeHenceThis exceeds $F_{3,3}(0.05)=9.28$ but not $F_{3,3}(0.01)=29.46$. Therefore the operator effect is now significant at five percent, but not at one percent. Adjustment for machines has removed substantial residual variation without altering the operator numerator. This illustrates why a balanced nuisance factor can change precision and significance even when it does not change the estimated treatment contrasts.

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 commands read the 52 cell values into a vector and build the matching two-factor layout. In expand.grid(country, goods) the first factor varies fastest, so the four countries are listed successively within each goods category. This matches row-by-row reading of the displayed table. Converting Goods to a factor is essential: its labels identify 13 categories and are not a quantitative trend. Country supplies the other categorical factor. The summaries suppress coefficient correlations, not coefficient standard errors.

The first fit is the additive two-factor normal linear modelThere are $1+12+3=16$ free coefficients and $52-16=36$ residual statistical degrees of freedom. This is two-way analysis of variance without an interaction term. For this complete balanced layout, ordinary least squares givesThese formulas follow from the row- and column-sum normal equations: fitted row and column totals equal their observed totals. The balanced-design coefficient standard errors are $s\sqrt{2/4}$ for a goods contrast, $s\sqrt{2/13}$ for a country contrast, and $s\sqrt{1/4+1/13-1/52}$ for the reference-cell intercept. These yield the common contrast standard errors and the different intercept standard error in each summary. In particular the intercept is the fitted Software–UK value, not the grand mean or the observed Software–UK entry.

Thus the first intercept predicts $149.8269$ million euros for Software in the UK. The Books coefficient adds $254.7$ million euros to the fitted value in any country, relative to Software; every other goods coefficient has this same arithmetic-difference interpretation. The country coefficients subtract $193.7$, $57.0308$ and $97.3769$ million euros from the UK prediction for France, Germany and the rest of Europe respectively, regardless of goods. The model therefore assumes common absolute country differences across goods categories. For Software–France its prediction iswhich is impossible for positive spending and indicates a limitation of this additive original-scale specification.

The residual standard error is $80.98$ million euros, obtained from $s^2=\mathrm{RSS}/36$; direct calculation gives $\mathrm{RSS}=236088.7435$. Each coefficient's printed statistic is $\hat{\theta }/\mathrm{se}(\hat{\theta })$, with a $t_{36}$ reference distribution under this normal linear model. The two-sided p-values compare each goods category with Software, or each other country with the UK, after adjustment for the other factor. For example France and the rest of Europe differ significantly from the UK at 5%, but Germany does not. The raw-scale significant goods differences are Books, Clothing, Electronic Goods, Groceries and Leisure Travel; these are individual reference-level comparisons, not an overall test of the goods factor.

The coefficient of determination is $R^2=1-\mathrm{RSS}/\mathrm{TSS}=0.7743$. The omnibus F-test compares all 15 nonintercept coefficients with zero:with reference $F_{15,36}$ and $p=1.2544\times 10^{-7}$. This says at least one factor contrast is nonzero under the assumed error model. There is one observation per cell, so an unrestricted goods-country interaction would saturate the model, leaving zero residual degrees of freedom. Without replication or a specified interaction structure, the residual variation cannot be separated into interaction and independent error.

The second fit applies a logarithmic transformation and fits the same factor design to $Z_{gc}=\log Y_{gc}$. The Gaussian log-response model isIts intercept exponentiates to a fitted Software–UK conditional median of $e^{4.5267}\approx 92.45$ million euros. Goods coefficients now give multiplicative ratios to Software, and country coefficients give multiplicative ratios to the UK. The estimated country ratios areFor example the Books ratio is $e^{1.6415}\approx 5.16$ and the Sports Equipment ratio is $e^{-1.8364}\approx 0.1594$, consistently across countries. The model assumes common relative, rather than common absolute, country effects. It also guarantees positive retransformed fitted values. Under normal log errors, the original-scale conditional mean is $e^{\eta +{\sigma }^2/2}$, whereas $e^{\eta }$ is the conditional median and geometric mean. Correcting the retransformation bias here multiplies the naive predictions by approximately $e^{0.127406/2}=1.0658$.

