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In an ordinary least squares regression with an intercept and a nonconstant response, $R^2=1-\mathrm{RSS}/{\sum }_i(Y_i-\overline{Y}{)}^2$. It measures the fitted proportion of sample variation relative to an intercept-only baseline. It increases when regressors are added, even if those regressors are unhelpful for prediction. A high value does not establish the assumptions of a normal linear model or rule out a systematic regression residual pattern.

For the five-point example in linear regression, the total centered sum of squares is $10.8$ and the residual sum of squares is $0.8$, so $R^2=25/27\approx 0.926$. This summarizes the fitted reduction in squared error. It does not record the order or shape of the regression residuals, which are worth examining in a residual-versus-fitted plot.

Here is a fully constructed illustration. Use $x=(-2,-1,0,1,2)$ and the same fitted line $3+x$ for both datasets. Add respectively the residual vectorsBoth sum to zero, both have zero inner product with $x$, and both have squared norm $0.56$. Their ordinary least squares fit is therefore exactly the same line, with the same $R^2=10/(10+0.56)=125/132\approx 0.947$. But the first residual is $0.2(x^2-2)$, a systematic quadratic shape, while the second changes direction across the predictor values. These examples show directly which information the scalar $R^2$ discards; neither constructed pattern proves a random-error model.

A deviance goodness-of-fit test compares a fitted generalized linear model with its saturated statistical model. With $n$ independent response groups, an identifiable rank-$p$ mean model and known common dispersion parameter $\varphi$, the candidate statistic is the scaled devianceIn an appropriate large-information regime, compare it with a chi-squared distribution on $n-p$ statistical degrees of freedom. For a normal linear model this is exact when $\varphi ={\sigma }^2$ is known: $D$ is the residual sum of squares. For Poisson regression, $\varphi =1$; adequate expected cell counts are still needed for a useful approximation.^[ref]

More individual Bernoulli observations do not automatically justify comparing their residual deviance with ${\chi }_{n-p}^2$. Small cell counts, dependence and overdispersion can invalidate that calibration. An unknown dispersion parameter also needs an explicit estimation and testing convention; estimating it from $D/(n-p)$ and then treating $D/\hat{\varphi }$ as an independent goodness-of-fit check is circular.

Consider the correctly specified fixed-design normal linear model $Y=X\beta +\epsilon$, with $X$ an $n\times p$ full-column-rank design matrix, $n>p$, and $\epsilon \sim N_n(0,{\sigma }^2{I}_n)$ with ${\sigma }^2>0$. Both $\beta$ and ${\sigma }^2$ are unknown, so the Akaike information criterion counts $p+1$ fitted parameters. If $\mathrm{RSS}=\Vert Y-X\hat{\beta }{\Vert }^2$ is the residual sum of squares, thenThe orthogonal-projection decomposition of a random variable with a multivariate normal distribution gives $\mathrm{RSS}/{\sigma }^2\sim {\chi }_{n-p}^2$. Substituting into $\mathrm{AIC}=-2\ell (\hat{\beta },{\hat{\sigma }}^2)+2(p+1)$ yieldsHere ${\chi }_{n-p}^2$ denotes a random variable, not a quantile. The maximum-likelihood estimator of the error variance uses divisor $n$; the unbiased residual variance estimator with divisor $n-p$ cannot be substituted into the maximized likelihood. If $n=p$ and $X$ is invertible, the fit has zero residual sum of squares, and the Gaussian likelihood is unbounded as ${\sigma }^2$ decreases to zero. There is then no finite positive-variance maximum-likelihood estimator to which the displayed Akaike information criterion law applies.

Suppose a normal linear model gives an unrestricted mean to each cell and has $m$ independent observations per cell, with common error variance ${\sigma }^2$. Each fitted cell mean is its sample mean, with variance ${\sigma }^2/m$; different fitted cell means are independent. Estimate ${\sigma }^2$ by the pooled within-cell residual sum of squares divided by $n-q$, where $q$ is the number of cells. Thus a cell mean has standard error $s/\sqrt{m}$, a difference of two cell means has standard error $s\sqrt{2/m}$, and a two-by-two difference-of-differences has standard error $2s/\sqrt{m}$.

