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Let $s_i$ be speed and $j_i\in \{1,2,3,4\}$ the tool type. Each fit is a normal linear model with independent $N(0,{\sigma }^2)$ errors. The three mean specifications areThe constraints are treatment coding with type 1 as reference. The parameter counts are $2,5,8$, respectively. Model 2 gives parallel lines; model 3 allows interaction terms and distinct slopes.

In the analysis of variance interaction row, adding three independent slope differences costs three degrees of freedom. Its extra sum of squares is the reduction in residual sum of squares, $118.53-91.51=27.02$. Its mean square is $27.02/3\approx 9.007$, and its F-test statistic isThus the four missing entries are $3$, $27.02$, $9.01$, and $1.18$, to the precision allowed by the rounded output. This tests $H_0:{\delta }_2={\delta }_3={\delta }_4=0$ against the alternative that at least one slope difference is nonzero. Under $H_0$ and the normal linear model assumptions, the statistic has distribution $F_{3,12}$. Its p-value $0.3579213$ gives no reason to reject at 5%.

The simpler common-line model is inadequate compared with the parallel-line model. Testing $H_0:{\tau }_2={\tau }_3={\tau }_4=0$ against at least one nonzero type effect, while retaining speed, gives the partial F-testIts p-value is far below $0.001$, so reject the common-line restriction. This reduction differs from the sequential type sum of squares in the displayed table, because that table adds type before speed. Recommend model 2: different intercepts and a common decreasing slope.

The selected coefficient estimates giveThe intercept estimates the expected lifetime of type 1 at speed zero; if zero speed is outside the data range, it is only an extrapolated intercept. Its standard error is $4.039533$, and its Student t-test ratio is $8.885$, with two-sided p-value $2.31\times 10^{-7}$. At any fixed speed, type 2 differs from type 1 by $-0.833408$ hours, with standard error $1.904707$, $t=-0.438$ and p-value $0.668$, giving little evidence of a difference. Types 3 and 4 exceed type 1 by $12.607660$ and $16.539121$ hours, with standard errors $1.777937$ and $1.876116$; their test ratios $7.091$ and $8.816$ and p-values $3.68\times 10^{-6}$ and $2.55\times 10^{-7}$ support positive differences. Each of these individual coefficient tests has null value zero, alternative nonzero, and null distribution $t_{15}$.

The speed coefficient means a reduction of $0.024585$ hours per extra revolution per minute, or $2.4585$ hours per additional 100 rpm, for every type. Its standard error is $0.004682$; the ratio $-5.251$ has null distribution $t_{15}$ when the common slope is zero, and two-sided p-value $9.77\times 10^{-5}$.

The residual standard error $2.811$ estimates the common noise standard deviation in hours using 15 residual degrees of freedom, since $20-5=15$. The coefficient of determination $0.9247$ means that about $92.47\%$ of the corrected lifetime variation is explained by this fit. The overall F-test statistic $46.08$ tests the simultaneous restriction ${\tau }_2={\tau }_3={\tau }_4=\beta =0$ against at least one nonzero coefficient; its null distribution is $F_{4,15}$ and its p-value $2.974\times 10^{-8}$ rejects an intercept-only mean. These conclusions remain conditional on suitable regression diagnostics for independent, homoscedastic, approximately normal errors.

The required sketch plots the four fitted lines. Its speed interval is illustrative because the observed speeds are not supplied; the ordering and vertical gaps are determined by the fitted coefficients.

For the gamma variance function, the Pearson dispersion estimator isusing $61-4=57$ residual degrees of freedom. Under a specified null dispersion ${\varphi }_0$, the scaled statistic based on squared Pearson residuals $57\hat{\varphi }/{\varphi }_0$ has an approximate ${\chi }_{57}^2$ distribution under the usual residual approximation.

Thus test1 corresponds to $H_0:\varphi =1$, equivalently gamma shape $\alpha =1$, the exponential distribution. test2 corresponds to $H_0:\varphi =1/3$, equivalently shape $\alpha =3$. The null hypotheses concern dispersion or shape, not whether the regression coefficients vanish.

