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The constant subspace has one statistical degree of freedom. Day contrasts have $5-1=4$, and the within-block ANOVA stratum has $50-5=45$. Because the orthogonal block design puts all four independent dose contrasts in that last ANOVA stratum, the residual has $45-4=41$ statistical degrees of freedom.

The corrected total has 49 statistical degrees of freedom; dose is tested against the within-day residual. The quantitative dose scale also permits the dose component to be split into linear, quadratic, cubic and quartic orthogonal polynomial contrasts, each with one statistical degree of freedom. This is optional and does not assume the response is linear in dose.

Start with six program sequences: $ABCABC$, $ACBACB$, $BACBAC$, $BCABCA$, $CABCAB$ and $CBACBA$. This gives two appearances of each program per row and per column. It also balances transitions: each ordered pair of distinct programs occurs five times across all adjacent sessions.

Apply restricted randomization by randomly assigning the six sequences to the six volunteers and independently permuting the three program labels. Using the supplied numbers, rank the first six from smallest to largest: the sequence indices are $(4,1,5,2,3,6)$. Assign these, in order, to volunteers 1 through 6. Rank the next three numbers attached to symbols $(A,B,C)$: their order is $(B,C,A)$. Assign actual program labels $(A,B,C)$ to these ordered symbols, so base $A\mapsto C$, base $B\mapsto A$, base $C\mapsto B$. The final ready-to-use randomized row-column design is:

Programs retain their real identities after this label permutation. The researcher should follow the table across chronological afternoons; arbitrary column permutations are deliberately excluded so that transition balance survives. Each program receives twelve sessions and occurs twice in every block. Under the additive block model, the analysis of variance has row, column, program and residual statistical degrees of freedom $5,5,2,23$, respectively, besides the grand mean. Transition balance is useful against differential carryover effects, but is not a proof that they are absent.

A completely randomized design uses $2m$ eligible animals, randomly selecting $m$ to receive $P$ and assigning the remainder to $Q$. Use comparable follow-up and assess the same response variable in both groups. Its advantage is simplicity and freedom from previous-treatment carryover effects; its disadvantage is that between-animal variation enters the residual and can make a treatment contrast imprecise.

A randomized complete block design with matched pairs first forms $m$ pairs using pre-treatment characteristics such as initial disease severity, age or breed. Independently choose which animal in each pair receives $P$, with its partner receiving $Q$. The experimental units are animals; pairs are blocks. The average within-pair difference estimates the treatment contrast, and positive within-pair similarity can reduce its variance. Its advantage is control of known heterogeneity; its disadvantage is the need for useful matching, with fewer residual statistical degrees of freedom and little gain if the matching variables are uninformative. Do not construct pairs using outcomes observed after assignment.

A two-period crossover design randomly assigns half the animals to sequence $PQ$ and half to $QP$. Each animal receives both treatments in separate periods, with a scientifically justified interval between them and the same outcome assessment after each period. Animal blocks remove persistent between-animal differences, while the two sequences balance treatment against period. Its advantage is potentially high precision from within-animal comparisons; its disadvantage is vulnerability to carryover effects, changing disease state and irreversible effects. It is suitable only when comparing the treatments in both periods remains meaningful and residual effects of the first treatment are adequately controlled. A two-period crossover design does not by itself disentangle arbitrary treatment-specific carryover effects.

Both factors are assigned at orchard level. Therefore the experimental units are the twelve orchards; the trees are observational units within them. The six combinations form a balanced factorial design, replicated twice. The orchard ANOVA stratum has $12-1=11$ statistical degrees of freedom. Spray uses $2-1=1$, pruning uses $3-1=2$, and their interaction term uses $(2-1)(3-1)=2$, leaving six for error.

Dividing each treatment sum of squares in ANOVA by its statistical degrees of freedom and using $1440/6=240$ as the denominator gives all missing entries:

The unrounded F-test statistics are $998/240=4.15833\dots$, $560/240=2.33333\dots$ and $202/240=0.841667\dots$. The within-orchard tree mean square in ANOVA, 180, is not the treatment error denominator: using it would confuse subsampling with independent replication. The tree ANOVA stratum has $12(30-1)=348$ statistical degrees of freedom; $1+11+348=360$ is the uncorrected total, and the corrected total is 359.

