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For a normal sample of variance ${\sigma }^2>0$, define $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$. The exact chi-squared distribution identity $(n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2$ and the central limit theorem for independent squared standard normal random variables implyThe delta method for $g(v)=\sqrt{v}$, with derivative $g^{\prime }({\sigma }^2)=1/(2\sigma )$, therefore gives the displayed limit for $S$. Exact independence of the normal sample mean and sample variance also makes $S$ independent of $\overline{X}$. The resulting joint limit supports combining location and scale estimates when a model relates their parameters.

If ${\mathbb{E}}_{0}W=d(1+b/n+O(n^{-2}))$, the Bartlett-corrected statistic has mean $d+O(n^{-2})$. It is commonly calibrated against a chi-squared distribution with $d$ degrees of freedom. Equivalently, to the displayed order, one may multiply $W$ by $1-b/n$.

Independent gamma distributions $S\sim \Gamma (\alpha ,r)$ and $T\sim \Gamma (\beta ,r)$ with a common rate yield a Beta distribution $S/(S+T)\sim \mathrm{Beta}(\alpha ,\beta )$. The change of variables $S=wu$, $T=w(1-u)$ has Jacobian determinant $w$, and factors the joint probability density function into a Beta distribution density in $u$ and a gamma distribution density in $w$. In particular, independent chi-squared distributions with $2\alpha$ and $2\beta$ degrees of freedom give the required ratio.

For $N$ independent groups each containing $m$ independent Bernoulli trials with common probability $\theta$, the group count has a binomial distribution. If $O_j$ groups have count $j$, the maximum-likelihood estimate is $\hat{\theta }={\sum }_{j}j{O}_j/(mN)$. Form the Pearson chi-squared statistic using $E_j=N\left(\genfrac{}{}{0ex}{}{m}{j}\right){\hat{\theta }}^j(1-\hat{\theta }{)}^{m-j}$. With all $m+1$ categories retained, an interior true parameter, and sufficiently large expected counts, its null limit is a chi-squared distribution with $m-1$ statistical degrees of freedom: one degree is removed by the fixed total and another by fitting $\theta$. Small expected counts require care with this approximation. If categories are pooled, the fitted parameter and calibration should correspond to the pooled statistical model; simply pooling a statistic after fitting the unpooled model does not automatically give the usual chi-squared limit.

An integer pattern must survive the stated measurement uncertainty. For an original simulated example, choose a reference electric charge $Q_{\ast }>0$, define $y_i=|q_i|/Q_{\ast }$, and stipulate state meansHere each ${\sigma }_i$ is an independent standard uncertainty of a state mean, before shared calibration effects. The generator constructs repeated noisy observations with these means; the numbers are deliberately chosen, not historical data. There is no conversion of $Q_{\ast }$ to coulombs and no measured value of the elementary charge in this example.

For a candidate spacing $a>0$ in units of $Q_{\ast }$ and proposed integers $n_i$, use weighted least squares:The last two expressions follow by differentiating the quadratic in $a$ and propagating the independent errors. They are conditional on fixed, correct integer assignments and the stated error model. They do not include uncertainty from choosing those assignments. For $\mathbf{n}=(2,3,5,4)$, they give $\hat{a}=1$, $u(\hat{a})\simeq 0.01089$, and ${\chi }^2=0.46875$. Such a small residual is compatible with this chosen error scale; it is not evidence that the errors vanish. A chi-squared distribution interpretation additionally needs the appropriate normal distribution error model and allowance for the fitted parameter; searching many assignments changes that interpretation.

The obstruction is identifiability: replacing $(a,n_i)$ by $(a/m,m{n}_i)$ for any positive integer $m$ leaves every predicted reading unchanged. In particular, an $e$-spaced pattern also fits an $e/2$ lattice using even indices. Even the coarsest well-fitting spacing need not be the physical unit: a dataset containing only even charge states can have twice that unit. Worse, if arbitrarily fine spacings and unrestricted integers are allowed, rounding each $y_i/a$ givesConsequently a finite fit cannot prove a minimal transferable electric charge. Reproducible smallest observed jumps supply additional evidence; interpreting them as single-carrier transfers requires independent physical support. One visibly abrupt transition could contain several unresolved transfers. In the simulation the successive differences are $0.93$, $2.06$, and $-1.03$, each with standard uncertainty $\sqrt{2}(0.08)\simeq 0.1131$. These support changes near $a,2a,-a$ conditional on that unit; they equally support $2,4,-2$ units of $a/2$. Neighboring differences share a state reading and are therefore correlated, even though the state means were stipulated independent.

A shared scale error can preserve every apparent integer relation. To first order, an independent fractional scale standard uncertainty $s$ adds $s^2{a}^2{n}_i{n}_j$ to the covariance matrix of the $y_i$. For the simulation, stipulating $s=0.02$ changes the conditional uncertainty of $\hat{a}$ to approximately $\sqrt{0.01089^2+0.02^2}=0.02277$, despite the same residual pattern. Repeating timings cannot average away this common scale contribution. In the uncorrected continuum model in elementary charge, holding the other measured inputs fixed gives $|q|\propto {\eta }^{3/2}$, so a small fractional error in air dynamic viscosity causes $3/2$ times that fractional error in inferred electric charge. The gas slip correction to Stokes drag requires its own propagation. Pattern agreement, drop-stability controls and absolute measurement calibration together make the inference persuasive within resolution.