The code computes lower-tail probabilities. Used as one-sided tests, the alternatives are $\varphi <1$ and $\varphi <1/3$, respectively, equivalently shapes larger than one and three. At the 5% level the first null is rejected because the lower-tail probability is $1.199504\times 10^{-7}$, while the second is not rejected because its probability is $0.3768748$. If the intended alternatives are two-sided, $\varphi \ne {\varphi }_0$, these displayed numbers must not be called two-sided p-values: doubling the smaller tail gives approximately $2.40\times 10^{-7}$ and $0.75375$, with the same decisions. The chi-squared distribution approximation is not an exact finite-sample gamma identity.

The dispersion-one assumption is strongly contradicted by the preceding calculation. Use the gamma analysis with estimated dispersion, namely the F table. A failure to reject shape three does not establish that the true shape equals exactly three; retaining the estimated dispersion parameter $0.3103711$ is appropriate.

In the analysis of deviance for nested generalized linear models, a deviance reduction for $r$ extra coefficients is divided by $r\hat{\varphi }$ when dispersion is estimated. The approximate null calibration is $F_{r,57}$. For the type term, the null is that the two type contrasts vanish, against at least one nonzero contrast. The statistic iswith approximate null distribution $F_{2,57}$ and p-value $0.02007$. For position, the null is that its coefficient vanishes after accounting for type; the statistic iswith approximate null distribution $F_{1,57}$ and p-value $0.34963$. At 5% there is evidence that component type affects the expected failure time, and no significant additional position effect. The table is sequential: the type comparison is to the intercept-only model, while the position comparison adjusts for type. The individual gamma coefficient tests also suggest that type 3 accounts for the clearer type difference. The inverse link function means its positive contrast corresponds to a lower fitted mean failure time, holding position fixed.

The three Poisson regression models have independent counts and a logarithmic link function: their mean predictors include both pc and urban in m1, only urban in m2, and only pc in m3. Nested comparisons through the full model are preferable to a direct comparison of the two nonnested one-predictor models.

To remove urban from m1, test $H_0:{\beta }_{\mathrm{urban}}=0$ against $H_1:{\beta }_{\mathrm{urban}}\ne 0$. The likelihood-ratio test statistic is the deviance differencewith approximate null distribution ${\chi }_1^2$. It is below $3.84$, so do not reject at 5%. The corresponding Wald p-value $0.588$ is consistent with this decision. To remove pc from m1, test $H_0:{\beta }_{\mathrm{pc}}=0$ against a nonzero coefficient, giving $559.17-290.64=268.53$ with approximate null distribution ${\chi }_1^2$; reject decisively. The Akaike information criterion also prefers m3, whose value $710.94$ is smaller than $712.65$ and $979.19$.

Among these three models choose m3, retaining building cover and omitting the urban indicator. This is a comparison within the stated Poisson family; its absolute fit must still be checked, as the next part does.

The first model's residual deviance is $671.27$ on $197$ residual degrees of freedom, a ratio of about $3.41$. This is far above the scale expected from a well-fitting dispersion-one grouped binomial regression. It suggests an inadequate mean function, overdispersion, or both. The linear age and exposure restrictions may miss nonlinear associations; the generalized additive model allows those shapes to be estimated through penalized cubic regression splines rather than assumed.

The very small exposure Wald p-value in the first fit suggests an association, but does not establish that a linear logit effect is adequate. Nor does the nonsignificant linear age term exclude every possible nonlinear age effect. Allowing estimated scale also addresses the excessive residual variation, which can arise from clustered data because several births belong to the same woman. The second model's estimated scale $3.3506$ confirms that permitting smooth means has not removed all extra variation. The reason to proceed is poor initial fit and possible nonlinear effects, with dispersion-aware inference, not merely the wish to obtain more significant tests. These observational fits alone do not establish causation by the disaster.

The original PDF's age smooth is nearly flat, its uncertainty band includes zero across the age range, and its approximate p-value is $0.859$. The exposure smooth is increasing and convex: it is fairly flat at low exposure and rises more steeply at high exposure. Its effective degrees of freedom $1.895$ suggest a modest departure from a straight line, rather than a highly oscillatory relationship.

A suitable simpler model is therefore the grouped quasibinomial regressionwith age omitted and dispersion estimated. A positive quadratic coefficient captures the convex exposure relationship; retaining the linear term respects the usual polynomial hierarchy. This model should be refitted, rather than assigning coefficients from the smooth plot. Compare its residual pattern and dispersion-adjusted fit with the additive model to check that the quadratic approximation is adequate. An exposure-only quadratic logit model is a justified parsimonious candidate, while the graph does not justify a sharp threshold or a zero-dispersion model.