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.

This produces a split-plot design: assign spray to whole orchards, with six sprayed and six unsprayed, then independently randomize ten trees to each pruning method inside every orchard. Orchards remain experimental units for spray; individual trees become experimental units for pruning. Pruning contrasts and the spray-by-pruning interaction term now lie in the within-block ANOVA stratum, while spray is tested between orchards.

The between-orchard ANOVA stratum has eleven statistical degrees of freedom, split into one for spray and ten for error. The within-orchard ANOVA stratum has 348, split into two for pruning, two for the interaction term and 344 for error. This pooling of within-orchard error is appropriate under the stated compound-symmetry covariance model; additional orchard-specific pruning variation would need its own variance component rather than this simplified error model.

The shared orchard effect cancels in a pruning difference within an orchard. Its variance is $2a/10$, so averaging across twelve orchards givesThe spray contrast still compares means of six orchards per group, each based on 30 trees, so its estimated variance remains $8/3$. For a difference of pruning differences between the two spray groups, each group's pruning difference has estimated variance $2\cdot 180/(10\cdot 6)=6$, and the resulting interaction contrast has estimated variance $12$, compared with $16$ in the original allocation.

Splitting pruning within orchards improves pruning and interaction precision without extra trees, and gives spray a less sparse error estimate; it does not reduce the spray contrast's variance. This option requires tree-level pruning to be practical without interference between neighboring trees. The numerical gains, like those in the other options, assume the current variance components remain applicable next year.

Allocate the extra orchards evenly, giving three independent orchards to each of the six combinations. Each orchard still has mean variance $8$ under the fitted model. Each pruning marginal now uses six orchards and each spray marginal nine, givingBoth are two-thirds of their original values. The orchard residual statistical degrees of freedom increase from $12-6=6$ to $18-6=12$.

Extra orchards improve independent replication, give a predicted one-third reduction in contrast variances, and improve error estimation. Both this option and using 45 trees require 540 tree measurements, but extra orchards offer the greater predicted statistical gain under the fitted model. They may cost more to recruit and administer; their suitability and comparability also matter.

Under the fitted compound-symmetry covariance model, an orchard mean based on $k$ trees has varianceWith the current estimates this falls from $v_{30}=2+180/30=8$ to $v_{45}=2+180/45=6$. The numbers of orchards contributing to each marginal remain unchanged, so both marginal treatment-contrast variances fall by a factor $6/8=3/4$: pruning differences have estimated variance $3$ and the spray difference has estimated variance $2$.

Adding trees gives a predicted 25% reduction in these variances, but no extra independent orchard replication. It requires 50% more tree measurements and leaves six orchard residual statistical degrees of freedom. Increasing $k$ reduces the $a/k$ term but cannot remove the shared orchard component $c$, so gains eventually diminish.

A chi-squared test compares discrepancies with their expected random size. For independent Gaussian observations with known standard deviations, a fitted model givesUnder a correct regular model with $p$ fitted parameters, its reference statistical degrees of freedom are usually $\nu =n-p$. Correlated errors instead require ${\chi }^2={\mathbf{r}}^T{C}^{-1}\mathbf{r}$ with the correct covariance matrix. The reduced statistic has mean about one and standard deviation $\sqrt{2/\nu }$, so it should not be forced to equal exactly one.

For binned counts, Pearson's statistic is ${\sum }_i(O_i-E_i{)}^2/E_i$. With a fixed total and $k$ populated categories, fitting $p$ distribution parameters generally gives $\nu =k-1-p$. The approximation needs adequately populated expected bins and the appropriate dependence assumptions. Sparse data may require a Poisson likelihood, exact calculation or simulated reference distribution rather than a casual chi-squared approximation.

The upper-tail probability $p_{\mathrm{value}}=P({\chi }_{\nu }^2\ge {\chi }_{\mathrm{obs}}^2)$ measures how unusual the discrepancy would be under the null model; it is not the probability that the model is true. A large statistic can indicate a wrong model, underestimated noise or unmodelled systematic effects; an unusually small one can indicate overestimated errors or correlations. Residual plots and checks of the noise model are needed alongside the numerical test.

