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A confidence interval for the success probability of a binomial distribution, based on the observed success count and number of trials. A nominal confidence level need not equal the actual finite-sample coverage probability; the distinction matters especially for small samples and proportions near zero or one.

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.

The coverage probability is the repeated-sampling probability that a random confidence interval contains the fixed true statistical parameter. A finite-sample level-$1-\alpha$ procedure has coverage probability at least $1-\alpha$ at every admissible statistical parameter; exact nominal coverage means equality. An asymptotically calibrated procedure approaches the nominal value as sample size grows, under its stated assumptions. Higher-order Edgeworth expansions can quantify the discrepancy from nominal coverage probability when their regularity conditions hold.

Observed coverage in finitely many repetitions is itself random. For $m$ independent repetitions of the exact normal mean confidence interval with known variance, the count $K$ of covering confidence intervals has binomial distribution $\mathrm{Binomial}(m,0.95)$. At $m=40$, its expectation is $38$, butConsequently, the 39 successes in the figure are compatible with exact 95% coverage probability. The guarantee follows from the pivotal quantity's probability distribution; the finite observed fraction is an illustration. Also distinguish a confidence interval for a fixed mean from a prediction interval for a future observation: the latter must accommodate the new observation's variation as well as estimation uncertainty.

A credible interval is an interval $C(d)$ containing a prescribed amount $1-\alpha$ of posterior probability for an unknown scalar statistical parameter $\Theta$. For a continuous posterior distribution with posterior density $p(\theta \mid d)$, its endpoints satisfyThe conditioning includes the chosen statistical model and prior distribution. For a discrete posterior distribution, exact equality may be impossible because of its probability masses; an interval can instead be required to contain at least the requested posterior probability. An interval compresses the full posterior distribution into a range whose uncertainty can be communicated in the parameter's own units. See MIT 18.05, Class 16: probability intervals.

For the eight-head, two-tail example in Bayesian statistics, the posterior density is $2860{\theta }^9(1-\theta {)}^3$ on $[0,1]$. Its cumulative distribution function, obtained by integration of this polynomial, isSolving $F(a)=0.025$ and $F(b)=0.975$ by the bisection method givesThese are numerical approximations to the exact quantiles defined by those equations; the endpoints were computed from the cumulative distribution function, rather than read off a plot. The resulting equal-tailed credible interval leaves 2.5% posterior probability on each side. Its shaded area is 95% of the total area under the posterior density.

Credible intervals are not unique. The one-sided credible interval $[0,F^{-1}(0.95)]\approx [0,0.887334219]$ also has 95% posterior probability, but leaves all 5% outside it on the right. A highest density region instead selects points above a density threshold; it need not be an equal-tailed credible interval, and when the posterior density has separated peaks it need not be one interval at all. Thus a reported credible interval should specify its construction.

A 95% credible interval makes a statement about the statistical parameter conditional on these observed data and the Bayesian model. A 95% confidence interval instead refers to the coverage probability of a procedure under repeated sampling at a fixed parameter value. These are different conditioning statements: 95% posterior probability does not by itself imply 95% coverage probability for every fixed parameter. The confidence interval article develops that sampling interpretation.

For an exponential distribution with mean $\theta$, the maximum-likelihood estimator is $\bar{Y}$ and $\sqrt{n}(\bar{Y}-\theta )/\theta$ converges to a standard normal distribution. Inverting the event that this pivot lies between $-z$ and $z$ gives the displayed confidence interval, provided $n>z^2$. Its two-sided coverage probability has error $O(n^{-1})$, because the first Edgeworth expansion term is even and cancels between the two tails.

Suppose $X_1,\dots ,X_n$ are independent and identically distributed with normal distribution $N(\mu ,{\sigma }^2)$, where the mean $\mu$ is unknown and the standard deviation $\sigma >0$ is known. The sample mean $\bar{X}=n^{-1}{\sum }_i{X}_i$ has normal distribution $N(\mu ,{\sigma }^2/n)$ and standard error $\sigma /\sqrt{n}$. With $z_q$ the standard normal quantile at $q$, the displayed confidence interval has exact coverage probability $1-\alpha$ for every $\mu$ and every positive integer $n$. No central limit theorem approximation is needed.

Take $\sigma =2$, $n=16$ and nominal confidence 95%. The standard error is $2/\sqrt{16}=0.5$, and $z_{0.975}=1.9599639845\dots$. ThereforeRounded for reporting, this is $\bar{x}\pm 0.98$. If $\bar{x}=1.2$, the realized confidence interval is approximately $[0.22,2.18]$. Exact 95% calibration refers to the unrounded standard normal quantile; using $1.96$ literally gives coverage probability $0.9500042097\dots$, almost but not exactly 0.95.

The confidence interval's width scales as $n^{-1/2}$: increasing sample size from 16 to 64 halves its half-width to $0.4899909961\dots$. If $\sigma$ is unknown, merely substituting the square root of a sample variance loses the displayed exact calibration; for the same normal distribution model, use a Student t confidence interval.

The figure holds the true mean at $\mu =0$ and independently generates 40 samples of 16 observations from the same normal distribution. NumPy's PCG64 generator with seed 42 gives 39 covering confidence intervals and one miss, at sample 32: observed coverage is $39/40=97.5\%$. Every bar uses the unrounded standard normal quantile. This finite collection illustrates the moving endpoints; it does not establish the analytic coverage probability or require exactly two misses.