Under the point hypothesis $\beta =0$, the test statistic $\sqrt{n}\overline{Y}$ has a standard normal distribution. The observed statistic is three, giving a two-sided p-value $2[1-\Phi (3)]\simeq 0.00270$. This is conventionally strong evidence against that point hypothesis. The narrow model $H_0$ is a continuous prior distribution around zero, rather than a point hypothesis.

For $n_0/n=100$ and $c=100$, Normal-normal conjugacy givesThe narrow-model posterior mean is strongly pulled toward zero and its standard deviation is about $0.0995/\sqrt{n}$. The wide-model posterior mean is about $2.9703/\sqrt{n}$, very close to the observation, with standard deviation about $0.9950/\sqrt{n}$. Each model produces a markedly different posterior, so choosing between them requires their predictive evidence.

The familywise error rate is the probability of rejecting at least one true null:A valid p-value is super-uniform under its null: $\mathbb{P}(p_i\le t)\le t$ for $0\le t\le 1$. The Bonferroni correction rejects $H_i$ if $p_i\le \alpha /m$. The union bound givesThis controls the familywise error rate for every configuration of true and false nulls, without any independence assumption on the p-values.

For a nonempty index set $I\subseteq \{1,\dots ,m\}$, the intersection hypothesis $H_I={\bigcap }_{i\in I}{H}_i$ asserts that every null indexed by $I$ holds. The closure of a family of statistical hypotheses consists of all these nonempty intersections, including the original hypotheses. For each intersection choose a local test ${\varphi }_I$ of size at most $\alpha$ under $H_I$. Such tests always exist: for example, within $I$ use the Bonferroni correction, rejecting $H_I$ if ${\min }_{i\in I}{p}_i\le \alpha /|I|$. More powerful local intersection tests can be substituted whenever valid.

The closed testing procedure rejects an elementary hypothesis $H_i$ only when every local test ${\varphi }_I$ with $i\in I$ rejects its intersection hypothesis. If $I_0\ne \varnothing$, the intersection $H_{I_0}$ is true. Rejecting any true $H_i$ requires rejection by its local test ${\varphi }_{I_0}$, soIf $I_0=\varnothing$, there is no possible false rejection. This proves closed-testing control of the familywise error rate. The same argument applies if intersection hypotheses themselves are reported as rejected: requiring all their supersets to reject includes $I_0$ whenever the reported intersection is true.

Finally, consider the given hierarchical family as a laminar family of sets. Call a set $I\in \mathcal{I}$ true when $I\subseteq I_0$, so its entire intersection hypothesis is true. Let $\mathcal{M}$ be the inclusion-maximal true sets in $\mathcal{I}$. Distinct members of $\mathcal{M}$ are disjoint: if they met, laminarity would make one contain the other, contradicting maximality. Consequently,Every true set $I$ lies in a maximal true $J$. If $H_I$ is rejected by the prescribed adjusted p-value rule, the maximum over its ancestors includes $J$, henceThus the event of any false rejection is contained in the union of these events over the disjoint maximal true sets. Using validity of each $p_J$ under its true intersection and the union bound,Equivalently, with probability at least $1-\alpha$ every rejection is correct. This proves familywise error control for a laminar hypothesis family. It requires neither independence nor the presence of all possible intersections or singletons in $\mathcal{I}$. Empty index sets are excluded, as required by the denominator $|J|$; if there are no true sets the result is immediate. Adjusted values exceeding one can be truncated at one without changing decisions for $\alpha <1$.

The ordinary least squares objective has gradient $2X^{T}X\beta -2X^{T}Y$. The normal equation therefore givesThe Gram matrix $X^{T}X$ is a positive-definite matrix because $X$ has full column rank. More explicitly, for any $b$,since $X^T(Y-X\hat{\beta })=0$. This proves that the displayed solution is the unique global minimum.

