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For $D\sim \mathrm{Bin}(n,p)$ with $n>0$, the sample binomial proportion is $D/n$. Its expectation is p and its variance is $p(1-p)/n$. The central limit theorem gives a normal approximation when both expected outcome counts are sufficiently large. These moments determine a standard error and model-based control limits.

For $\theta ={\beta }_1{\beta }_2$, use $\hat{\theta }={\hat{\beta }}_1{\hat{\beta }}_2$. Under regular regression assumptions, the delta method gives standard error $\hat{s}=(d^{\top }\hat{V}d{)}^{1/2}$, where $d=({\hat{\beta }}_2,{\hat{\beta }}_1{)}^{\top }$ and $\hat{V}$ estimates the covariance of the regression estimate. Resample the observational units, refit, and compute both ${\hat{\theta }}^{\ast }$ and ${\hat{s}}^{\ast }$. A percentile bootstrap confidence interval uses the .025 and .975 quantiles of ${\hat{\theta }}^{\ast }$. A bootstrap-t confidence interval uses quantiles of $({\hat{\theta }}^{\ast }-\hat{\theta })/{\hat{s}}^{\ast }$ and reverses them when solving for $\theta$. If both true coefficients vanish, the first derivative is zero and the standard first-order justification is unavailable; product-specific nonregular inference is then needed.

Replace the unknown sampling distribution by the empirical distribution, generate independent bootstrap samples, and recompute the statistic. The sample standard deviation of those recomputed values estimates the conditional bootstrap standard deviation, which approximates the statistic's sampling standard error when the bootstrap is consistent. It is different from the Monte Carlo standard error of the estimated bootstrap mean.

For each bootstrap sample, form $T^{\ast }=({\hat{\theta }}^{\ast }-\hat{\theta })/{\hat{s}}^{\ast }$, using that sample's estimated standard error, possibly from a nested bootstrap. If $q_p$ are conditional quantiles of $T^{\ast }$ and $\hat{s}$ is the original estimated standard error, invert the studentized inequalities to obtain the displayed interval. The quantile order is reversed by subtraction. Its coverage is approximate and requires a nondegenerate statistic and suitable bootstrap regularity.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with expected value $\mu =\mathbb{E}[X_1]$ and finite, strictly positive variance ${\sigma }^2=\mathbb{E}[(X_1-\mu {)}^2]$. Write $S_n={\sum }_{j=1}^n{X}_j$ and ${\overline{X}}_n=S_n/n$ for their sum and sample mean. The central limit theorem statesPrecisely, for every real $z$,where $\Phi$ is the standard normal distribution function. The conclusion is convergence in distribution to the standard normal distribution. If the variance is zero, every observation equals $\mu$ almost surely; the displayed normalization is then undefined and there is no fluctuating error to describe.

The striking feature is that the observations need not have a normal distribution. They may be binary, bounded, or strongly skewed. Their detailed probability distribution disappears from this limit except through the expected value and variance. The subtraction removes predictable growth: $\mathbb{E}[S_n]=n\mu$. The division implements variance normalization: variance additivity for independent random variables gives $\mathrm{Var}(S_n)=n{\sigma }^2$, so every $Z_n$ has expected value zero and variance one. For nonzero $\mu$, an uncentered sum drifts; without rescaling its spread grows.

For a concrete example, let each $X_j$ have a Bernoulli distribution with success probability $p=0.2$. Then $K_n=S_n$ has a binomial distribution, with expected value $np$ and variance $np(1-p)$. Its exact probability mass function and its normalized count areHere $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is a binomial coefficient. Individual trials are conspicuously skewed, yet their centered, rescaled counts approach the same standard normal distribution as sums of fair trials.

At $n=100$, $K_n$ has expected value $20$ and standard deviation $4$. The exact binomial distribution gives $\mathbb{P}(16\le K_n\le 24)=0.740141\dots$. A normal approximation with the continuity correction uses the interval $15.5$ to $24.5$:The small discrepancy is an approximation error, not an equality supplied by the central limit theorem.