The log-scale residual standard error is $0.3569$, with the same 36 residual degrees of freedom, corresponding to a typical multiplicative error multiplier $e^{0.3569}\approx 1.429$. France and the rest remain significant reference-country contrasts; Germany remains nonsignificant at 5%. The log-scale significant reference-goods contrasts are Books, Music, Videos/DVDs, Clothing, Sports Equipment, Electronic Goods, Groceries and Leisure Travel. Event Tickets, Toys, Video Games and Housewares do not individually differ significantly from Software at that level. These p-values remain subject to the usual multiple testing qualification.

The log-scale $R^2=0.9381$ is computed using variation in the logarithms, so it cannot by itself establish better fit on the original response scale. The change is plausible because spending varies widely in magnitude and relative discrepancies are more meaningful than constant absolute discrepancies, but it should be assessed with regression diagnostics, especially a residual-versus-fitted plot and a normal Q-Q plot. The minimum residual near $-1.01$ merits inspection; the command summaries alone do not verify independent normal errors. Moreover these responses are forecasts, so the inferential p-values have a literal sampling interpretation only if an appropriate stochastic model for their uncertainty is justified.

The log-scale analysis of variance adds Goods and then Country. In this balanced design their effect spaces are orthogonal after removing the intercept, so the sequential sums of squares equal the corresponding adjusted factor sums and are unchanged by reversing the two factors:The residual sum of squares is $4.58662$, giving $M{S}_E=4.58662/36=0.127406$. ThusTheir tiny upper-tail p-values show strong overall goods and country effects under the log-error model; a globally significant country factor is compatible with the nonsignificant individual Germany contrast. The omnibus fit statistic is $36.3433$ on $(15,36)$ degrees of freedom, whose p-value is approximately $3.14\times 10^{-17}$. The printed zero reflects numerical formatting, not an exactly zero probability.

If two cell entries were missing, retain their correct factor labels and mark those responses unavailable, rather than shifting the remaining scan values into different cells. Fit the additive models to the 50 observed cells, assuming the missingness mechanism permits those observations to be analyzed without selection bias. All factor levels remain represented, and the row-column incidence graph remains connected: any two goods rows still share an observed country, since only two edges of the complete layout have been removed, and every country still has many observed goods. By identifiability of an incomplete two-factor additive design, the design therefore still has rank $13+4-1=16$. HenceThe coefficients and their covariance must be recomputed by the observed-cell ordinary least squares equations. The balanced row-mean-plus-column-mean formulas and common within-factor standard errors no longer apply. Goods and Country effects are generally no longer orthogonal, so sequential ANOVA sums can depend on order; factor tests adjusted for the other factor should use nested model comparisons. Missing cells do not restore replication or make an unrestricted interaction test possible: that larger model still fits every observed cell exactly. If missingness depends on the unavailable spending values, additional modelling of that mechanism would be needed.

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.

Multivariate analysis of variance extends analysis of variance to vector responses. For example, a treatment may affect several correlated physiological measurements jointly. Separate univariate tests do not use the correlation geometry and require adjustment if they are interpreted as one multiple-testing procedure. A joint test asks whether group mean vectors differ in any direction.

In a one-way model, take independent $Y_{ri}\sim N_p({\mu }_r,\Sigma )$, $r=1,\dots ,g$, $i=1,\dots ,n_r$, with a common positive-definite covariance. Put $N={\sum }_r{n}_r$, and let ${\bar{Y}}_r$ and $\bar{Y}$ be group and overall sample means. The within-group and between-group scatter matrices areExpanding around each group mean makes the cross terms vanish, giving total scatter $E+H$. To test $H_0:{\mu }_1=\cdots ={\mu }_g$, use the Wilks lambda statisticwhere ${\theta }_j$ solve the generalized eigenvalue problem $Hv=\theta Ev$ and $E$ is nonsingular. Under the alternative, maximize the normal likelihood at the group means and covariance $E/N$; under the null, use the overall mean and covariance $(E+H)/N$. Their likelihood ratio is ${\Lambda }^{N/2}$, so small $\Lambda$ rejects equal means. Under the null, orthogonal projections of the Gaussian data give independent $E\sim W_p(\Sigma ,N-g)$ and $H\sim W_p(\Sigma ,g-1)$, supplying its null calibration. Here $E$ is nonsingular almost surely when $N-g\ge p$. Merely having more total observations than responses is not enough if group fitting consumes too many statistical degrees of freedom.