Use the sample mean to estimate each magnetic susceptibility, and the unbiased sample variance with denominator $n-1=4$ to estimate the repeatability of individual readings. In the numerical scale supplied by the paper,If the repeated measurements have independent random errors, their mean estimates have standard errorsThe difference and its estimated standard uncertainty are thereforeThe separation is about $2.53$ standard errors. Under an approximately normal distribution model with no experimenter-dependent bias, Welch's t-test gives about $7.34$ statistical degrees of freedom and a two-sided p-value of $0.038$. Thus there is modest evidence against equal underlying susceptibilities at a conventional $5\%$ significance level; overlap of the ranges of individual readings is not an adequate comparison of the means. Five readings per sample provide limited information about the error distribution.

Crucially, material and student exhibit complete confounding: each student measured only one sample. An experimenter-dependent systematic error could mimic the difference. Paramagnetism also depends on conditions such as temperature, so equal material alone does not guarantee equal readings without controlling those conditions. The defensible physical conclusion is the data suggest a difference, but do not establish different materials. Both students should measure both samples in randomized or alternating order, with calibration and temperature controlled, to separate a sample effect from an operator effect. No unit or scale factor absent from the original table has been invented.

A Gaussian distribution with mean $\mu$ and standard deviation $\sigma$ has probability densityIts symmetry, finite variance and simple behavior under linear combinations make it useful for repeated measurements. Roughly $68.3\%$, $95.4\%$ and $99.7\%$ of observations lie within one, two and three true standard deviations of the mean. These are coverage probabilities for a specified distribution, not guarantees that every experimental error is normal.

Why it appears. The central limit theorem explains why many sums of small independent finite-variance contributions approach a Gaussian law after centering and scaling, provided no single contribution dominates. Detector noise, small mechanical disturbances and multiple electronic fluctuations can combine approximately this way. The theorem does not remove a common calibration bias, and it does not justify assuming normal tails for rare large disturbances or arbitrarily correlated errors. A large-mean Poisson distribution also approaches a Gaussian with variance equal to its mean, making normal approximations useful for high-count photon or particle data; low counts require the discrete model.

A classical particle at thermal equilibrium in a one-dimensional harmonic potential has a Gaussian position distribution directly from the Boltzmann distribution:Thus a Gaussian law can arise from the energy dependence of equilibrium probability, rather than only from an informal appeal to many independent errors.

Precision of repeated measurements. For $N$ independent observations with the same mean and variance, the sample mean has variance ${\sigma }^2/N$. For Gaussian observations this mean is exactly Gaussian:Consequently quadrupling the number of independent readings halves the random standard error. When the true variance is unknown, a small-sample confidence interval for a normal mean uses a Student t-distribution with $N-1$ statistical degrees of freedom; replacing that by a normal quantile can overstate precision. Shared systematic errors and correlations do not decrease by the same averaging rule.

Combining results and fitting models. Independent Gaussian estimates $x_i$ of one quantity, with known variances ${\sigma }_i^2$, combine through the inverse-variance weighted meanFor a model $y_i=g_i(\theta )+{ϵ}_i$ with independent zero-mean Gaussian noise, maximizing the likelihood is equivalent to minimizing the weighted least-squares statisticFor a correct linear model with known independent Gaussian errors and $p$ fitted parameters of full rank, the minimized statistic has a chi-squared distribution with $N-p$ degrees of freedom. Nonlinear fits only inherit that familiar approximation under additional regularity and adequate-data assumptions. Known correlated Gaussian errors require the full inverse covariance matrix in the quadratic form.

These properties support confidence intervals, consistency checks, calibration and error propagation. They also give practical diagnostics: inspect residuals for trends, changing spread, correlations or outliers before trusting a Gaussian fit. Gaussian models connect physical fluctuations to quantitative uncertainty, but their assumptions must be checked rather than inferred from the shape of a convenient bell curve.

A chi-squared test compares residuals with the random scatter predicted by a model. For independent measurements $y_i$ with known Gaussian standard deviations ${\sigma }_i$ and predictions ${\mu }_i(\theta )$, formIf the mean model and error assumptions are correct, a full-rank linear fit of $p$ parameters gives a chi-squared distribution with $\nu =N-p$ statistical degrees of freedom. With a fully specified prediction and no fitted parameters, $\nu =N$. Nonlinear fits usually make this an approximation. Independent external calibration is not the same as fitting a parameter to these test data.