The fitted values are $\hat{Y}=X\hat{\beta }=HY$, where the hat matrix isThe inverse of the symmetric matrix $X^{T}X$ is symmetric, so $H^T=H$. Multiplication gives $H^2=X(X^{T}X{)}^{-1}(X^{T}X)(X^{T}X{)}^{-1}{X}^T=H$. Thus $H$ is the orthogonal projection onto the column space of the design matrix.

The regression residual vector is $e=Y-\hat{Y}=GY$ with $G=I_n-H$. Consequently $G^T=G$, $G^2=I_n-2H+H^2=G$, and the residual sum of squares isBy fitted-residual orthogonality, $GH=0$. Using $\mathrm{Cov}(Y)={\sigma }^2{I}_n$, the cross-covariance matrix isBoth vectors are jointly normal, so they are also independent by independence of uncorrelated jointly normal variables. The zero covariance calculation itself only needs the common error variance and lack of error correlations.

For the first wind fit, let $v_i$ be wind velocity and $Y_i$ electrical output. The simple linear regression is $Y_i=\alpha +\beta v_i+{\epsilon }_i$, with independent ${\epsilon }_i\sim N(0,{\sigma }^2)$. It has two fitted mean parameters and $25-2=23$ residual degrees of freedom. The missing analysis of variance entries are therefore velocity degrees of freedom $1$, velocity mean square $8.9296$, and residual degrees of freedom $23$. The residual mean square is $1.2816/23\simeq 0.05572$, and $8.9296/(1.2816/23)\simeq 160.25$, consistent with the printed value after rounding.

Because the model includes a regression intercept, the coefficient of determination is the explained sum of squares divided by the total centered sum of squares:A large coefficient of determination does not rule out a wrong mean function. In the PDF's first residual-versus-fitted plot, residuals are negative at both ends and positive in the middle. This curved pattern agrees with the visibly flattening output-versus-velocity relationship and motivates a polynomial regression with a quadratic term.

The second wind model is $Y_i=\alpha +{\beta }_1{v}_i+{\beta }_2{v}_i^2+{\epsilon }_i$, again with independent common-variance normal errors. Its fitted mean is $-1.155898+0.722936v_i-0.038121v_i^2$. The line marked (A) is a two-sided Student t-test of $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$, conditional on retaining the intercept and linear term. Under the null hypothesis,Its two-sided p-value is $6.59\times 10^{-8}$. This is far below $0.05$, so reject a purely linear mean in favour of the quadratic fit. Equivalently, the extra-term nested-model F-test has $F=t^2\simeq 63.15$ and null law $F_{1,22}$.

The third fit is a reciprocal-predictor regression, $Y_i=a+b/v_i+{\epsilon }_i$, with estimated mean $2.9789-6.9345/v_i$. It has two mean parameters rather than the quadratic model's three. Its residual standard error is smaller, $0.09417$ versus $0.1227$, and its coefficient of determination is larger, $0.9800$ versus $0.9676$. The corresponding residual sums of squares are approximately $23(0.09417{)}^2=0.20396$ and $22(0.1227{)}^2=0.33122$. The reciprocal fit is preferable on both these fit measures and parsimony. The two models are not nested, so an ordinary extra-term F-test between them is inappropriate. On the common normal-error likelihood, the difference ${\mathrm{AIC}}_3-{\mathrm{AIC}}_2=25\log (0.20396/0.33122)-2\simeq -14.12$ also favours the third model.

Neither lower residual sum of squares nor a higher coefficient of determination establishes adequate assumptions. The quadratic residual-versus-fitted plot removes the original pronounced curvature; the reciprocal plot also has no comparably obvious mean trend. The low-output residuals appear somewhat more spread out, so check scale-location plots and residuals against velocity for heteroscedasticity. Q-Q plots assess normality; residuals against observation order assess serial dependence; regression leverage and Cook's distance identify influential observations. Further cross-validation, replicate observations at comparable velocities, and prediction errors would help choose between their extrapolation behaviours. Both fitted shapes should be judged principally over the observed positive-velocity range.

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

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

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

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

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

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

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

Let $Y_i$ be the number of maintenance jobs, $t_i$ average temperature and $p_i$ average precipitation. The fitted Poisson regression assumes independent conditional responses withThe logarithm is the Poisson canonical link, so fitted means are always positive. The estimated regression coefficients areAt fixed precipitation, one unit of temperature multiplies the fitted mean by $e^{-0.006162}\simeq 0.99386$, a decrease of about $0.614\%$. The fitted precipitation multiplier is $e^{-0.002922}\simeq 0.99708$ per unit, but its large p-value gives little evidence for that effect. These are conditional associations, not established causal effects.

