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Let $I_0$ be the indices of the true null hypotheses, let $m_0=|I_0|$, and let $V$ count false rejections. The familywise error rate isThe Bonferroni correction rejects a null exactly when its p-value is at most $\alpha /m$. The union bound givesThis controls the familywise error rate for every configuration of true and false nulls, with arbitrary dependence between their p-values. Marginal super-uniformity, $\mathbb{P}(P_i\le t)\le t$, would also suffice.

For the Holm step-down procedure, inspect the ordered p-values in ascending order and stop at the first failed comparison. If the first comparison fails, reject none; this is the convention $k=0$ when the defining set is empty. Ties can be ordered by any fixed rule.

If $m_0=0$ there can be no false rejection. Otherwise let $j$ be the rank of the first true null. Since at most $m-m_0$ false nulls precede it, $j\le m-m_0+1$ and thus $m-j+1\ge m_0$. If any true null is rejected, the step-down rule must have rejected this first true null, which requiresUsing the union bound on the true-null p-values provesThis first true null argument for Holm control also requires no independence. Holm controls the familywise error rate at level $\alpha$ under arbitrary dependence.

Now let $R$ count all rejections. The false discovery rate is the expected proportion of rejections that are false, with zero assigned when nothing is rejected:It differs from the familywise error rate: several false rejections can still represent a small proportion of a large collection of discoveries.

The BH procedure is a step-up rule. SetIf $K=0$, reject nothing; otherwise reject all hypotheses with $P_i\le \alpha K/m$. Exactly $K$ hypotheses are rejected: if more than $K$ p-values were below that threshold, the next ordered value would also satisfy its own larger threshold, contradicting maximality. Thus $R=K$. Unlike the step-down rule, a failed early comparison does not make this procedure stop.

For the proof, assume that each true-null p-value is uniform on $[0,1]$ and independent of the entire vector of the other p-values. Joint independence of all p-values is a sufficient condition; the false-null marginal distributions can be arbitrary. Mere uniformity of the true-null marginals without a dependence condition is insufficient for this argument or for general unmodified BH control.

Fix a true null $i$. Replace its p-value by zero and let $R^{(i)}$ be the number of rejections made by the Benjamini-Hochberg procedure on the modified vector. This variable depends only on the other p-values, and $R^{(i)}\ge 1$. The Benjamini-Hochberg leave-one-out identity isTo prove it, suppose $i$ is rejected with $R=r$. Decreasing its p-value to zero leaves every ordered value above rank $r$ unchanged, since it was already among the first $r$ values. Those higher ranks still fail their thresholds, while rank $r$ still succeeds; hence $R^{(i)}=r$. Conversely, suppose $R^{(i)}=r$ and $P_i\le \alpha r/m$. Restoring $P_i$ still leaves at least $r$ values at most $\alpha r/m$, so $R\ge r$. Increasing one p-value cannot increase the maximal successful rank, so $R\le R^{(i)}=r$. This gives equality and rejection of $i$.

Independence and uniformity now giveSumming over the $m_0$ true nulls proves the exact false discovery rate under independent null p-values:If the independent true-null p-values are only super-uniform, the same calculation gives the inequality instead of equality. The distinction between the two types of error control and the dependence conditions is essential: Bonferroni and Holm have the preceding guarantees without independence, while the BH proof here explicitly uses it.

Under the usual normal linear model for independent orchard means with common error variance, the treatment ratios have null F-distributions with denominator six statistical degrees of freedom. Their upper-tail P-values are approximatelyUse the unrounded ratios when evaluating these P-values. Thus none of spray, pruning or their interaction term is significant at a 5% significance level. Spray gives modest evidence if a 10% significance level was chosen in advance, but this is not strong evidence.

Failure to reject does not establish absence of treatment effects. Only two orchards per combination and six residual statistical degrees of freedom leave considerable uncertainty and potentially low statistical power. The table alone cannot give the direction or magnitude of individual effects: that requires treatment means. These model-based F-tests also rely on appropriate orchard allocation and comparable residual variation; they should not be interpreted as unconditional conclusions from the table alone.

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.

For regression diagnostics, inspect regression residuals against fitted means and hardwood percentage for omitted curvature, a scale-location plot for nonconstant variance, and a quantile-quantile plot against the normal distribution for departures from the error assumption. Check unusual observations using regression leverage and Cook's distance, and examine residuals in collection or batch order if dependence is plausible. Independence and a common variance require substantive justification as well as these plots; a small p-value for a polynomial term does not itself check the error model.

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.