// Target: probability-and-statistics.bigb

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.

To pool factor levels in a normal linear model, impose equality of their fitted means while retaining any needed interaction terms. Compare the resulting nested model with the full model using a nested-model F-test. For two cells each of size $m$, pooling their fitted means adds $(m/2)({\overline{Y}}_1-{\overline{Y}}_2{)}^2$ to the residual sum of squares. If pooling two pairs separately within two other-factor levels, there are four restrictions.

Fit the positions of successive, consistently defined troughs by linear regression, $V_n=b+nd+{ϵ}_n$. A constant intercept $b$ absorbs the contact potential difference, collection phase and initial energy effects; $d$ estimates the repeated loss expressed as a voltage. Subtracting successive positions cancels $b$. A line fit uses every position and leaves residuals that can reveal curvature or a misidentified trough. It still assumes that the phase is constant over the fitted range and that all troughs belong to the same loss channel.

For an original numerical example, choose $n=1,2,3,4,5$ and synthetic positions $V_n=(6.62,11.46,16.48,21.30,26.14)\mathrm{V}$. Their successive spacings are $(4.84,5.02,4.82,4.84)\mathrm{V}$. The ordinary least squares formulas giveThe fitted residuals are $(-0.004,-0.052,0.080,0.012,-0.036)\mathrm{V}$, with zero sum and zero index-weighted sum. Their residual sum of squares is $0.01056{\mathrm{V}}^2$. If independent, equal-variance, zero-mean errors describe these positions, the residual variance estimate is $s^2=0.01056/(5-2)=0.00352{\mathrm{V}}^2$ and the slope standard error is $s/\sqrt{10}=0.01876\mathrm{V}$. Thus the inferred model loss is $4.888\mathrm{eV}$ with conditional standard error $0.01876\mathrm{eV}$. These errors summarize only the chosen position scatter; a voltage-scale error or a drifting collection phase is not included. Adjacent spacings share a position error and are therefore correlated even when position errors are independent.

The offset has a sharper identifiability limitation. Even if a constant-phase model were exact, an initial beam kinetic energy $E_0$ shifts the troughs by $-E_0/e$. For the symmetric illustrative curve, $b=V_c-E_0/e+V_r/2$. The transformation $V_c\mapsto V_c+\alpha$, $E_0\mapsto E_0+e\alpha$ leaves every trough position unchanged. Measuring $b$ alone therefore cannot determine the contact potential difference. Relabeling $n\mapsto n+m$ also changes $b$ by $-md$ without changing $d$. An independent offset calibration and a justified collection model are needed before turning a first trough into an excitation threshold. Varying $V_r$ tests this model's predicted half-voltage shift of the trough centers; it is a diagnostic prediction of this chosen curve, not a universal correction for experimental minima.

Linear regression models a response's conditional expectation as a linear combination of specified predictors. In matrix form,where the design matrix $X$ describes the observations, $\beta$ contains the regression coefficients, and $\epsilon$ contains the errors. The fitted values $X\hat{\beta }$ estimate the response means. The model is linear in its regression coefficients; its columns may include powers, interactions, or other transformed predictors, so polynomial regression can fit curved relationships using the same machinery.

For a concrete example, take five constructed observationsAn ordinary least squares fit with an intercept gives the line $\hat{y}=0.8+x$. Its regression residuals are $(0.2,0.2,-0.8,0.2,0.2)$ and its residual sum of squares is $0.8$. A one-unit increase in the predictor corresponds to a one-unit increase in the fitted response. This is a description of this fitted association; interpreting a change as a causal effect requires the assumptions of a causal inference problem.

The geometric reason for the fit is simple. The possible mean vectors $X\beta$ form the column space of $X$. Ordinary least squares chooses the orthogonal projection of the data vector onto that vector subspace, so the remaining regression residual is orthogonal to every permitted fitted direction. The least-squares normal equations express this orthogonality as $X^T(Y-X\hat{\beta })=0$. The data's explained component and unexplained component are therefore separated by the same geometry used in a Hilbert space.