The test is invariant under a nonsingular linear change of response coordinates: both determinants acquire the same factor $|A{|}^2$. The directions associated with large ${\theta }_j$ identify mean separation relative to residual variability. With one response it becomes the ordinary ANOVA F-test. With two groups, writing $S_p=E/(N-2)$ and $d={\bar{Y}}_1-{\bar{Y}}_2$, the version using Hotelling's T-squared statistic iswith $N>p+1$. This formula demonstrates explicitly how the covariance weights a group difference.

The sketch shows hypothetical groups with strongly positively correlated responses and mean separation along their low-variance contrast. Marginal comparisons can look weak even when the joint comparison is informative. The ellipses represent individual-response covariance, not confidence regions for the means. Independence of observations, approximate multivariate normality and common covariance are substantive assumptions; outliers and unequal covariance can invalidate the usual calibration. Follow-up contrasts can identify the affected responses or groups, with appropriate multiplicity control. MANOVA uses shared covariance to test a multivariate mean difference, rather than testing unrelated coordinates in isolation.

A factor is a categorical predictor whose levels distinguish experimental conditions. Two factors have a statistical interaction when the effect of changing one depends on the other's level. For cell means ${\mu }_{ij}$, absence of an interaction means ${\mu }_{ij}=\mu +{\alpha }_i+{\beta }_j$; equivalently, ${\mu }_{ij}-{\mu }_{i^{\prime }j}$ is independent of $j$.

Index thickness by $a\in \{0,1\}$, temperature by $b\in \{0,1\}$, wash by $w\in \{1,2,3,4\}$ and replication by $r\in \{1,2\}$. The full factorial design has the normal linear modelThe errors are independent with a common positive variance. Sum-to-zero constraints in each factor index, or the displayed treatment coding, identify the mean coefficients. This full model allows an arbitrary mean in each of the sixteen cells.

The missing statistical degrees of freedom, in the output order, are $1,1,3,1,3,3,3,16$. The first seven entries total fifteen; with the regression intercept there are sixteen mean coefficients among thirty-two observations, leaving sixteen residual statistical degrees of freedom. Balanced factorial orthogonality makes the term sums of squares in ANOVA independent of their ordering.

All terms involving wash have nonsignificant individual F-tests, and it is useful to test their removal jointly. The reduced model omits twelve mean coefficients. Its residual sum of squares is $57.09875$, compared with $32.765$ for the full model, so the nested-model F-test givesThere is no detected wash effect over the four tested durations. Removing its main effect and all its interaction terms respects the principle of marginality. Retaining the very large thickness-by-temperature interaction is essential: the thickness effect changes substantially with temperature.

With numeric indicators $a,b\in \{0,1\}$, treatment coding expresses the reduced fitted mean asThus the thin/low, thick/low, thin/high and thick/high fitted means are $4.225,5.6875,17.425,29.950$, respectively, at every wash level. Each is the sample mean of eight observations. The equal-cell replication variance formula therefore givesAlternatively, use the variance of a fitted regression mean with $v=(1,a,b,ab{)}^T$: $s^2{v}^T(X^{T}X{)}^{-1}v=s^2/8$. Individual coefficient standard errors cannot simply be added; their covariances enter the fitted-mean calculation.

Increasing temperature raises fitted lustre by $13.2$ for thin film and $24.2625$ for thick film. Increasing thickness raises it by $1.4625$ at low temperature and $12.525$ at high temperature. Their difference, $11.0625$, is the estimated interaction contrast; its standard error is $1.0098$, giving $t\simeq 10.956$ on twenty-eight statistical degrees of freedom. The interaction plot shows this strong departure from parallel mean lines, while the wash panel shows the much smaller variation across wash durations. Neither the nonsignificant wash test nor the plot proves that its true effect is exactly zero.