Under the reference distribution, the mean statistic is $\nu$ and its variance is $2\nu$. The upper-tail p-value $\mathrm{Pr}({\chi }_{\nu }^2\ge Q_{\mathrm{obs}})$ quantifies how unusual a discrepancy at least this large would be under the assumed model. A small value suggests model failure or incorrect uncertainties; it is not the probability that the model is true. An unusually small statistic can also reveal inflated uncertainties or correlations. A rejection threshold should be specified in advance.

For independent Poisson distribution counts, the model variance is its expected count ${\mu }_i$. The Pearson chi-squared statistic is $\sum (y_i-{\mu }_i{)}^2/{\mu }_i$. The estimated-error alternative ${\sigma }_i\simeq \sqrt{y_i}$ is common for large counts but can differ substantially when counts are small. In that regime the Poisson deviance or a direct likelihood calculation is preferable, with simulation if an accurately calibrated goodness-of-fit probability is required.

A chi-squared test compares squared deviations with the size expected from the assumed random errors. For independent observations with a normal distribution $y_i$ with known standard deviations ${\sigma }_i$ and predictions ${\mu }_i(\theta )$, useFor a correct fixed model, the standardized independent residuals are unit-variance normal distributions, so $Q$ has a chi-squared distribution with $\nu =N$. Fitting $p$ independent linear parameters removes $p$ residual directions, giving $\nu =N-p$. This distribution is exact for a full-rank linear model with known errors with a normal distribution; it is an approximation for many nonlinear fits. With known correlated errors with a multivariate normal distribution, replace the sum by $Q={\mathbf{r}}^{\mathsf{T}}{\Sigma }^{-1}\mathbf{r}$, where $\Sigma$ is the covariance matrix and its rank and fitted constraints determine the available statistical degrees of freedom.

Under the null hypothesis, $\mathbb{E}Q=\nu$ and $\mathrm{Var}Q=2\nu$. The reduced statistic $Q/\nu$ should therefore be near one, with standard deviation $\sqrt{2/\nu }$, rather than equal to one exactly. An upper-tail p-value isCompare it with a significance threshold chosen in advance. A small value indicates tension with the model together with its error assumptions; it can arise from a wrong mean model, underestimated uncertainties or neglected correlations. An unusually small statistic can suggest overestimated uncertainties, correlations or overfitting. A p-value is not the probability that the model is true, and failure to reject is not proof of the model.

For counts in categories, the Pearson form is $Q={\sum }_{j=1}^k(O_j-E_j{)}^2/E_j$. Its chi-squared distribution is an asymptotic approximation requiring adequate expected counts. For a fixed-total multinomial sample there is one normalization constraint, so the usual statistical degrees of freedom are $k-1-p$ after fitting $p$ other parameters. Independent Poisson bins instead start with $k$ statistical degrees of freedom before parameters are fitted. Sparse counts require an appropriate count likelihood or a calibrated exact or simulation-based test; plugging tiny expected counts into the asymptotic formula is unreliable. The error model, binning, fitted parameters and chosen test must therefore be reported with the result.

Assume independent normally distributed angle errors with known standard deviation ${\sigma }_y=0.002$. The relevant chi-squared test uses the sum of squared standardized residuals,For the model through the origin, the predictions and residuals, expressed in milliradians, areConsequently ${\chi }_0^2=30.0280$. There are $4-1=3$ statistical degrees of freedom. The upper-tail P-value is approximately $1.36\times 10^{-6}$; equivalently the value greatly exceeds the $5\%$ critical value $7.815$. The zero-intercept model is strongly rejected under the stated errors.