The optimization definition makes sense for any observed data. Quantifying the uncertainty of the regression coefficients needs additional assumptions about the sampling and errors. With fixed $X$ of full column rank and mean-zero errors with covariance matrix ${\sigma }^{2}I$, ordinary least squares is unbiased and has covariance ${\sigma }^2(X^{T}X{)}^{-1}$; the Gauss-Markov theorem compares it with other linear unbiased estimators. A normal linear model additionally makes Student t-distribution intervals and tests exact in finite samples. Model adequacy and predictive assessment remain relevant even when these calculations are available.

Two historical settings help explain the subject's appeal. Adrien-Marie Legendre's Nouvelles méthodes pour la détermination des orbites des comètes (1805) includes an appendix on the method of least squares: observations give more equations than unknowns, and squared errors provide a criterion for reconciling them. Its astronomical setting shows the motivation for combining imperfect measurements. ^[ref] Francis Galton's Regression Towards Mediocrity in Hereditary Stature (1886) examines the association between parental and offspring heights and the tendency of group averages toward the population average. This is a historical use of the regression name; the modern linear model is a general statistical formulation. ^[ref]

Two linear regression models are nested when the possible mean vectors of the smaller model form a linear subspace of those of the larger one. With design matrices $X_0,X_1$, this is $\mathrm{col}(X_0)\subseteq \mathrm{col}(X_1)$. Adding a quadratic predictor to an intercept-and-linear model is an example when both use the same observations.

Orthogonal projection then gives ${\mathrm{RSS}}_1\le {\mathrm{RSS}}_0$. If $r_j=\mathrm{rank}(X_j)$, $r_1>r_0$, and $n>r_1$, independent errors with a normal distribution and common positive variance yieldunder the smaller model. The F-test compares a gain in fit with its extra degrees of freedom; simply obtaining a smaller residual sum of squares is automatic and is not itself evidence that the larger model generalizes better.

The descriptive summaries give 1784 observed ages and diversity scores, with no missing values. Age ranges from 18 to 89 years, with mean 43.035 and sample standard deviation 16.862 years. Diversity ranges from 0 to 48, with mean 19.592 and sample standard deviation 10.240 points. The scale combines online and offline consumption sources; the output does not say that one point equals one source, so that unit should not be invented. Comparing the numerical standard deviations across these different units does not establish which variable is more variable substantively.

A simple linear regression posits $\mathbb{E}(Y\mid x)={\beta }_0+{\beta }_{1}x$, where Y is diversity and x is age. Ordinary least squares fits that conditional mean by minimizing the residual sum of squares. If the coefficient estimates are trusted, the fitted line isIt predicts about 23.259 diversity points at age 18 and 12.861 at age 89. At the mean age it predicts $25.89474-0.14645(43.035)\approx 19.59226$, agreeing with the rounded outcome mean. This is the mean-fitting identity expected of least squares with an intercept. It does not validate every other output field. Individual scores can deviate substantially from these fitted means, and a cross-sectional age contrast is not an estimate of each person's change as they age.

The intercept of about 25.895 is the fitted mean at age zero. Zero lies outside the observed age range, so that direct interpretation is an extrapolation. Its test is not a test of whether age predicts diversity. The source's intercept row is internally inconsistent: $25.89474/6.30222\approx 4.10883$, not the displayed 4.13. At 1782 residual statistical degrees of freedom, these imply two-sided p-values of about $0.0000416$ and $0.0000380$, respectively, neither of which is the printed 0.00012. These differences exceed the displayed rounding precision, although each isolated p-value would still indicate a nonzero intercept at conventional levels. More importantly, its printed standard error is incompatible with the supplied predictor summary and residual error scale, as checked below.

