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Use the logit link $g(\pi )=\log \{\pi /(1-\pi )\}$ and independent counts $R_{ij}\sim \mathrm{Bin}(n_{ij},{\pi }_{ij})$. This gives a grouped-binomial logistic regression with 24 cells and 13 free coefficients: an intercept, 11 centre contrasts, and one period contrast. Reference treatment coding makes the intercept the log odds for Centre 1 in the earlier of the two included periods. The two factors are assumed to have been coded as specified in the question.

The binomial log-likelihood, apart from known constants, isMaximize by Fisher scoring or iteratively reweighted least squares. Using response proportions $r_{ij}/n_{ij}$ with binomial trial weights $n_{ij}$ is equivalent to using the count response $(r_{ij},n_{ij}-r_{ij})$; omitting those weights would give the wrong likelihood and standard errors. The coefficient covariance matrix is approximately the inverse of $X^T\mathrm{diag}\{n_{ij}{\hat{\pi }}_{ij}(1-{\hat{\pi }}_{ij})\}X$.

The fitted reference log odds are $-0.809585$, and the period change is $-0.426758$, givingThese are fitted probabilities under a common period effect, so they need not equal the separate empirical proportions at Centre 1. The estimated period odds ratio isThus the later-period death odds are estimated to be about 35% lower at each centre, with strong evidence of a period effect: the normal Wald statistic is $-5.415$. The displayed “t value” is a coefficient divided by its standard error; with fixed binomial dispersion its usual large-sample calibration is standard normal, not a Student distribution with an estimated Gaussian-error variance. The odds ratio is not a mortality probability ratio.

Each centre coefficient is a log odds ratio relative to Centre 1, adjusted for period. All its point estimates are negative, so all other centres have lower fitted mortality probabilities in either period. For example, Centre 11's odds ratio is $e^{-1.079474}=0.3398$, or about $2.94$ times greater odds at Centre 1 than at Centre 11. Individual unadjusted two-sided Wald tests find significant differences at the 5% level for centres 2, 4, 6, 7, 8, 9, 11 and 12; the coefficients for 3, 5 and 10 do not individually reach that threshold. These separate tests are correlated comparisons against a common reference and should not be presented as 11 independent discoveries without considering multiple testing. They adjust only for time period, not for differing patient case mix.

For fit assessment, the residual binomial deviance is $15.73891$ on $24-13=11$ statistical degrees of freedom. Its ${\chi }_{11}^2$ upper tail is approximately $0.151$, so there is no appreciable evidence of lack of fit to these binomial cell counts. In particular, comparing with a saturated centre-by-period model gives this same deviance and 11 additional parameters; the data do not strongly require a centre-period interaction. The moderate deviance residuals are consistent with that assessment, although a residual plot by centre and period would be more informative than the five-number summary alone.

The reduction from null to fitted deviance isIt tests the 12 centre-and-period coefficients jointly against an intercept-only model, with reference ${\chi }_{12}^2$ and $p\approx 1.24\times 10^{-16}$. It is not a centre-only test adjusted for period. For that test, fit the period-only model and compare its deviance with the full fit: the pooled binomial probabilities in each period give deviance $90.34578$, soThus centre heterogeneity persists after allowing for period. Dispersion one is the stipulated binomial variance, and the five scoring iterations describe numerical convergence rather than statistical evidence. The fitted associations alone do not establish a causal difference in surgical quality.

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.

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.

factor(day) treats the nine days as categorical rather than imposing a numerical time trend. rep(400,27) supplies the number of trials for every dish, and prop is the observed surviving proportion. The contrast setting uses treatment coding with day 1 as reference; the ordered-factor polynomial option is unused. Supplying these proportions to glm with the binomial family and weights $400$ gives precisely the grouped-binomial logistic regression likelihood for the counts. The weights represent trial counts, not additional independent dish observations. The binomial family's default link is the logit link and its dispersion is fixed at $1$.

Let $Y_{ij}$ be survivors in dish $j=1,2,3$ on day $i=1,\dots ,9$. Both fitted models assume independent $Y_{ij}\sim \mathrm{Bin}(400,p_i)$. Model 1 specifiesso it has one common probability and one fitted coefficient. Its estimated probability is the total surviving count divided by the total trial count, $\hat{p}=3539/10800\simeq 0.327685$.

Model 2 specifiesIt has nine coefficients and allows a separate probability for each day, while retaining a common probability for the three dishes within that day. The MLEs are the day totals divided by $1200$. In day order they areThe intercept is the first day's logit, and each other coefficient is its day's logit minus that reference. These are additive effects on log odds, not additive changes in survival probability.