This is an analysis of covariance: achievement is the response, aptitude is a quantitative covariate, and teaching method is a categorical explanatory variable. The usual normal linear model assumes independent errors with common variance ${\sigma }^2$.

read.table reads the data with a header row; attach makes its columns directly accessible, and the two printed vectors confirm their ordering. rep(1:3, each=7) supplies the three group labels, and factor makes them categorical. The contrast option selects treatment coding for unordered factors, with method A as the first-level reference. The polynomial option would apply to ordered factors and has no role here.

The formula using aptitude * Group includes both main effects and their interaction. Writing $I_B,I_C$ for the method indicators, its fitted mean has the formThus A, B and C have separate intercepts and slopes. There are six coefficients and $21-6=15$ residual statistical degrees of freedom. The analysis of variance table uses sequential sums of squares: adding aptitude to an intercept-only model removes $36.57548$ from the residual sum of squares; subsequently adding the two group indicators removes $16.93200$; adding the two slope interactions removes only $0.66714$. The remaining sum of squares is $9.63490$, giving mean square $9.63490/15=0.64233$.

Each displayed F-test divides the added term's sum of squares by its statistical degrees of freedom and then by this full-model residual mean square. In particular the interaction test isThere is no evidence that the slopes differ, so the common-slope model is preferable among the fitted models. The small group $p$-value concerns differences after including aptitude, whereas the first aptitude row is sequential and does not adjust for Group. It should not be interpreted as the covariate test after treatment adjustment.

The additive formula gives the parallel-lines modelIt has four coefficients and $17$ residual statistical degrees of freedom. Removing the interaction raises the residual sum of squares to $9.63490+0.66714=10.30204$, but lowers the residual mean square to $10.30204/17=0.60600$. The sequential aptitude and group sums of squares are unchanged. The adjusted group test is now $F=(16.93200/2)/0.60600=13.97024$ on $(2,17)$ statistical degrees of freedom, with $p=0.0002579$: after allowing for initial aptitude, the methods are not all equally effective.

summary(..., cor=F) displays the fitted coefficients, their standard errors, the individual $t$ statistics, and two-sided $p$-values; it omits the coefficient-correlation matrix. The intercept $2.8367$ is the fitted A mean at aptitude zero, which is outside the observed range and is mainly a parametrization device. The common slope $0.7429$ predicts about $0.74$ extra achievement units per aptitude unit, holding method fixed. Its adjusted $t$ test is $5.2267$ on $17$ statistical degrees of freedom, with $p$ about $0.0001$. The method coefficients are adjusted differences from A, not the B and C intercepts themselves. The three fitted lines areThe B-minus-A estimate is $2.1878$ with standard error $0.4545$ and $t=4.8139$; the C-minus-A estimate is $1.8612$ with standard error $0.4240$ and $t=4.3901$. Both comparisons are strongly significant.

The B-minus-C comparison needs its own treatment contrast; its significance cannot be read off either comparison with A. Reconstructing the coefficient covariance from the design givesThus there is no detectable difference between B and C. This is consistent with pooling their effects in a further reduced model, but does not establish exact equality.

The residual standard error is $\sqrt{0.60600}=0.7785$. The five-number residual summary shows deviations on both sides of zero, with the largest magnitude about $1.74$; it does not by itself verify constant variance or normality. The coefficient of determination $R^2=0.8386$ means about $84\%$ of the achievement variation about its overall mean is explained. The overall $F=29.43$ compares all three nonintercept coefficients with zero, on $(3,17)$ statistical degrees of freedom; its $p$-value is $5.889\times 10^{-7}$. Printed coefficient $p$-values rounded to 0.0000 are small, not literally zero.

The following original sketch shows the separate-slope model and the preferred parallel-line fit.

At the same initial aptitude, B and C give higher predicted achievement than A, with no convincing difference between B and C. Interpreting this adjusted association as a causal teaching effect additionally relies on the experimental allocation and error assumptions.