The slope row is much more internally coherent: $-0.14645/0.01396\approx -10.4907$, agreeing with the displayed $t=-10.49$. There are $1784-2=1782$ residual statistical degrees of freedom, matching the output. Under the specified linear-model assumptions, this slope test strongly rejects $H_0:{\beta }_1=0$; its calculated two-sided p-value is about $5.0\times 10^{-25}$, consistent with the displayed upper bound $2\times 10^{-16}$. The F test with 1 and 1782 degrees of freedom, $F\approx 110$, has about the same tail probability and tests the same zero-slope hypothesis. The significance stars merely encode thresholds; these are not independent pieces of evidence.^[ref]

Conditional on the printed slope and its standard error, a 95% confidence interval is approximatelyThat interval is a conditional reading of the coherent slope fields, not a raw-data refit. A p-value is not the probability of a causal age effect, and the confidence interval relies on the sampling and error model. Even a well-estimated observational association does not isolate age from other measured or unmeasured determinants of media use.

The printed coefficient of determination, $R^2=0.15816$, would mean that the fitted line accounts for 15.816% of the sample's centred diversity sum of squares, relative to an intercept-only fit. It would not mean that age causes 15.816% of diversity or correctly predicts that proportion of respondents. But the simple-regression output identities requireHere the supplied $R^2$ implies $F\approx 334.792$, not 110, whereas $F=110$ implies $R^2\approx 0.05814$. The printed slope and the two standard deviations independently give correlation $\hat{r}={\hat{\beta }}_1{s}_x/s_y\approx -0.24116$, whose square is about 0.05816. These mutually compatible fields suggest an explained fraction around 5.8% if they are trusted, rather than the printed 15.816%. No unique replacement value is claimed.

The adjusted coefficient of determination is also inconsistent. Using the printed $R^2$ and sample size gives $1-(1-0.15816)(1783/1782)\approx 0.1576876$, which cannot round to the printed 0.15763. Its conceptual purpose is to penalize the unadjusted fit measure for estimating the extra predictor, not to supply a separate effect estimate or significance test. It cannot repair the more substantial F–$R^2$ conflict.

The residual standard error of 9.94 points estimates unexplained response spread, not uncertainty in the slope and not a guaranteed error bound for every prediction. It approximately agrees with the smaller implied $R^2$: $1-(9.94^2\times 1782)/(10.240^2\times 1783)\approx 0.05826$. The printed $R^2=0.15816$ would instead imply residual standard error about 9.398. Rounding on those fields cannot bridge the difference. Thus uncertainty and fit must be read in internally compatible groups rather than mixed indiscriminately.

In a simple ordinary least-squares model, $S_{xx}=(n-1)s_x^2$ andWith the printed $s=9.94$ and predictor summary, these give slope standard error about 0.0139605, agreeing with the displayed 0.01396, but intercept standard error about 0.64524, not 6.30222. This is a further diagnostic inconsistency, not authority to guess where a decimal point should have been. The Student t test for a simple-regression intercept explains the relevant variance formula.

The residual quantiles extend from about $-23.26$ to $30.31$, with median about $-0.447$ and central quartiles about $-6.799$ and $5.509$. An intercept fit has zero residual mean, not necessarily zero residual median. These five quantiles do not establish a normal error distribution, constant conditional spread or absence of influential observations. A residual-versus-fitted plot could assess curvature and changing spread, and a normal probability plot could assess error shape; neither can be reconstructed from the supplied summary. An individual prediction interval includes residual variation and is wider than a confidence interval for a mean response.

Conventional t and F calibration assumes an appropriate linear mean, independent mean-zero errors conditional on age, and homoscedasticity; normal errors give the usual exact finite-sample theory. The bounded response scale and observational design call for assessing those assumptions rather than treating significance as model validation. The coherent coefficient/slope/F fields support a negative association with considerable unexplained individual variation. The printed explanatory-fit and intercept-uncertainty fields require correction before a single definitive numerical summary of the entire model is justified.

The descriptive statistics report 2466 observed ages and efficacy scores, with no missing values for either variable. Age ranges from 14 to 97 years, with sample mean 52.689 years and sample standard deviation 19.064 years. Efficacy ranges from 0 to 16, with sample mean 9.348 and sample standard deviation 3.176 scale points. The standard deviations use different units and should not be compared as though their numerical ratio measured relative substantive variability. The efficacy mean retains the conflict with Question 1's group summaries; that conflict is not resolved by this separate regression output.