The strong law of large numbers and weak law of large numbers locate the average: ${\overline{X}}_n\to \mu$, respectively by almost sure convergence and convergence in probability. The central limit theorem magnifies that shrinking error and describes its shape. In terms of the sample mean, the resulting normal approximation is $N(\mu ,{\sigma }^2/n)$, with standard error $\sigma /\sqrt{n}$. This explains why four times as many independent observations halve the standard error. When $\sigma$ is known, ${\overline{X}}_n\pm 1.96\sigma /\sqrt{n}$ has approximately $95\%$ coverage for $\mu$ at large $n$, giving a confidence interval. This coverage refers to repeated samples, not a guarantee for an individual interval.

A useful application is the Monte Carlo method: estimate a success probability by the proportion of successful simulated trials. Nicholas Metropolis and S. Ulam's original paper, "The Monte Carlo Method" (1949), illustrates this with games of solitaire and estimation of a region's volume by randomly selected points, and discusses why estimates require probabilistic error control. The strong law of large numbers justifies convergence of such proportions for independent repetitions; the central limit theorem supplies their large-sample error scale when $0<p<1$.^[ref]

There are three useful limits to this picture:

A historical original-paper reference is J. W. Lindeberg's "Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung" (1922), published in Mathematische Zeitschrift 15, pp. 211–225. The Lindeberg condition and Lindeberg-Feller central limit theorem below describe a version for sums whose summands need not share a probability distribution.^[ref]

A brief proof outline uses the characteristic function of $Y=(X_1-\mu )/\sigma$. Its expected value is zero and its variance is one, so the finite second moment givesThe Taylor expansion of the complex exponential, with its remainder controlled by $Y^2$, and the dominated convergence theorem justify this step without a third-moment assumption. Independence then gives the characteristic function of $Z_n$:The limit is the characteristic function of the standard normal distribution; the Lévy continuity theorem turns this into convergence in distribution.

Partition a sample into two disjoint groups $A,B$ with fixed positive sizes $n_A,n_B$, and let $w=n_A/(n_A+n_B)$. The overall sample mean is the weighted meanSubtracting gives the displayed identity. The subgroup and overall sample means share observations, so they are not independent. If the two disjoint groups are sampled independently, thenThus treating the subgroup and the combined sample as two independent samples uses the wrong standard error. Equality of the subgroup and overall population means is equivalent to equality of subgroup and complement means when both groups have positive weight, but the sampling distribution must respect the shared observations.

A confidence interval is a random interval constructed from a statistical sample to cover a fixed unknown statistical parameter with a prescribed coverage probability. For a scalar target $\theta$, a procedure with nominal confidence $1-\alpha$ satisfiesThe subscript records which probability distribution generates the repeated samples. The endpoints are statistics, so they vary from sample to sample; the target is fixed. Equality gives exact nominal coverage probability, while the inequality also permits conservative procedures. For a general estimand $\psi (\theta )$, replace the target inside the interval by $\psi (\theta )$.

A realized confidence interval communicates the precision of an estimate in the target's own units. Reporting $1.2$ alone conceals the difference between an estimate with standard error $0.5$ and one with standard error $0.05$. A narrow confidence interval indicates greater precision for the stated statistical model, rather than a guarantee that its assumptions describe the experiment correctly.

Once $x$ is observed, $C(x)$ either contains the fixed target or does not. The 95% statement concerns the random procedure $C(X)$, not a posterior probability of 0.95 that this particular interval contains $\theta$. A credible interval instead assigns specified posterior probability to a region under Bayesian inference. These interpretations can differ even when the numerical endpoints agree. The repeated-sampling interpretation is explained in NIST's confidence-limits teaching reference.

Jerzy Neyman's 1937 paper formulates the fixed-target, random-endpoint construction in Section II, especially equation (20) on page 348. Its confidence coefficient must hold across the allowed statistical parameters, rather than only at the value suggested by one sample. See Neyman, Outline of a Theory of Statistical Estimation Based on the Classical Theory of Probability and the original paper scan.