Write $q=n-p$. In this normal linear model, the ordinary least squares criterion is $\Vert Y-Xb{\Vert }^2$. Differentiating with respect to $b$ gives the normal equations for linear least squares, $X^{T}Xb=X^{T}Y$. Full column rank of the design matrix makes $X^{T}X$ invertible, and completing the square proves that the unique minimum isSince $\hat{\beta }=\beta +(X^{T}X{)}^{-1}{X}^Tϵ$, its sampling distribution is the multivariate normal distributionLet $H=X(X^{T}X{)}^{-1}{X}^T$ be the hat matrix. It is symmetric and idempotent with rank $p$, and $HX=X$. The regression residual vector is $(I-H)Y=(I-H)ϵ$, soAn orthogonal projection of a Gaussian vector onto the residual space gives $\mathrm{RSS}/{\sigma }^2\sim {\chi }_q^2$. The residual projection and fitted projection have zero cross-covariance, hence are independent because they are jointly normal. In particular, $s^2$ is an unbiased estimator of ${\sigma }^2$ and is independent of $\hat{\beta }$. The maximum-likelihood estimator of ${\sigma }^2$ instead divides by $n$ and is not unbiased.

For the null hypothesis ${\beta }_1=0$, put $v_{11}=((X^{T}X{)}^{-1}{)}_{11}$. Under that null hypothesis, ${\hat{\beta }}_1/(\sigma \sqrt{v_{11}})$ is standard normal and independent of $q{s}^2/{\sigma }^2$. Therefore the Student t-test usesA two-sided level-$\alpha$ test rejects when $|T|>t_{q,1-\alpha /2}$. The denominator is the estimated standard error of the regression coefficient.

The fitted linear regression includes an intercept and three covariates:Thus $n=25$, $p=4$ and the residual degrees of freedom are $21$. The fitted conditional mean isEach slope measures a conditional association with the response while the other two covariates are held fixed. Increasing wind speed by one mile per hour corresponds to a fitted decrease of $0.527410$ in the concentration units used. Its standard error is $0.224904$, giving $t=-2.345$ and a two-sided p-value of $0.0289$: evidence of a negative conditional association at the 5% level. The temperature slope is $0.124991$ per degree Fahrenheit, with standard error $0.075173$, $t=1.663$ and p-value $0.1112$. The insolation slope is $-0.005259$ per langley per day, with standard error $0.006637$, $t=-0.792$ and p-value $0.4370$. Neither of these last two conditional effects is significant at 5%; that does not establish that either true slope is zero. These are associations, not a demonstration of causal effects.

The intercept estimates the mean at zero values of all three covariates. Its value $3.784916$ has a large standard error $7.979647$, giving $t=0.474$ and p-value $0.6402$. The zero-covariate combination may be outside the observed range, so this test may have little practical relevance. Every coefficient test uses a Student t-distribution with $21$ degrees of freedom.

The reported residual standard deviation $s=1.844$ measures unexplained variation in concentration units; the subsequent table gives $\mathrm{RSS}=71.428$, consistent with $s=\sqrt{71.428/21}$ after rounding. The five-number residual summary describes the spread, from $-2.3052$ to $3.4033$, with median $-0.4990$. An intercept forces the residual sum to zero, not the residual median. These five numbers alone cannot establish normal errors, constant variance or independence; residual-versus-fitted and quantile-quantile plots would be more informative, and successive days also warrant checking serial dependence. The coefficient of determination $R^2=0.6533$ means that 65.33% of the sample response sum of squares about its mean is explained by this fit. Its adjusted version iswhich accounts for the number of estimated regression coefficients.

The stepwise selection by the Akaike information criterion searches between the intercept-only model and the specified model with all three main effects. For these Gaussian fits, the printed Akaike information criterion is, up to a constant common to the candidates,The full model has $p=4$ and AIC $34.246$. Removing insolation increases the residual sum of squares by $2.136$ but decreases AIC to $32.982$, so that removal is chosen. Removing temperature or wind instead would give AIC $35.337$ or $38.060$ and is worse. In the next step the candidate models include deletions and restoration of insolation. Keeping wind and temperature has the smallest AIC: removing temperature gives $33.389$, restoring insolation gives $34.246$, and removing wind gives $37.960$. Hence the search stops, givingThe accompanying F-tests compare a larger model with a one-parameter reduction: $F=\Delta \mathrm{RSS}/s^2$, where $s^2$ is estimated from the larger model. For example, removing insolation gives $2.136/(71.428/21)=0.628$, equal to the corresponding squared test statistic. Deletions from the second-stage model use $73.564/22$ in the denominator, whereas restoring insolation uses the full-model denominator. These tests describe the comparisons; they do not determine the AIC choice. In particular, temperature is retained despite its deletion-test p-value $0.15016$. AIC balances fit and complexity, and a stepwise optimum need not be a global optimum over a richer model class. Ordinary coefficient inference after choosing a model does not account for the selection itself.