For the alternative linear regression, minimize ${\sum }_i(y_i-c^{\prime }{x}_i-d{)}^2$. Setting both partial derivatives to zero gives the normal equations, whose solution isFor these data $\bar{x}=2508.83699{\mathrm{m}}^{-2}$, $\bar{y}=0.0475$, and $S_{xx}=4.830232\times 10^6{\mathrm{m}}^{-4}$. This yieldsThe new predictions are $(66.6548,55.4171,41.4315,26.4966)\mathrm{mrad}$, with residuals $(-0.6548,-0.4171,2.5685,-1.4966)\mathrm{mrad}$. ThereforeThe affine model is compatible with the data. Allowing for its additional fitted parameter, the decrease $\Delta {\chi }^2=27.668$ has a chi-squared distribution with one statistical degree of freedom under the nested zero-intercept hypothesis, corresponding to $p\simeq 1.44\times 10^{-7}$. Thus the model with a fitted intercept provides the better description, rather than merely lowering the residual by adding a parameter.

With the measurement variance treated as known, the coefficient uncertainties areHence $d=(13.11\pm 2.49)\mathrm{mrad}$, about $5.26$ standard deviations above zero. A positive constant systematic error in the angular zero, or an additional nearly constant torque, could account for it. Its presence is evidence against the ideal zero-intercept model; four points alone do not identify the physical cause uniquely. A calibration error in $\kappa$ only rescales the inferred gravitational constant and cannot remove this residual pattern.

Using the fitted slope after allowing for this offset would give $G^{\prime }=\kappa {\hat{c}}^{\prime }/(mML)=(7.27\pm 0.51)\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, with the period and slope errors combined in quadrature. This is conditional on the constant-offset model.

Write $S^2=Q/(n-1)$ and $S=\sqrt{S^2}$, with $n\ge 2$. Two exact facts about a normal sample areand independence of the normal sample mean and sample variance holds. A chi-squared distribution with $n-1$ statistical degrees of freedom is a sum of $n-1$ independent ${\chi }_1^2$ variables, each with mean one and variance two. The central limit theorem therefore givesFor $g(x)=\sqrt{x}$, the derivative at ${\mu }^2$ is $1/(2\mu )$ because $\mu >0$. The delta method gives the asymptotic normality of a normal sample standard deviation:The exact independence of $\overline{X}$ and $S$ makes their limiting normal distributions independent. Consequently,ThusThis has the same leading asymptotic variance as $\hat{\mu }$. The weights $1/3$ and $2/3$ are those in the optimal asymptotic combination of a normal sample mean and standard deviation. This does not require $S$ to be an exactly unbiased estimator of $\mu$.

Test homogeneity of the four category probabilities among the three samples. The null hypothesis is $p_{iA}=p_{iB}=p_{iC}$ for every category $i$; the alternative allows at least one difference. This is a Pearson chi-squared test of homogeneity, assuming independently sampled grains with consistent classification.

The row totals are $(195,52,64,16)$, the sample totals $(131,135,61)$ and the grand total $N=327$. Under the null hypothesis, $E_{ij}=O_{i+}{O}_{+j}/N$, giving the expected value tableFor example, $E_{\mathrm{undulatory},C}=52(61)/327=9.7003$, so that cell contributes $(4-9.7003{)}^2/9.7003=3.34974$. The complete cell contributions areso $X^2=6.87754$. There are $(4-1)(3-1)=6$ statistical degrees of freedom. The supplied $5\%$ critical value is $12.592$, and $6.87754<12.592$, so do not reject homogeneity at $\alpha =0.05$. The approximate upper-tail p-value is $0.3323$.

One of twelve expected cells is below five, with none below one; the asymptotic approximation is broadly defensible by a common expected-count guideline but is not exact. An exact/conditional simulation could check it, especially if grain counts are clustered. Do not pool categories solely to obtain a desired result. The conclusion is that these counts do not demonstrate different category proportions; they do not prove that the rocks share one geological formation. Different formations can share such proportions, and real grains within one rock can fail the independent-sampling assumption.

TODO: Attachment C is absent from the held release. Recover it to check any supplied critical-value table, significance convention or additional instructions. This leaf remains incomplete against that missing input. The printed summary statistics nevertheless allow the following conditional numerical answer, using a two-sided $5\%$ Student's t-test and independent observations with common variance.

Test $H_0:{\mu }_W={\mu }_S$ against $H_1:{\mu }_W\ne {\mu }_S$. The pooled sample variance isThe test statistic and statistical degrees of freedom areUsing the Student t-distribution directly gives a two-sided p-value $p\simeq 0.00250$ and a $5\%$ critical value $1.96076$. Thus reject equal seasonal means at $5\%$ and $1\%$; the winter mean is higher by $0.36$ concentration units. This does not establish a large physical effect or its cause.