A simple linear regression models $\mathbb{E}(Y\mid x)={\beta }_0+{\beta }_{1}x$, where $Y$ is efficacy and $x$ is age in years. Ordinary least squares chooses the coefficients to minimize the residual sum of squares. From the printed coefficients the fitted line isHolding to this model, ten additional years correspond to $-0.03192$ fitted scale points. For example, the fitted mean is about 9.453 at age 20 and 9.261 at age 80, a difference of about $-0.192$ points. These are fitted means, not exact predictions of individual scores. The line does not imply that every person's efficacy falls at that rate, nor that cross-sectional age differences identify within-person changes over time.

The intercept is the predicted mean at age zero, about 9.517. Since zero is outside the observed 14–97 age range, its direct interpretation requires extrapolation. The printed standard error of 0.188043 and $t\approx 50.608$ yield the extremely small displayed p-value for a zero intercept. That test does not test whether age is a useful predictor. Recentring age at its observed sample mean would put the intercept near $9.516515-0.003192(52.689)\approx 9.34833$, the fitted mean at a substantively relevant reference age. This agrees with the rounded overall efficacy mean.

The age standard error is 0.003356 points per year. The coefficient divided by this standard error gives $-0.003192/0.003356\approx -0.9511$, agreeing with the printed t value. Estimating an intercept and one regression slope from 2466 observations leaves $2466-2=2464$ residual statistical degrees of freedom. Under a correctly specified normal linear model, the Student t-distribution provides the stated two-sided p-value. At 5%, or 1%, it does not reject $H_0:{\beta }_1=0$. The p-value is not a 34.2% probability that age has no effect, and nonrejection does not establish a practically negligible effect size.^[ref]

A 95% confidence interval calculated from the printed estimate and standard error is approximatelyor about $[-0.09773,0.03389]$ points per decade. It includes zero and gives the scale of slopes compatible with these summaries and assumptions. Assessing a practically negligible effect would require an explicit substantive tolerance and an appropriate equivalence test, not simply the fact that $p>0.05$. A nonlinear association could also coexist with a near-zero overall linear slope.

The coefficient of determination is $R^2=0.000367$, meaning that the fitted line explains about $100R^2=0.0367\%$ of the sample's centred efficacy sum of squares compared with an intercept-only fit. This is not 36.7%, a fraction of people whose efficacy is predicted correctly, or the proportion of efficacy caused by age. The adjusted coefficient of determination, approximately $-0.00003873$, is slightly negative because its penalty for estimating the extra coefficient exceeds the tiny improvement in fit. It need not be constrained to be nonnegative and does not mean a negative amount of explained variation.

The overall $F\approx 0.9045$ on 1 and 2464 statistical degrees of freedom, with $p\approx 0.3417$, tests the same zero-slope hypothesis as the age t test because there is only one predictor. The simple-regression output identities give $F=t_{\mathrm{slope}}^2$ and $F=(2464)R^2/(1-R^2)$; the printed numbers agree allowing their displayed precision. The signed Pearson correlation coefficient is consequently about $-\sqrt{R^2}\approx -0.0192$. These are related summaries of the same weak linear association, not independent corroborating tests.

The residual standard error of 3.177 points estimates the typical unexplained error scale under the fitted model. It is close to the efficacy sample standard deviation, as expected when the line explains almost none of the variation. It is not a standard error of the slope or a guarantee that each individual's prediction error lies within 3.177 points. The residual summary spans roughly $-9.46$ to $6.75$, with sample median about 0.573 and central quartiles about $-1.437$ and $2.579$. An ordinary least-squares fit with an intercept has zero mean residual, not necessarily zero median residual. These few quantiles cannot establish normality, constant conditional spread, absence of outliers or linearity. The apparent last-decimal discrepancies among rounded error scales and other summary fields do not force an additional contradiction: their rounding intervals admit compatible regression summaries.

The conventional uncertainty calculations assume independent, mean-zero errors conditional on age, an appropriate linear mean, and homoscedasticity; normal errors give the usual exact finite-sample t and F calibration. Large samples can relax some distributional requirements but do not remove confounding, selection bias or dependence. A residual-versus-fitted plot could reveal curvature or changing spread, and a normal probability plot could assess error shape; neither plot can be reconstructed from the printed residual quantiles. A prediction interval for an individual includes residual variation and is wider than a confidence interval for the mean response. The supplied output supports a very weak, imprecisely signed linear association with little predictive gain from age alone, while leaving nonlinear, causal and design-based questions unresolved.