Many quadratic and likelihood-ratio statistics converge under a regular null hypothesis to a chi-squared distribution. Using its quantiles for a finite sample is accurate when the sample is large relative to the model dimension and expected category counts are not too small.

Suppose $N_0$ is independently calibrated and counts $y_i$ have predicted means ${\mu }_i=t_i{N}_0{e}^{-b{n}_i}$ under a specified candidate $b$. For independent sufficiently large Poisson counts, the Pearson chi-squared statistic is $Q={\sum }_i(y_i-{\mu }_i{)}^2/{\mu }_i$. With no parameters fitted to these test counts, its approximate reference chi-squared distribution has one degree of freedom per observation. Fitting a continuous $b$ to the same data instead removes one residual degree of freedom. Replacing the denominators by observed counts is a common estimated-error approximation but can behave badly at small counts.

Calibration uncertainty ${\sigma }_0$ also induces correlated prediction errors. First-order propagation of uncertainty givesA generalized quadratic form uses this covariance matrix. For sparse counts use the Poisson deviance or the full count likelihood, with simulation if a calibrated goodness-of-fit probability is required. Choosing a range after inspecting residuals also affects formal test calibration; a physically specified signal-dominated range is preferable when applying a pure attenuation model.

With fixed inverse-variance weights and weighted mean $\hat{\mu }$, $Q={\sum }_i{w}_i(y_i-\hat{\mu }{)}^2$. Under a common-effect normal model with $m$ independent estimates and known variances it has a chi-squared distribution with $m-1$ degrees of freedom. Estimated variances make this calibration approximate.

For a continuous predictor $x$ and a categorical factor, an additive Poisson regression with log link has $\log {\mu }_j(x)=a_j+bx$. The log-mean curves are parallel and ${\mu }_j(x)/{\mu }_k(x)=e^{a_j-a_k}$ is constant in $x$. Adding a predictor-by-factor interaction gives $\log {\mu }_j(x)=a_j+b_{j}x$, so this ratio becomes $e^{a_j-a_k+(b_j-b_k)x}$. With $J$ factor levels, the common-slope restriction has $J-1$ independent constraints. A likelihood-ratio test compares the two deviances with an asymptotic chi-squared distribution on $J-1$ degrees of freedom. Evidence against common slopes establishes varying effects on the log scale; it does not by itself determine their signs or the ordering of the mean curves.

If two independent approximately Gaussian mean lifetime estimates have known standard deviations $s_1,s_2$, equality of their true lifetimes implies that their difference has mean zero and variance $s_1^2+s_2^2$. Its squared standardized difference therefore has a one-degree-of-freedom chi-squared distribution, giving the statistic in the title. A chosen significance threshold converts its upper-tail P-value into a compatibility decision.

Testing a fitted decay curve against a fixed numerical lifetime is a different question: that test does not include the external measurement uncertainty. Comparing the two experiments requires both uncertainties or a joint likelihood with a common lifetime. Correlated estimates replace the denominator by $s_1^2+s_2^2-2\mathrm{Cov}({\hat{\tau }}_1,{\hat{\tau }}_2)$.

An ellipsoidal confidence region defined by a positive-definite matrix $A$ and a calibrated threshold $c$. For a known-covariance multivariate normal distribution sample, $A=n{\Sigma }^{-1}$ and the $1-\alpha$ quantile of the chi-squared distribution with $d$ degrees of freedom give exact coverage for the $d$-dimensional mean.

For a suitably regular generalized linear model, a residual deviance $D$ can be compared approximately with a chi-squared distribution on $n-p$ degrees of freedom to assess fit relative to the saturated model. Sparse or very small expected cell counts can invalidate this reference distribution. A large p-value means no detected lack of fit, not proof of correctness; dependence, wrong mean structure or overdispersion still require diagnostics.

In a normal linear model with residual degrees of freedom $\nu \ge 2$, an internally studentized residual satisfies $|{\eta }_i|\le \sqrt{\nu }$ and ${\eta }_i^2/\nu \sim \mathrm{Beta}(1/2,(\nu -1)/2)$, provided its leverage is less than one. Choose an orthonormal basis of the residual space with its first direction proportional to the projection of the $i$th coordinate vector. The Gaussian coordinates in this basis are independent standard normal random variables $Z_1,\dots ,Z_{\nu }$, and ${\eta }_i=\sqrt{\nu }{Z}_1/({\sum }_j{Z}_j^2{)}^{1/2}$. The beta distribution follows from the independent chi-squared distributions of $Z_1^2$ and ${\sum }_{j=2}^{\nu }{Z}_j^2$. The bounded residual is not exactly distributed according to Student's t-distribution; it approaches the standard normal distribution as $\nu \to \infty$.