For positive responses, the Box–Cox transformation is $T_{\lambda }(y)=(y^{\lambda }-1)/\lambda$ for $\lambda \ne 0$, and $T_0(y)=\log y$. Its profile likelihood refits the linear regression and variance at each $\lambda$ and includes the transformation Jacobian. The graph peaks at roughly $\lambda =0.3$–$0.4$. The approximate 95% likelihood-ratio confidence interval contains values a little below zero through approximately one, using the cutoff ${\ell }_{\max }-{\chi }_{1,0.95}^2/2$. Thus a logarithmic transformation ($\lambda =0$) and a square-root transformation ($\lambda =1/2$) are plausible, while leaving the response on its original scale ($\lambda =1$) is at or very near the upper confidence boundary. The broad profile supports considering a mild power transformation; it does not determine a sharply estimated power. The raster does not justify more precise endpoints or a decisive assertion about which side of the boundary $\lambda =1$ lies.

Let $m={\overline{Y}}_{+++}$, $r_i={\overline{Y}}_{i++}$ and $c_j={\overline{Y}}_{+j+}$. The proposed least-squares estimators satisfy the sum-to-zero constraints because ${\sum }_i{r}_i=Im$ and ${\sum }_j{c}_j=Jm$. Define $e_{ijk}=Y_{ijk}-r_i-c_j+m$. Direct summation givesNow write any other admissible parameters as $\mu =m+u$, ${\alpha }_i=r_i-m+a_i$, ${\beta }_j=c_j-m+b_j$, where ${\sum }_i{a}_i={\sum }_j{b}_j=0$. Expanding the ordinary least squares criterion, the cross-products with $e$ vanish by the preceding identities. The cross-products between $u$, $a_i$ and $b_j$ vanish by their zero sums. ConsequentlyEvery added term is nonnegative and all vanish only at $u=0$, $a_i=b_j=0$. This proves the unique constrained minimum, withThe additive-model residual sum of squares and intercept-only residual sum of squares are thereforeThe same orthogonality gives the analysis of variance identityFitting factor B alone yields fitted values $c_j$ and residual sum ${\mathrm{RSS}}_0-S{S}_B$. Fitting A alone gives residual sum ${\mathrm{RSS}}_0-S{S}_A$. Adding B after A then reduces it to ${\mathrm{RSS}}_1$, a reduction of $S{S}_B$ again. The reduction due to B is $S{S}_B$ in either order. This is balanced factorial orthogonality; it generally fails for unequal cell replication.

There are $12$ observations. The additive model estimates $1+(3-1)+(2-1)=4$ independent parameters, so the missing degrees of freedom are A: 2, B: 1, residuals: 8. The individual requested residual sums are below.

To check for an interaction, extend the normal linear model to $\mu +{\alpha }_i+{\beta }_j+{\gamma }_{ij}$, with each row and column sum of $\gamma$ zero. Equivalently, fit the six unrestricted cell means. Its residual sum of squares is the within-cell sumThe interaction adds $(I-1)(J-1)=2$ parameters; the cell-means residual has $IJ(K-1)=6$ degrees of freedom. Under the no-interaction null hypothesis and the independent equal-variance normal error model,The numerator and denominator arise from orthogonal Gaussian projections, proving the exact F-test. Reject for a sufficiently large value. The printed additive analysis of variance table does not supply ${\mathrm{RSS}}_{\mathrm{cell}}$, so it cannot determine this test statistic numerically; the replicated cell data would do so.