The intercept-only model has $n-1=23$ residual degrees of freedom, and the full model has $n-3=21$. Both imply $n=24$ months.

For an approximate deviance goodness-of-fit test of the full Poisson regression, compare residual deviance $23.527$ with ${\chi }_{21}^2$. Its upper-tail p-value $0.3165361$ supplies no evidence of lack of fit. The ratio $23.527/21\simeq 1.120$ also gives no striking indication of overdispersion, though deviance per degree of freedom is only a rough dispersion check. Fitted count means around fifty make the usual chi-squared approximation plausible. Independence between months, the conditional variance-equals-mean assumption and the absence of residual structure still need diagnostics. Failure to reject is not proof that the model is correct.

The null deviance $37.969$ on $23$ degrees of freedom has $p=0.02566753$, suggesting the constant-mean model is inadequate. In the sequential analysis of deviance for nested generalized linear models, adding temperature to that model reduces deviance by $14.204$ on one degree of freedom, giving $p=0.000164$. Adding precipitation after temperature reduces it by only $0.238$, with $p=0.625677$. Therefore retain temperature; these data give no reason to retain precipitation after temperature. The temperature-only residual deviance is $23.765$ on $22$ degrees of freedom. This prediction-oriented reduced model should be refitted before reporting its coefficients: the printed temperature coefficient belongs to the full model. The Wald test results, $p=0.000833$ and $0.625582$, broadly support the same conclusion, while the sequential deviance tests answer the stated nested-model questions.

The command cox.zph checks the proportional hazards assumption test through time dependence of scaled Schoenfeld residuals. Under constant coefficients, their expected trend against transformed event time should be flat. A significant covariate-specific test would suggest that its coefficient varies with time; the global test checks the coefficients jointly.

The p-values are $0.461$ for age, $0.538$ for gender and $0.882$ for group, with global $p=0.846$. Thus these tests find no evidence against proportional hazards. They do not prove the assumption, establish the correct age functional form, or test the censoring mechanism.

The separate diagnostic plot targets the broader fitted survival distribution. The code evaluates a fitted cumulative hazard function at each observed time and multiplies by the subject's fitted hazard multiplier, forming Cox–Snell residuals:The curve returned without newdata by survfit is a reference-profile curve, not necessarily the all-zero-covariate baseline. The multiplier must use the same centering as that reference curve; the fitted linear predictors use the model's reference convention. With this consistency, $-\log {\hat{S}}_{\mathrm{ref}}$ times the multiplier is the fitted subject-specific cumulative hazard.

Part (a) shows why complete transformed event times should be approximately unit exponential if the Cox proportional-hazards model is correct. The Q–Q plot compares ordered fitted residuals with an independent random exponential sample. Most points are near the diagonal, with some upper-tail departures, so it gives no obvious indication of a gross distributional failure. A random reference sample adds simulation noise, and estimated parameters and tied recorded times prevent this from being an exact distribution-free test.

Importantly, four observed times are censored. Their transformed times retain the censoring indicators and are not complete exponential observations. Simply plotting all $100$ residuals as uncensored therefore gives only a rough visual check. A more appropriate Cox–Snell residual survival diagnostic fits a Kaplan–Meier estimator or Nelson–Aalen estimator to $(r_i,D_i)$, retains the censoring flags, and checks whether the estimated cumulative hazard is close to $r$, equivalently whether estimated survival is close to $e^{-r}$. Taken together, the supplied checks are broadly compatible with the fitted model, while none resolves potentially informative censoring from giving up.

The statistician first models the age- and gender-dependent mean by linear regression. Its fitted value is $27.01675-0.10483\mathrm{age}+1.47031\mathrm{gender}$. At fixed gender, a year of age is associated with a decrease of $0.10483$ BMI units; the group coded gender $1$ has a fitted mean $1.47031$ units above the group coded $0$ at the same age. This question does not state which gender receives which code. The intercept refers to age zero in the reference group, outside a typical adult-patient range, so its substantive interpretation is limited. Both slope Student t-tests have small p-values, and the overall F-test supports an age/gender-dependent mean. The coefficient of determination is only $0.18$, leaving considerable individual variation to investigate.