Write $a=\nu /2$. Against an exponential distribution proposal of mean $\nu$, the chi-squared distribution density has envelope constant $M=a^a{e}^{1-a}/\Gamma (a)$. The density ratio divided by $M$ is $\exp ((a-1)(\log (y/\nu )-y/\nu +1))$. Thus generate $Y=-\nu \log U$ and accept when $\log V\le (a-1)(\log (Y/\nu )-Y/\nu +1)$ for independent uniform distributions. The inequality $\log t\le t-1$ validates the acceptance probability. The accepted probability density function is the target because rejection sampling weights the proposal by its target-to-envelope ratio.

For independent $N(0,{\sigma }^2)$ observations, testing a fixed variance against larger variances uses the sum of squares $S={\sum }_i{X}_i^2$. The likelihood ratio for any simple larger variance is strictly increasing in $S$. The Neyman-Pearson lemma therefore gives the displayed rejection rule, where $F_n$ is the chi-squared distribution function with $n$ degrees of freedom. Its threshold is independent of the selected larger variance, so it is a uniformly most powerful test for the one-sided composite alternative. Its power function is $1-F_n(({\sigma }_0^2/{\sigma }^2)F_n^{-1}(1-\alpha ))$, increasing continuously from zero to one and equaling $\alpha$ at the null. A known nonzero mean is handled by replacing each observation by its difference from that mean.

For covariance eigenpairs $({\lambda }_k,{\varphi }_k)$, the $K$-coordinate FPCA mean statistic is $n{\sum }_{k=1}^K⟨\overline{X},{\varphi }_k{⟩}^2/{\lambda }_k$. Under a zero-mean null and standard estimation conditions, its plug-in version converges to a chi-squared distribution with $K$ degrees of freedom.

When a Poisson distribution is observed in cells $0,1,\dots ,r-1,\ge r$, the last cell probability is a tail sum, not the probability of exactly $r$. With $p_i(\theta )=e^{-\theta }{\theta }^i/i!$ and $Q_r(\theta )=1-{\sum }_{i<r}{p}_i(\theta )$, telescoping derivatives give $Q_r^{\prime }=p_{r-1}$. Thus an interior maximum-likelihood estimate solves ${\sum }_{i<r}{n}_i(i/\theta -1)+n_r{p}_{r-1}(\theta )/Q_r(\theta )=0$. Substitution of the estimate into these cell probabilities supplies fitted expected counts for a likelihood-ratio test. With $r+1$ cells and one regular fitted parameter, the asymptotic chi-squared distribution has $r-1$ degrees of freedom. Replacing every grouped tail observation by $r$ changes the data model and generally does not give the grouped maximum likelihood estimate.

For $K$ independent strata, each containing an $r\times c$ contingency table, test independence of the two table variables within every stratum. Under adequate expected counts and the usual multinomial distribution sampling assumptions, each stratum's Pearson chi-squared statistic has asymptotic chi-squared distribution with $(r-1)(c-1)$ statistical degrees of freedom. Their sum has asymptotic chi-squared distribution with $K(r-1)(c-1)$ statistical degrees of freedom, by the additivity of independent chi-squared distributions. This is a test of conditional independence, not of equality of a common effect across strata; the latter needs an interaction contrast or another specified homogeneity test.

Let $X_i=\theta +E_i$ for $n\ge 2$ independent and identically distributed observations, with $E_i$ having an exponential distribution of unknown scale $\beta >0$. Test the null hypothesis $\theta =0$ within $\theta \ge 0$. For nondegenerate data, the likelihood function isMaximizing over the scale gives $\hat{\beta }(\theta )=\overline{X}-\theta$, and the resulting profile likelihood increases with $\theta$. Thus the unrestricted maximum-likelihood estimates are $\hat{\theta }=X_{(1)}$ and $\hat{\beta }=\overline{X}-X_{(1)}$, while the null estimate is $\tilde{\beta }=\overline{X}$. The likelihood ratio is consequently

Under the null, put $A=n{X}_{(1)}$ and $D={\sum }_i(X_i-X_{(1)})$. The memoryless property gives independent gamma distributions of shapes $1$ and $n-1$ and common scale $\beta$. For a direct verification, fix which observation is the minimum and write its value as $z$ and the other observations as $z+e_j$. Their joint density is proportional to $e^{-nz/\beta }{\prod }_j{e}^{-e_j/\beta }$, which factors into an exponential density for the minimum and independent exponential densities for the excesses. The minimum's index does not affect the resulting laws of $A,D$.

By beta sampling by gamma ratios, $D/(A+D)\sim \mathrm{Beta}(n-1,1)$. ThereforeAn exact test size $\alpha$ is obtained by rejecting for $R\le {\alpha }^{n/(n-1)}$. Equivalently,Although the alternative adds one endpoint parameter, a regular one-parameter Wilks theorem calibration is incorrect: the support depends on the endpoint, and the null lies on its boundary. In particular, $-2\log R$ tends to a chi-squared distribution with two statistical degrees of freedom, not one.

For nested null and alternative parameter spaces, the likelihood-ratio test statistic isUnder the regularity conditions of Wilks theorem, its null limit is a chi-squared distribution whose degrees of freedom equal the difference in model dimensions.