The two variances are similar, and Welch's t-test gives essentially the same result: $t=3.02344$, $\nu \simeq 2958.36$, $p\simeq 0.00252$. Both calculations assume the counts represent independent observations. Serial dependence, repeated measurements and strong non-normality can invalidate nominal p-values; the summary table alone cannot diagnose them. The unequal means are a conclusion under these statistical assumptions, pending recovery of Attachment C.

For a one-factor analysis of variance, let $n_j,{\bar{x}}_j,s_j$ be the four seasonal sizes, means and standard deviations. The weighted grand arithmetic mean isThis agrees with the rounded annual mean $5.07$. The between-season and within-season sums of squares areThere are $4-1=3$ between-season statistical degrees of freedom, because the four fitted seasonal means are compared subject to their weighted grand-mean constraint. There are $N-4=6882$ within-season statistical degrees of freedom, because one mean was estimated in each group. The total has $N-1=6885$, equal to their sum. HenceUnder $H_0:{\mu }_W={\mu }_{Sp}={\mu }_{Su}={\mu }_F$, compare with the F-distribution on $(3,6882)$ statistical degrees of freedom. The $5\%$ critical value is about $2.60620$, with p-value approximately $8.9\times 10^{-63}$ from the rounded summaries. Reject equality of all four means very strongly under the stated model. The omnibus hypothesis test establishes that at least one mean differs; it does not identify every pairwise difference.

The summer variance is substantially lower than the other three, so the exact common-variance ANOVA assumptions deserve checking, as does temporal independence. Raw observations would permit diagnostics and a heteroscedastic or time-dependent analysis. As a cross-check on arithmetic, $S{S}_B+S{S}_W=64499.774183$ implies annual sample standard deviation $3.06074$, matching the table's $3.06$. The numerical significance must not be mistaken for verification of the unobserved Attachment C or of the sampling model.

A regression factor is a categorical variable encoded by indicator columns or treatment contrasts in a statistical model. A factor with $k$ observed levels normally contributes $k-1$ statistical degrees of freedom when an intercept is present. The reference level in a regression factor has coefficient zero under treatment contrasts. Coding a year as a factor permits arbitrary year effects rather than imposing a linear trend.

Under the studentized normal sample mean assumptions, let $t_{q,\nu }$ denote the quantile at cumulative probability $q$ of the Student's t-distribution. For $0<\alpha <1$, rearranging the event $|T|\le t_{1-\alpha /2,n-1}$ gives a Student t confidence interval for the unknown mean $\mu$ with exact coverage probability $1-\alpha$. ^[ref]

Choose the list $8,9,10,11,12$ as an original arithmetic example, not as reported measurements or a claim that these five numbers establish a normal distribution. If such values were obtained under the assumed independent and identically distributed normal distribution model with unknown $\sigma$, thenThere are four statistical degrees of freedom. For a 95% confidence interval, $t_{0.975,4}=2.776445105\dots$, soUsing the standard normal quantile $z_{0.975}=1.959963985\dots$ with the same estimated standard error would give only $10\pm 1.385903824\dots$. That underestimates the correct margin by about 29.4%. Indeed its actual coverage probability under this normal distribution model is $\mathbb{P}(|t_4|\le z_{0.975})\simeq 0.8784$, rather than 0.95.

The observations, the sample mean and the confidence interval all have different meanings. The bar estimates the population mean; it is not an interval intended to contain 95% of individual observations. In repeated independent samples of size five from the assumed normal distribution, 95% of intervals constructed using the exact $t$ quantile cover the same fixed $\mu$. This is not a 95% posterior probability that the fixed statistical parameter lies in this particular realized interval. Such a probability requires a separate Bayesian inference model.

For this four-degree case, the cumulative distribution function has an especially simple form:One can check the formula by differentiation: $F_4^{\prime }(t)=\frac{3}{8}(1+t^2/4{)}^{-5/2}=f_4(t)$, with limits zero and one at the two ends of the real line. Solving the increasing equation $F_4(t)=0.975$ by the bisection method gives the quantile used above.