The residual diagnostics check whether mean adjustment is plausible and whether a single normal error law is adequate. The residual-versus-fitted plot has no strong curved mean trend, although its smooth curve is not perfectly flat. The scale-location plot suggests some decline in spread as fitted BMI increases, so common residual variance is a working approximation rather than an established fact. The normal Q-Q plot has systematic departures, notably shorter tails than the normal reference, suggesting the residual law is not exactly normal. The regression leverage plot does not display an obviously extreme leverage point, but the marked observations and Cook's distance should still be checked for influence. Independence between patients cannot be diagnosed from these four plots alone.

Next the statistician extracts the regression residuals and fits an intercept-only normal model as a baseline residual distribution. With an intercept in the original ordinary least squares model, the residuals sum to zero by the normal equation; the near-zero fitted residual mean and its $p=1$ are therefore automatic, not new evidence of good fit. The $2.698$ standard error in this refit uses $499$ degrees of freedom, whereas the original $2.704$ uses $497$ after estimating three regression coefficients. The reported log-likelihood for the single-normal residual model is $-1205.25$.

The histogram suggests a shape worth exploring beyond one normal density, with broad shoulders and some asymmetry, without showing unambiguous separated clusters. The statistician fits two- and three-component finite Gaussian mixtures with a common variance by the expectation-maximization algorithm, starting from separated means and positive weights. Successive likelihoods increase and the displayed final iterations stabilize, as expected from EM likelihood monotonicity. This is evidence of numerical convergence from those starts, not proof of a global maximum.

The two-component fit assigns weights $(0.3317,0.6683)$, residual means $(-2.7844,1.3819)$ and common variance $3.4175$. The three-component fit assigns weights $(0.2051,0.3926,0.4023)$, residual means $(-3.7306,-0.4417,2.3329)$ and common variance $2.1448$. The code correctly passes the square roots of those variances to dnorm and overlays the resulting weighted densities. Both curves broadly follow the histogram and are very similar. In particular, a two-component mixture need not have two distinct visible modes.

These fits explore residual mixture clustering: groups differ in BMI relative to the same age/gender-adjusted mean, rather than simply in raw BMI. Membership can be summarized by the fitted mixture responsibilities, retaining uncertainty instead of asserting a certain label for each patient. The common-variance assumption is economical, but should be checked; the common regression slope assumption and constancy of mixture proportions across age and gender are also substantive. A residual mixture alone does not establish genuine biological subpopulations; a flexible continuous distribution, missing predictors, nonlinear mean effects or heteroscedasticity could explain a similar marginal shape.

There is also a two-stage residual mixture fitting limitation. Fitted regression residuals are not exactly independent identically distributed observations: even under a normal regression, their covariance matrix is ${\sigma }^2(I-H)$, and estimating the mean adjustment introduces uncertainty. With $500$ subjects and three initial regression coefficients, treating them as independent is a useful approximation for exploring shape, but ordinary mixture likelihoods omit the first-stage uncertainty. A joint mixture regression with shared slopes, or a suitable bootstrap of the whole analysis, would give a firmer basis for parameter uncertainty and cluster selection. Examine component membership versus age/gender and repeat the algorithm from multiple starts before making substantive clustering claims.

Under the null hypothesis the three outcome probabilities are the same in the two independent samples. The pooled row totals are $50,20,30$, and both column totals are $50$, out of a total of $100$. Thus the expected counts in each column are $25,10,15$. These exceed the usual small-count threshold for the asymptotic Pearson chi-squared test of homogeneity.

The Pearson chi-squared statistic for contingency tables isUnder the null, its asymptotic chi-squared distribution has $(3-1)(2-1)=2$ degrees of freedom. Equivalently, the two independent three-category samples have four probability parameters unrestricted and two under the common-probability null. The $1\%$ upper-tail critical value is $9.21$.

Since $9.12<9.21$, do not reject the equal-effect hypothesis at the 1% level. For two degrees of freedom, the approximate p-value is $e^{-9.12/2}\simeq 0.01046$, just above $0.01$. This is a decision at the specified significance level, not evidence that the two outcome distributions have been proved identical.