For one two-sided t procedure with observed estimate $\hat{\Delta }$, positive standard error $s$, null value ${\Delta }_0$ and reference Student t-distribution with $\nu$ statistical degrees of freedom, the test statistic and p-value must satisfyIts symmetric confidence interval at coverage probability $1-\alpha$ is $[\hat{\Delta }-t_{\nu ,1-\alpha /2}s,\hat{\Delta }+t_{\nu ,1-\alpha /2}s]$ and therefore has midpoint $\hat{\Delta }$. When the same standard error and reference distribution are used, excluding ${\Delta }_0$ is equivalent to $|t|>t_{\nu ,1-\alpha /2}$ and hence $p<\alpha$ for a continuous reference distribution. This proves the test–interval equivalence and supplies several independent checks on a printed output. An approximate Welch's t-test uses its same approximate statistical degrees of freedom throughout. Rounding can blur a boundary decision but cannot explain a confidence-interval midpoint far from the reported estimate.

Design of experiments chooses experimental units, treatments, replication, blocks in experimental design and randomization so that scientifically useful treatment contrasts can be estimated with meaningful standard errors. Allocation determines which measurements supply independent treatment information.

For two independent samples of Bernoulli random variables, the difference of their sample proportions estimates the difference of population probabilities. Since the samples are independent,Substitution of the sample proportions gives a standard error for a large-sample Wald confidence interval. This interval is a descriptive comparison of the specified groups; a causal effect needs further assumptions. Sparse counts and probabilities near zero or one can make the normal approximation poor. Testing equality can instead use a null-pooled estimate of the common probability for the test's standard error.

The elaboration model compares a pooled statistical association with the associations conditional on a third variable. Replication describes a similar pattern within the strata. Specification describes a relationship whose size or direction varies between strata. Attenuation can suggest explanation by an antecedent confounder, or interpretation through an intervening causal mediator. These causal readings require justified temporal and causal assumptions; conditional tables alone do not identify the third variable's role.^[ref]

Assess the patterns through conditional probabilities or comparable effect sizes. Different p-values also reflect different sample sizes and standard errors, so significant and nonsignificant results need not differ significantly.

Suppose a normal linear model gives an unrestricted mean to each cell and has $m$ independent observations per cell, with common error variance ${\sigma }^2$. Each fitted cell mean is its sample mean, with variance ${\sigma }^2/m$; different fitted cell means are independent. Estimate ${\sigma }^2$ by the pooled within-cell residual sum of squares divided by $n-q$, where $q$ is the number of cells. Thus a cell mean has standard error $s/\sqrt{m}$, a difference of two cell means has standard error $s\sqrt{2/m}$, and a two-by-two difference-of-differences has standard error $2s/\sqrt{m}$.

// Target: probability-and-statistics.bigb

An equivalence test asks whether an effect lies strictly inside a prespecified negligible-effect range, such as $(-\delta ,\delta )$ for $\delta >0$. Its null hypothesis is that the effect lies outside that range, including its boundary; its alternative hypothesis is $-\delta <\theta <\delta$. This differs from a test with a zero-effect null hypothesis: failing to reject zero does not establish that the effect is sufficiently small.

One construction performs two level-$\alpha$ one-sided tests, rejecting $\theta \le -\delta$ and rejecting $\theta \ge \delta$, and declares equivalence only if both reject. If the equivalence null hypothesis is true, at least one component null is true. The probability of rejecting both cannot exceed the probability of rejecting that true component, which is at most $\alpha$. Thus the combined procedure controls Type I error at level $\alpha$. In a continuous symmetric t-based location model with a common standard error, this is equivalent to putting the central $100(1-2\alpha )\%$ confidence interval wholly inside $(-\delta ,\delta )$. At $\alpha =0.05$ that is a 90% interval, not the 95% interval used for the usual two-sided zero-effect test. The margin must be chosen for substantive reasons rather than selected afterward to accommodate an observed interval.

Prepare a fresh copy of the same quantum state for each trial and measure the same observable. If preparations and readouts are independent, the outcomes $X_i$ are independent and identically distributed random variables with the Born rule probability distribution. Their sample mean is an unbiased estimator of the expectation value:The law of large numbers explains convergence of the sample mean, while the standard error describes its finite-sample spread. It takes four times as many independent trials to halve that standard error. This reduces estimation error; it does not reduce the quantum uncertainty of a single prepared copy.

