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In a normal linear model with an intercept, $n$ observations and $p$ independent mean coefficients, adjusted $R^2$ isIt compares the residual variance estimate with the unmodelled sample variance, adjusting the raw coefficient of determination for fitted dimension. It can be negative and is not guaranteed to increase when another regressor is added. In penalized regression a related convention uses the fit's effective degrees of freedom instead of $p$; the precise smoothing-software variance convention should be specified.

A likelihood-based analogue of the coefficient of determination compares the fitted and intercept-only log-likelihoods ${\ell }_1$ and ${\ell }_0$:For a Cox proportional-hazards model, these are maximized partial likelihood logarithms with and without the covariates. The expression measures improvement in fit, rather than the fraction of variation in event times explained. Its attainable maximum can be below one. It does not test the proportional hazards assumption or establish predictive calibration.

Multiple linear regression models a scalar response by an intercept and several predictors. It is linear in the regression coefficients, even when predictors include transformed variables or interactions. In an observational study, adjustment can address measured confounding, but does not automatically identify a causal effect. Improvements in fitted coefficient of determination require external predictive assessment rather than being treated as proof of improved future prediction.

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.

The fitted simple linear regression for the numerical vaccination intention scale isHere age is measured in years. The fitted mean increases by about 0.01649 score points per year, or 0.16489 points per decade. The regression coefficient is an association across respondents, not an estimate of how an individual would change on ageing or of a causal effect of age.

The standard error of the age regression coefficient is $0.006806$. Dividing gives $t\approx 2.423$, with 2,133 residual statistical degrees of freedom, corresponding to $2135-2$ fitted residual degrees. The reported two-sided p-value is $0.0155$, so the zero-slope null hypothesis is rejected at a prespecified 5% level, but not at 1%. An approximate 95% confidence interval isIts positive endpoints are consistent with the test decision; they do not make the slope large.

The coefficient of determination is $R^2=0.002744$, so the fitted age trend accounts for only 0.2744% of the observed score variation, compared with an intercept-only prediction. Its adjusted value is $0.002276$. In an intercept-containing simple linear regression, the positive slope implies a positive Pearson correlation coefficient $r=\sqrt{R^2}\approx 0.05238$. The reported overall $F$ statistic, $5.869$ on one and 2,133 degrees of freedom, agrees with the squared slope test statistic up to printed rounding.

The residual standard deviation estimate is $5.618$ score points, much larger than the fitted decade contrast. At age 50 the fitted value is about $13.4697$, and at 60 it is about $13.6346$. Across the observed ages 19 to 91, the fitted mean changes by only $1.1872$ points. Statistical detection with 2,135 observations therefore coexists with weak individual prediction.

The intercept predicts $12.645291$ at age zero, outside the observed age range; it should not be given a substantive infant interpretation. Centring age at its observed sample mean, $50.174$, giveswhich agrees with the reported score sample mean of $13.473$ after rounding. The bounded scale and upper quartile at its maximum motivate attention to a possible ceiling effect; they do not establish homoskedasticity, a normal error distribution or a satisfactory linear conditional mean. Five residual quantiles cannot replace a residual plot or checks of the sampling assumptions.

There is also a source consistency problem. The output reports that all 2,135 records have age and score present, that the attained minimum score is one, and that ages lie between 19 and 91. The positive fitted slope therefore places every fitted value between $12.958582$ and $14.145790$. By residual extrema when response extrema are attained, the minimum regression residual must satisfyAt the observation with score one, its residual is already at most $1-12.958582=-11.958582$. The printed minimum residual, $-10.014$, lies outside this necessary interval. Those summaries cannot all describe the same complete sample and fitted model. The discrepancy is too large to be explained by the displayed rounding. It is present in the original PDF; the individual records would be needed to establish which printed item is wrong.

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 model's explained-variance value is absent. For an ordinary least-squares linear regression with an intercept, the coefficient of determination is $R^2=1-S{S}_{\mathrm{residual}}/S{S}_{\mathrm{total}}$. Its value multiplied by one hundred is the corresponding percentage of sample variation explained by that fitted model. Read the actual reported $R^2$ and distinguish it from the adjusted coefficient of determination; neither is a verified prediction-performance estimate on new data. Without the missing summary or sums of squares, no numerical percentage can be recovered.

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.

The association is strongly statistically significant under the reported regression assumptions, but age alone has only moderate predictive ability. The reported confidence interval for the age regression coefficient excludes zero by a wide margin. If it is the usual two-sided 95% interval from a normal linear model with an intercept and one predictor, there are $206-2=204$ degrees of freedom. Taking $t_{0.975,204}\simeq 1.97$ givesThus the two-sided p-value against zero slope is far below $0.001$. This conclusion concerns a statistical association in the sampled population; it does not establish an individual growth rate. Clustering by practitioner, nonconstant variance or nonindependent observations would require an appropriate standard error rather than blind reliance on this calculation.

The scatter plot has substantial vertical spread at each age. A narrow slope confidence interval describes uncertainty about the average trend, whereas a prediction interval for a new person's ear length must also include the residual variance. The reported interval even permits an approximate numerical assessment: write $S_{xx}={\sum }_i(x_i-\bar{x}{)}^2$ and let $s_e^2=\mathrm{RSS}/(n-2)$. In simple linear regression, $t^2={\hat{\beta }}^2{S}_{xx}/s_e^2$, while the fitted and residual sums of squares are ${\hat{\beta }}^2{S}_{xx}$ and $(n-2)s_e^2$. Hence the coefficient of determination isUnder these assumptions, age explains only about 27% of the sample variation; roughly 73% remains unexplained. The calculation is approximate because the printed slope and interval are rounded. For a new individual aged $x_0$, the usual prediction interval is based onThe leading one represents individual residual variation. The provided summaries do not determine $s_e$ and $S_{xx}$ separately, so they do not supply a numerical prediction interval.