The analysis of deviance adds the eight day indicators after an intercept-only model. Its residual deviance compares a fitted model with the saturated statistical model assigning one separate probability to each of the $27$ dishes. For fitted dish probabilities ${\hat{p}}_{ij}$, the binomial deviance isThe binomial scale is one, so this is also scaled deviance. The reported decrease measures the extra fit provided by day. There are $27$ observational units for these residual statistical degrees of freedom, not $10800$ separately fitted Bernoulli responses.

Let $Y_{ijkl}$ be the flight distance in millimetres for paper level $i$, launch-angle level $j$, design level $k$ and replicate $l$, where $i,j,k\in \{1,2\}$ and $l\in \{1,2\}$. The full factorial design is fitted by the normal linear modelThe assumptions are independent errors with a common variance and normal distribution, and the displayed factor-combination mean structure. The corner-point constraints areand $(PAD{)}_{ijk}=0$ whenever at least one index equals $1$. Thus the eight free mean coefficients are $\mu ,P_2,A_2,D_2,(PA{)}_{22},(PD{)}_{22},(AD{)}_{22},(PAD{)}_{222}$. These treatment coding coefficients are comparisons at the other factors' reference levels, rather than necessarily averaged main effects.

To test the three-factor interaction term, compare the full model with the model retaining all main effects and all two-factor interactions. The null hypothesis is $(PAD{)}_{222}=0$, against a nonzero value. Removing it increases the residual sum of squares by $21025$ and removes one parameter. With eight residual degrees of freedom, the nested-model F-test givesThere is no evidence that this three-factor interaction is needed. This does not establish that all two-factor interactions vanish: the paper-design interaction is substantial.

The stepAIC command performs stepwise selection by the Akaike information criterion. Its upper scope is the full factorial model and its lower scope is the intercept-only model. Starting from the full model, it compares admissible additions and deletions, preserving the hierarchy of interaction terms, and chooses changes that reduce the Akaike information criterion. With the supplied scope its default search permits both directions. For these Gaussian fits the displayed criterion, up to constants common to all models, is $16\log (\mathrm{RSS}/16)+2k$, where $k$ counts mean coefficients. The first deletion changes it from $226.72$ to $224.76$: the small increase in residual sum of squares is outweighed by removing a parameter. This is a greedy model selection procedure, not an exhaustive search or a sequence of fixed-level hypothesis tests. The command's arguments are described in the MASS documentation.

For the selected model, let $p=1_{\{i=2\}}$ and $d=1_{\{k=2\}}$. Thenindependently between flights. It has no angle term. The maximum-likelihood estimates of its mean coefficients are also obtained by ordinary least squares:For example, the four fitted means are just the four paper-design cell means, each averaging four flights over the angle levels. With $s^2=\mathrm{RSS}/12$, the variance-covariance matrix of the coefficient estimates is $s^2(X^{T}X{)}^{-1}$. Its diagonal gives the displayed standard errors; their ratios are $s/2,s/\sqrt{2},s/\sqrt{2},s$, with $s=892.6$ mm.

The final line of summary(model2.lm) is the joint nested-model F-test of $H_0:{\beta }_P={\beta }_D={\beta }_{PD}=0$, comparing this four-parameter model with a common mean. Its statistic iswith F-distribution $F_{3,12}$ under the fixed-model null and P-value $0.001057$. Thus the selected paper-design terms jointly improve the fit substantially. The coefficient of determination $0.7272$ means they account for about $72.7\%$ of the observed centred variation. These nominal inferential summaries do not adjust for the preceding model selection on the same data.

Using the unrounded fitted means givesin millimetres. Hence choose the high-performance design with the lighter paper; the fitted model leaves both launch angles tied, with expected distance about $5377$ mm. A preference for one angle is not supported by this selected model. The strong paper-design interaction reverses the design comparison: the simple design is better on heavy paper, whereas the high-performance design is better on light paper. Consequently a single design effect averaged over paper types would conceal the main result. There is little evidence here for an angle effect, and considerable flight-to-flight variability remains.

Before fitting, inspect raw-distance dot plots for all eight factor combinations, and interaction plots of cell means against design, with separate lines for paper and panels for angle. With only two replicates per cell, showing individual observations is more informative than elaborate box plots. These reveal reversals, spread and possible unusual flights. Afterwards inspect a residual-versus-fitted plot, a Q-Q plot of residuals, a scale-location plot, residuals against each factor and, if recorded, flight order, and Cook's distance for influential observations. Residual patterns by angle would question its deletion, while unequal spread or departures from the normal distribution would question the error assumptions rather than merely demand another mean term.