For the spin expectation value, let $n_{+}$ and $n_{-}$ be the two outcome counts, with $N=n_{+}+n_{-}$. ThenThe binomial distribution gives the exact sampling model. For $\theta =\pi /3$, the standard error is $\hslash \sqrt{3}/(4\sqrt{N})$: approximately $0.1083\hslash$ for $N=16$ and $0.03827\hslash$ for $N=128$. A run can fluctuate away from the predicted mean even as its typical spread decreases.

Repeatedly measuring one surviving system is a different protocol. Under the ideal Lüders rule, outcome $a_r$ leaves $P_r|\psi ⟩/\sqrt{p_r}$ when $p_r>0$. With no intervening change of observable or dynamics moving the state out of that eigenspace, the next projective measurement returns $a_r$ with certainty. Such a record is correlated and does not sample the original Born rule distribution independently. The quantum nondemolition measurement discussion describes this repeatability; fresh quantum state preparations supply the ensemble needed for the estimator above.

An estimation error is the difference between an estimator and the statistical parameter it estimates. It is a random variable before observing the sample. Its mean is the bias of the estimator, and its mean square is the sum of the variance and squared bias. Standardizing an estimation error by its standard error often yields an approximate standard normal distribution, allowing construction of a Wald confidence interval.

Take the original observations $(1,2,4)$, with sample mean $\overline{x}=7/3$. All $3^3=27$ ordered bootstrap samples have probability $1/27$. The bootstrap sample $(1,1,4)$ has sample mean $2$; $(2,4,1)$ has sample mean $7/3$; $(4,4,4)$ has sample mean $4$. These are illustrative possibilities, not three simulated measurements used to estimate the distribution.

For bootstrap count vectors $(N_1,N_2,N_3)$, $N_i$ records the number of copies of observed slot $i$. Their multinomial distribution isand ${\overline{X}}^{\ast }=(N_1+2N_2+4N_3)/3$. Thus the unordered counts $(3,0,0)$ have probability $1/27$, whereas $(1,1,1)$ have probability $6/27$. Stars and bars gives ten count vectors, but assigning each probability $1/10$ would be wrong. Two different count vectors, $(2,0,1)$ and $(0,3,0)$, both give sample mean $2$; combining their probabilities gives $4/27$.

The conditional bootstrap variance of a sample mean gives a compact check on the enumeration. Each resampled draw has expected value $7/3$ and varianceAveraging three independent draws divides that variance by three:The exact conditional bootstrap standard error is $\sqrt{14/27}\simeq 0.720$. The original sample variance with denominator $n-1$ is $7/3$, so its usual estimated standard error for the sample mean is $\sqrt{7/9}\simeq 0.882$. Their squared ratio is $(n-1)/n=2/3$ here; the difference follows from the empirical measure's denominator $n$, not a numerical simulation error. Neither number is the known true sampling standard error of the unknown source probability distribution.

For uniform sampling without replacement from a population of $n>1$ objects, the finite population correction accounts for the reduced spread caused by drawing distinct objects. For the marked-object count $K$ in a hypergeometric distribution, put $p=c/n$. ThenRelative to the corresponding binomial distribution, the variance is multiplied by $f^2=(n-r)/(n-1)$ and the standard deviation by $f$. The square-root convention for the correction is used in Philip Stark's discussion of sample sums.^[ref]

The factor has a direct interpretation. For $r=1$, replacement makes no difference and $f=1$. For $r=n$, the entire population is observed and the marked count is exactly $c$, so $f=0$. In the ten-object, five-draw example, $f^2=5/9$ and $(6/5)(5/9)=2/3$. The exceptional population size $n=1$ is deterministic and should be handled directly rather than inserting it into a zero denominator.

A short explanation uses an indicator random variable $I_j$ for a marked object at draw $j$. Each has expected value $p$ and variance $p(1-p)$. For distinct draw positions $i,j$,Since $K={\sum }_{j=1}^r{I}_j$, the variance of a sum includes $r$ individual terms and $r(r-1)$ ordered covariance terms, giving the displayed correction. Linearity of expectation still gives $rp$ despite this dependence.