The response is a device count, while the predictor is a skill score. A simple linear regression models $\mathbb{E}(Y\mid x)={\beta }_0+{\beta }_{1}x$, where $Y$ is the device count and $x$ the skill score. Ordinary least squares chooses the intercept and regression slope to minimize the residual sum of squares. The regression slope describes a conditional association, not the causal effect of training or acquiring a device.

The descriptive statistics put the device count between $0$ and $8$, with sample mean about $4.494$ and sample standard deviation about $2.128$ among $2053$ records. Skill scores range from $0$ to $24$, with sample mean about $13.546$ and sample standard deviation about $5.989$ among $2051$ records. Two missing skill values account for the stated two deleted records; because device counts have no missing values, the regression complete-case sample has $2051$ observations. Estimating two regression coefficients leaves $2051-2=2049$ residual statistical degrees of freedom, matching that part of the output. The units of the two standard deviations differ, so comparing $5.989$ with $2.128$ does not establish that skill is more variable in a substantive sense.

If the coefficient estimates are trusted, the fitted values follow $\hat{y}=0.392122+0.209240x$. The intercept predicts about $0.392$ devices at a skill score of zero; zero is within the stated score range, but precision and the number of observations near zero are not shown. A five-point increase in score corresponds to about $1.0462$ additional fitted devices and a ten-point increase to about $2.0924$. At a score of ten the fitted value is about $2.4845$, and at 24 it is about $5.4139$. Expected counts may be fractional even though individual counts are integers. These are conditional interpretations of the coefficient fields, not reconstructed raw observations or validated individual predictions.

A coefficient's standard error estimates its sampling uncertainty, whereas the residual standard error describes unexplained response variation. For the intercept, $0.392122/0.114570\approx 3.42255$, agreeing with the printed $t=3.423$ and rounded $p=0.000633$. Conditional on the model assumptions, this would reject a zero intercept at conventional significance levels. The intercept test is not a test of whether skills predict devices. For the slope, the printed $p=0.013672$ would reject $H_0:{\beta }_1=0$ at 5% but not 1%; it does not assign a probability to the null hypothesis or measure the usefulness of prediction. The significance stars simply encode thresholds and add no independent evidence.^[ref]

The slope row fails its own arithmetic check:At $2049$ statistical degrees of freedom the ratio gives a two-sided p-value about $0.00845$, while the printed $t=2.882$ gives about $0.003993$; neither is $0.013672$. Thus the choice of which source field is trusted changes the apparent strength of evidence, including the 1% decision. A 95% confidence interval built conditionally from the printed estimate and standard error would be about $[0.0536,0.3649]$, but it is not a verified interval from a coherent fitted model. In fact, using the printed residual standard error $1.793$, the complete predictor sample size and predictor standard deviation gives a slope standard error near $0.006612$, not $0.079377$.

The printed coefficient of determination $R^2=0.2893$ would mean that the fitted line explains $28.93\%$ of the complete-case sample's centred device-count sum of squares relative to an intercept-only fit; $71.07\%$ remains unexplained by that comparison. It is not the percentage of devices caused by skill, the fraction of correctly predicted households, or a measure of future predictive accuracy. The adjusted coefficient of determination $0.289$ is consistent with that $R^2$ and the given sample size, allowing rounding. The small adjustment reflects fitting one slope among many cases; it does not validate the model.

For a one-predictor ordinary least squares fit with an intercept and the usual common-variance standard errors, the simple-regression output identities requireHere the printed slope t gives $F\approx 8.305924$, the printed F field says $83.2$, and the printed $R^2$ instead implies $F\approx 834.073$. These cannot all hold. Moreover, $F=83.2$ on $1$ and $2049$ statistical degrees of freedom has p-value about $1.72\times 10^{-19}$, not $0.001978$. Reading the printed F p-value alone would give a 1% significant test of the zero-slope model, but its reported calibration is incompatible with its statistic. No significance label resolves these identities.