Let $W_{aj\ell }$ be recall for age level $a$, processing level $j$, and replicate $\ell =1,\dots ,10$. Write $u_a=1$ for Younger and $0$ for Older. With Older and A as the reference levels in a regression factor, the additive two-factor normal linear model under treatment coding iswhere the errors are independent $N(0,{\sigma }^2)$, with common positive variance. The design matrix is fixed; subjects supply independent observations; the specified conditional mean is correct; and the absence of an interaction term means the age difference is constant across processing levels. These are model assumptions, not consequences of random assignment. The treatment allocation supports independence between processing assignment and background characteristics, but age itself was not randomized.

An older subject in A has $u_a=0$ and ${\gamma }_A=0$, so the estimated mean is $11.35$ words. Its standard error is the intercept's $0.7632$. There are $100-6=94$ residual degrees of freedom, giving the 95% confidence interval for this mean:This is a mean-response confidence interval, not a prediction interval for a new individual's recall; the latter also includes the new observation's error variance.

The second two-factor normal linear model adds age-by-processing interaction terms:Its ten free mean parameters are equivalent to one mean for each of the ten cells. To compare the fits, the null hypothesis is ${\delta }_B={\delta }_C={\delta }_D={\delta }_E=0$. The full residual sum of squares is $722.30$ on $90$ degrees of freedom. Adding interaction reduces the additive model's residual sum of squares by $190.30$, so the reduced value is $912.60$. The nested-model F-test statistic isIts p-value is $0.0002793$. Reject additivity and retain the interaction model. The significant interaction means that a single age effect averaged over all processing methods does not adequately summarize the data.

The regression intercept $11.0$ is the older-A mean. The age coefficient $3.8$ is the younger-minus-older contrast specifically in A. The processing coefficients $(-4.0,2.4,1.0,-4.1)$ compare B, C, D and E with A specifically among older subjects. The interaction coefficients $(-4.3,0.4,3.5,-3.1)$ are differences between the age contrast in each of those methods and its value in A. Adding the relevant coefficients givesThus A, C and especially D favour younger subjects in their estimated means, whereas B and E have little estimated age difference. In the younger group, D has the largest estimated recall, followed by C and A; B and E are much lower. Among older subjects C is largest, D and A are intermediate, and B and E are lowest. Comparing these orders is a description of estimates, not a claim that every pairwise difference is significant.

Each coefficient's printed standard error estimates its uncertainty; its $t$ statistic divides the estimate by that error, and its two-sided p-value uses $t_{90}$ under the relevant zero-contrast null hypothesis. For example, the older B-versus-A and E-versus-A contrasts have $p=0.00217$ and $0.00170$. The C-versus-A and D-versus-A older contrasts have $p=0.06139$ and $0.43201$. The B interaction has $p=0.01846$, while the D interaction is borderline at $0.05387$ and the other two individual interaction tests are not significant at $5\%$. These individual tests do not override the joint interaction F-test, and multiple comparisons require care. A simple younger-versus-older contrast outside A combines coefficients and must use their estimated covariance, rather than adding their marginal errors.

The residual standard error $2.833$ estimates $\sigma$. The coefficient of determination $0.7293$ is the fraction of centered sample recall variation explained by the ten-cell model. The overall $F=26.93$ with null law $F_{9,90}$ tests all nine non-intercept coefficients jointly against zero, giving extremely strong evidence that the cell means are not all equal. The sequential analysis of variance separates age, process, and their interaction; the balanced design makes the main-effect sums orthogonal, whereas the coefficient table uses the specified reference-level contrasts.

To assess pooling into three processing types, retain age-by-type interaction because the previous test rejects additivity. The reduced model $W\sim \mathrm{Age}\ast \mathrm{Type}$ has six cell means. Its null hypothesis says A and C have equal means within each age, and B and E have equal means within each age: four restrictions. Compare it with the ten-cell model usingIn fact the supplied fitted cell means and equal cell sizes determine the increase without raw observations. Pooling two ten-person cells with means $m_1,m_2$ adds $5(m_1-m_2{)}^2$ to the residual sum of squares. Here the four differences are $2.4,2.8,-0.1,1.1$, soTherefore $F_{\mathrm{pool}}=2.3083$ and $p\simeq 0.0640$. At $5\%$, the data do not reject the three-type reduction, though the result is borderline and is not proof that the pooled means are identical. This factor-level pooling test keeps the previously supported interaction structure; reducing the categories and removing all interaction simultaneously would test a different set of restrictions.

A two-factor normal linear model describes independent responses with factor-specific means and common normal error variance. The additive mean $\mu +{\alpha }_a+{\gamma }_j$ assumes no interaction term; the full mean $\mu +{\alpha }_a+{\gamma }_j+{\delta }_{aj}$ permits the first factor effect to vary with the second. Constraints or treatment coding identify the parameters. In a balanced two-by-five design the full model has ten free cell means.