Record sex, height or another measure of body size, ancestry, family characteristics and practitioner or measurement method. Use these covariates in a multiple linear regression, allowing scientifically justified nonlinear age effects or interactions, and assess prediction error on held-out observations. A guessed overall sex proportion cannot replace individual sex measurements or identify sex-adjusted effects.

There is a source inconsistency: the PDF's numerical ear-length summary is ten times the scale of its plotted measurements and fitted line. Indeed, the fitted line at the reported mean age gives about $67.7$ mm, whereas the printed mean is about ten times larger. Missing decimal points in the summary are a plausible explanation. The significance and approximate $R^2$ above use the mutually consistent slope, interval and graph; they do not treat the inconsistent summary as valid data.

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 $Y_{ijkl}$ 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.

The additive model gives useful evidence for both predictors. To test for no adjusted sex effect, use $H_0:{\beta }_M=0$ against $H_1:{\beta }_M\ne 0$. The statistic $T={\hat{\beta }}_M/\mathrm{se}({\hat{\beta }}_M)$ has a Student t-distribution with 97 statistical degrees of freedom under the normal linear model and the null hypothesis. Here $T=3.016$ and $p=0.00327$. Similarly, $H_0:{\beta }_h=0$ against a nonzero common slope gives $T=4.397\sim t_{97}$ under the null hypothesis, with $p=2.81\times 10^{-5}$. Thus, at a fixed height males have a higher fitted mean weight, and within sex greater height is associated with greater weight.

The interaction model adds $\delta M(x-165)$. Its female height slope is $0.6696$ and its male height slope is $0.6696-0.0977=0.5719$. Test $H_0:\delta =0$ against $H_1:\delta \ne 0$ using $T=-0.344\sim t_{96}$ under the null hypothesis, or equivalently $T^2\sim F_{1,96}$. The $p$-value is 0.73139. The additive model is a reasonable parsimonious choice; the data give no evidence for different height slopes between sexes. This does not establish exact equality of population slopes. The additive model also has the higher adjusted coefficient of determination, 0.405 versus 0.3996, and a slightly smaller residual standard error.

Ignoring sex gives a height slope of 0.8769, appreciably above the adjusted slope 0.6211. This is omitted-variable bias in the descriptive regression: a taller group with a larger intercept makes the pooled height association steeper. Algebraically, with $z=x-165$,The printed estimates imply a positive height-sex covariance. Pooled and adjusted slopes therefore describe different comparisons; the difference is not evidence that adding sex changed a physical effect of height. For predictions, retain both sex and height, check regression diagnostics, and assess performance on new students. The residual standard error near 5.7 remains substantial and the coefficient of determination near 0.42 does not promise precise individual prediction.

For the normal linear model, the log-likelihood isThe full column rank of the design matrix makes $X^{T}X$ invertible. Differentiating in $\beta$ gives the normal equations $X^{T}X\hat{\beta }=X^{T}Y$. With the minimized residual sum of squares substituted, differentiation in ${\sigma }^2$ givesThese are the maximum-likelihood estimators almost surely; $RSS>0$ almost surely when ${\sigma }^2>0$ and $n>p$. The linear transformation of the multivariate normal distribution yieldsDefine the hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$. The fitted values are $\hat{Y}=HY=X\hat{\beta }$, the regression residuals are $e=Y-\hat{Y}=(I-H)Y$, and $RSS=e^{T}e$. The orthogonal projection $I-H$ has rank $n-p$ and annihilates $X\beta$. By Cochran's theorem,Consequently $E({\hat{\sigma }}^2)=(n-p){\sigma }^2/n$, with bias $-p{\sigma }^2/n$. The unbiased estimator isIts square root, the reported residual standard error, estimates $\sigma$; the assertion of unbiasedness applies to the variance, not generally to its square root.

For the paper-strength analysis, let $h_i$ be the measured percentage and $z_i=h_i-\overline{h}$. Both normal linear models use independent errors of common variance ${\sigma }^2$. Their mean functions are ${\beta }_0+{\beta }_1{z}_i$ for lm1, and ${\beta }_0+{\beta }_1{z}_i+{\beta }_2{z}_i^2$ for lm2, with separately fitted coefficients. The reported residual standard errors are $11.82$ and $4.42$, respectively.

For lm1, the estimated conditional expectation at a new percentage $x$ isThe original data mean $\overline{h}$ must be used for centering the new percentage; its numerical value is not supplied in the excerpt. Because ${\sum }_i{z}_i=0$, the two columns of this design matrix are orthogonal, and $\mathrm{Cov}({\hat{\beta }}_0,{\hat{\beta }}_1)=0$. Hence the coefficient standard errors in the output giveThis estimates uncertainty in the mean. Predicting an individual future batch would additionally require the new-error variance, estimated by $11.82^2$.

To compare the nested normal linear models, test $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$ within the quadratic model. The Student t-test statistic isIts two-sided p-value is $1.89\times 10^{-8}$. Equivalently $T^2$ is a partial F-test statistic with null distribution $F_{1,16}$. Reject the linear restriction and prefer lm2 at the 5% level. Its residual spread is much smaller; the increase in the coefficient of determination from $0.3054$ to $0.9085$ supports the same conclusion, although the test is the relevant complexity-adjusted comparison.