There is also a contradiction that does not depend on tail probabilities. Ordinary least squares with an intercept makes the arithmetic mean of fitted values equal the arithmetic mean of observed responses on the analysed records. The printed coefficients and predictor mean give $0.392122+0.209240(13.546)\approx 3.22649$. Deleting two device counts anywhere in the printed range $[0,8]$, and allowing the reported overall mean to be rounded, would leave the complete-case device mean between approximately $4.49008$ and $4.49888$. The mean bounds after deleting bounded observations therefore rule out reconciling $3.22649$ with the response summary merely by citing those two missing predictor values. This is not the mistaken assumption that full-sample and complete-case sample means must always be identical.

If its own field were trusted, residual standard error $1.793$ estimates the unexplained response standard deviation, in device-count units. It is neither the standard error of the slope nor a guarantee that every prediction is accurate to $1.793$ devices. The residual five-number summary has extremes near $-5.206$ and $5.086$ and median near $-0.060$, with central quartiles near $-1.372$ and $1.252$. The roughly central median does not establish normally distributed errors, homoscedasticity, independence or a linear conditional expectation. A residual-versus-fitted plot could check curvature and spread; a normal probability plot could assess an error-shape approximation. Neither diagnostic can be recovered from five residual quantiles alone. Individual prediction intervals include response variation as well as coefficient uncertainty and are wider than confidence intervals for the conditional expectation.

For conventional finite-sample t and F calibration, a normal linear model assumes independent mean-zero normal errors of common variance with an appropriate predictor specification. Mean-zero errors conditional on skill and adequate sampling remain essential even when large-sample approximations relax normality. A bounded count response can have nonconstant conditional variance, so residual diagnostics and a scientifically suitable response model matter; count data do not become normally distributed just because the sample is large. Poisson regression is one possible alternative to investigate, but its own conditional mean–variance assumptions and the bounded device range require assessment, not automatic substitution. Complete-case selection can also bias the fitted association if it depends on the response in a relevant way. The only safe summary is conditional: the printed slope is positive, but the numerical effect, its standard error, t/F evidence and explanatory fit require a corrected, coherent output before a single definitive regression claim is made.

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

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

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

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

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

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.

For the initial normal linear model, the design matrix has rows $(1,x_1,x_2,x_3)$. The eight factorial corners have balanced signs and orthogonal coordinate columns; the six centre points contribute only to the intercept. ConsequentlyUsing the least-squares normal equations gives the least-squares estimatorIt is a linear transformation of independent normal responses, so its sampling distribution isThe four estimated coefficients are independent, since they are jointly normal with diagonal covariance matrix.

Adding the axial settings produces a central composite design: eight factorial corners, six axial points and six replicated centre points. The axial distance $1.682$ is the rounded value of $8^{1/4}=1.6817928\dots$. With that exact distance it is a rotatable design for a full quadratic model. Indeed, sign symmetry annihilates odd moments, all coordinates have the same second moment, and the fourth moments satisfyRotational symmetry requires the first to be three times the second, giving ${\alpha }^4=8$. Using the printed rounded distance makes rotatability approximate, rather than mathematically exact.

The full quadratic regression mean has ten coefficients:The expanded design matrix has rank ten. To verify this rather than just count parameters, suppose such a quadratic vanishes at every setting. Its centre value sets its constant term to zero; subtracting the values at opposite axial points sets every linear coefficient to zero, and adding them then sets every squared-coordinate coefficient to zero. On the factorial corners, orthogonality of the three pair-product columns forces their coefficients to zero as well. Thus all ten columns are independent, leaving $20-10=10$ residual statistical degrees of freedom.