For arbitrary fixed real population values, use $N>1$ for population size and $1\le n\le N$ for sample size. Write the finite population mean $\bar{Y}=N^{-1}{\sum }_i{y}_i$, the finite population variance $S^2=(N-1{)}^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$ and ${\sigma }_N^2=N^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$. Under simple random sampling without replacement, the sample mean satisfiesThus $1-n/N$ is the variance factor relative to $S^2/n$, while $(N-n)/(N-1)$ is the exact variance ratio to independent uniform sampling with replacement, whose sample mean has variance ${\sigma }_N^2/n$. The standard error uses the square root of the corresponding factor. These formulas agree because ${\sigma }_N^2=(N-1)S^2/N$; the denominator convention must be specified. The sample mean under simple random sampling gives a derivation and an exact six-value example.

Fit the positions of successive, consistently defined troughs by linear regression, $V_n=b+nd+{ϵ}_n$. A constant intercept $b$ absorbs the contact potential difference, collection phase and initial energy effects; $d$ estimates the repeated loss expressed as a voltage. Subtracting successive positions cancels $b$. A line fit uses every position and leaves residuals that can reveal curvature or a misidentified trough. It still assumes that the phase is constant over the fitted range and that all troughs belong to the same loss channel.

For an original numerical example, choose $n=1,2,3,4,5$ and synthetic positions $V_n=(6.62,11.46,16.48,21.30,26.14)\mathrm{V}$. Their successive spacings are $(4.84,5.02,4.82,4.84)\mathrm{V}$. The ordinary least squares formulas giveThe fitted residuals are $(-0.004,-0.052,0.080,0.012,-0.036)\mathrm{V}$, with zero sum and zero index-weighted sum. Their residual sum of squares is $0.01056{\mathrm{V}}^2$. If independent, equal-variance, zero-mean errors describe these positions, the residual variance estimate is $s^2=0.01056/(5-2)=0.00352{\mathrm{V}}^2$ and the slope standard error is $s/\sqrt{10}=0.01876\mathrm{V}$. Thus the inferred model loss is $4.888\mathrm{eV}$ with conditional standard error $0.01876\mathrm{eV}$. These errors summarize only the chosen position scatter; a voltage-scale error or a drifting collection phase is not included. Adjacent spacings share a position error and are therefore correlated even when position errors are independent.

The offset has a sharper identifiability limitation. Even if a constant-phase model were exact, an initial beam kinetic energy $E_0$ shifts the troughs by $-E_0/e$. For the symmetric illustrative curve, $b=V_c-E_0/e+V_r/2$. The transformation $V_c\mapsto V_c+\alpha$, $E_0\mapsto E_0+e\alpha$ leaves every trough position unchanged. Measuring $b$ alone therefore cannot determine the contact potential difference. Relabeling $n\mapsto n+m$ also changes $b$ by $-md$ without changing $d$. An independent offset calibration and a justified collection model are needed before turning a first trough into an excitation threshold. Varying $V_r$ tests this model's predicted half-voltage shift of the trough centers; it is a diagnostic prediction of this chosen curve, not a universal correction for experimental minima.

A funnel plot displays study effect estimates against a measure of their precision, often standard error increasing down the vertical axis. Under a common-effect model, imprecise estimates spread more widely around the effect. Asymmetry is evidence of a small-study effect and can be consistent with publication bias, but it has other possible explanations.

For the Kaplan–Meier estimator, the estimated variance isThe formula follows by summing approximate variances of the logarithms of the conditional survival factors and applying the delta method to their product. Its square root is the estimated standard error. The same expression uses the actual delayed-entry risk sets for the Kaplan–Meier estimator with delayed entry. When $0<\hat{S}(t)<1$ and every contributing $r_j>d_j$, an approximate log-scale confidence interval is $\exp \{\log \hat{S}(t)\pm z_{1-\alpha /2}\widehat{\mathrm{se}}(\hat{S}(t))/\hat{S}(t)\}$, with the upper endpoint capped at one. This interval can be inaccurate with very small risk sets.