There are $15$ distinct settings: eight corners, six axial points and the centre. The six centre replicates give the only pure error, with $6-1=5$ degrees and sum of squares $859.33$. The saturated setting-mean model has $15$ parameters, so the lack of fit beyond the quadratic model has $15-10=5$ degrees. For any setting $r$ with $n_r$ replicates, the identitysplits the residual sum of squares into pure error and lack of fit; the cross term vanishes because the deviations from the setting mean sum to zero. Summing givesUnder a correct quadratic mean with independent errors of constant normal variance, the pure-error and lack-of-fit projections are orthogonal, hence their sums of squares are independent with respective distributions ${\sigma }^2{\chi }_5^2$ and ${\sigma }^2{\chi }_5^2$. The Lack-of-fit F-test therefore usesThis is below the given $10\%$ and $5\%$ critical values $3.45$ and $5.05$. There is no significant lack of fit at either level. That is failure to reject this quadratic mean, not proof that it is exact or that a maximum has already been located. The pure-error denominator also assumes the error variance at the centre represents the common variance throughout the experiment.

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.

Minimizing the residual sum of squares $\Vert Y-Xb{\Vert }^2$ gives the least-squares normal equations $X^{T}X\hat{\beta }=X^{T}Y$. Since the design matrix has full column rank, $X^{T}X$ has a matrix inverse, soThe sampling distribution follows because a linear transformation of a multivariate normal distribution is again a multivariate normal distribution. Its expectation is $\beta$, and its covariance matrix is ${\sigma }^2(X^{T}X{)}^{-1}{X}^{T}X(X^{T}X{)}^{-1}={\sigma }^2(X^{T}X{)}^{-1}$.

The hat matrix is $H=X(X^{T}X{)}^{-1}{X}^T$. It is the orthogonal projection matrix onto the column space of $X$: $H^T=H$, $H^2=H$ and $HX=X$. Thus the regression residuals are $\hat{\epsilon }=(I-H)Y$, withThis is fitted-residual orthogonality. The vectors are jointly normal, so independence of uncorrelated jointly normal variables strengthens both zero-covariance statements to independence.

For the quadratic linear regression, the columns of the design matrix are $1,x,x^2$. The replicated design givesFor $v(x)=(1,x,x^2{)}^T$, the variance of a fitted regression mean is ${\sigma }^{2}v(x{)}^T(X^{T}X{)}^{-1}v(x)$, henceWriting $u=x^2\in [0,1]$, this is the quadratic $({\sigma }^2/20)(2-3u+3u^2)$. Its derivative vanishes at $u=1/2$; it decreases before that point and increases afterwards. Therefore the maximum is ${\sigma }^2/10$ at $x=-1,0,1$, and the minimum is ${\sigma }^2/16$ at $x=\pm 1/\sqrt{2}$.

At $100^{\circ }\mathrm{C}$, $x=0$ and the expected yield is $\alpha$. The unbiased residual variance estimate is $s^2=2.43/(30-3)=0.09$. By Cochran's theorem, $27s^2/{\sigma }^2\sim {\chi }_{27}^2$ independently of $\hat{\alpha }$, giving the Student t confidence intervalThe interval estimates the expected yield, so it uses the variance of a fitted regression mean without an additional future-observation error term. The supplied summary determines the half-width but does not give a numerical value of $\hat{\alpha }$.

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.

Let $\mathrm{RSS}={\sum }_i(Y_i-\hat{\alpha }-\hat{\beta }{x}_i{)}^2$. Differentiating the log-likelihood in $v={\sigma }^2$ givesFor $\mathrm{RSS}>0$ this changes from positive to negative at $v=\mathrm{RSS}/n$, soTo find its bias, substitute the model and use the orthogonal fit directions $1$ and $\mathbf{x}$, obtainingThe three expectations are $n{\sigma }^2$, ${\sigma }^2$, and ${\sigma }^2$, respectively: $\mathrm{Var}(\bar{\epsilon })={\sigma }^2/n$ and $\mathrm{Var}({\sum }_i{x}_i{\epsilon }_i)={\sigma }^2{S}_{xx}$. Hence $\mathbb{E}\mathrm{RSS}=(n-2){\sigma }^2$, giving the unbiased estimatorWith $n>2$, a zero residual sum of squares is a probability-zero event in the model. For that exceptional dataset the likelihood is unbounded as ${\sigma }^2\downarrow 0$, rather than having a maximum at a positive variance. In the intercept-only case $S_{xx}=0$, the corresponding unbiased multiplier is $n/(n-1)$ for $n>1$.