The beauty of the construction is that a familiar standard normal distribution calibrates every possible unknown mean at once. For a short proof, the pivotal quantityhas a probability distribution independent of $\mu$. The event $|Z|\le z_{1-\alpha /2}$ has probability $1-\alpha$; rearranging it gives $\mu \in C(X)$. The right panel shows the equivalent event $|\bar{X}|\le 0.9799819923\dots$ when $\mu =0$.

The parameter space of a statistical model is the set of allowed statistical parameter values indexing its probability distributions. For an exponential distribution parametrized by its mean, $\Theta =(0,\infty )$; for a Bernoulli distribution, $\Theta =[0,1]$. Intersecting a confidence interval with the known parameter space cannot remove the true statistical parameter and therefore preserves its coverage probability.

Use the Probabilists' Hermite polynomials, rather than the differently scaled physicists' convention:Thus $H_2=x^2-1$, $H_3=x^3-3x$, and $H_5=x^5-10x^3+15x$. The cumulant-generating function of an exponential variable of mean $\theta$ is $-\log (1-\theta t)$ near zero, so its $r$th cumulant is $(r-1)!{\theta }^r$. Hence the standardized third and fourth cumulants are ${\rho }_3=2$ and ${\rho }_4=6$; the fourth is a cumulant, not the fourth raw central moment.

Write $F_n$ for the cumulative distribution function of the pivot in part (i). Its Edgeworth expansion, valid here because the exponential law is smooth, nonlattice and has exponential moments, isat the fixed endpoints $t=\pm z$. The coverage is exactly $F_n(z)-F_n(-z)$. Since $H_2$ and $\varphi$ are even, the $n^{-1/2}$ terms cancel; the two odd Hermite terms double. Consequently the symmetric Edgeworth coverage cancellation givesIn particular its coverage probability error is $\boxed{O(n^{-1})}$. This is a fixed-confidence-level expansion; it is not a uniform statement for normal cutoffs increasing with $n$.

For the mean parameter $\theta =1/\lambda >0$, the log-likelihood is $\ell (\theta )=-n\log \theta -{\sum }_i{Y}_i/\theta$. Its derivative is $n(\bar{Y}-\theta )/{\theta }^2$, positive below $\bar{Y}$ and negative above it. Therefore $\boxed{\hat{\theta }=\bar{Y}}$ is the unique maximum-likelihood estimator.

The exponential distribution has mean $\theta$ and variance ${\theta }^2$. The central limit theorem gives the pivotLet $z={\Phi }^{-1}(1-\alpha /2)$ for fixed $0<\alpha <1$. For $n>z^2$, the event $-z<T_n<z$ is exactlyBoth denominators are then positive, so the algebra neither reverses the ordering nor changes the parameter domain. Taking probabilities and using continuity of the limiting normal distribution proves the exponential-mean confidence interval by pivot inversion has coverage probability tending to $\boxed{1-\alpha }$. This is inversion of the pivot with the true mean in its denominator; it is not the ordinary symmetric Wald interval centered at $\bar{Y}$. For a small sample with $n\le z^2$, the displayed finite positive upper endpoint is not valid, although the large-sample claim is unaffected.

At sample size 100, keep essentially the same numerical confidence interval, but treat its calibration as approximate. For independent heights with finite, nonzero variance, the central limit theorem makes the sample mean approximately normally distributed, and the sample variance consistently estimates the population variance. Hence the studentized sample mean is approximately standard normal even without a normal distribution for individual heights.

Using the permitted cutoff 2 givesA Student t-distribution cutoff gives a similar approximation for large sample size, but its justification at fixed sample size is no longer exact. Extreme skewness or influential outliers could make 100 observations insufficient for accurate coverage probability; abandoning normality alone supplies no specific numerical correction. The sampling with replacement assumption still precludes a finite population correction.

Let $z={\Phi }^{-1}(1-\alpha /2)$, where $\Phi$ is the standard normal distribution's cumulative distribution function. Statistical consistency and Slutsky's theorem allow replacement of $I_1(\theta )$ by $I_1(\hat{\theta })$:Thus the probability that this standardized estimation error lies between $-z$ and $z$ tends to $1-\alpha$. Rearranging gives the Wald confidence intervalThe coverage probability is approximate, with limiting value $1-\alpha$; it is a repeated-sampling statement about the random confidence interval.

Given observed data $x$, a credible interval $[L(x),U(x)]$ has posterior probability $1-\alpha$:For a continuous posterior distribution with probability density function $\pi (\theta \mid x)$, this is ${\int }_{L(x)}^{U(x)}\pi (\theta \mid x)d\theta =1-\alpha$. The probability is conditional on the observed data and concerns the statistical parameter under its posterior distribution; it does not assert frequentist coverage probability.

Let $z=z_{1-\alpha /2}$ be the $(1-\alpha /2)$ standard normal quantile. Since $\hat{\mu }$ is a consistent estimator, Slutsky's theorem givesIts estimated standard error is $\hat{\mu }/\sqrt{3n}$, giving the approximate $(1-\alpha )$ Wald confidence intervalFor each fixed $\mu >0$, this procedure's coverage probability tends to $1-\alpha$. Intersecting with the positive parameter space preserves coverage of the true positive value.

The standard error suggested by part (iii) is $\theta /\sqrt{n}$. Substituting the consistent estimator $T_2$ gives $T_2/\sqrt{n}$; Slutsky's theorem givesUsing the central standard normal quantiles $\pm 1.96$, an approximate 95% Wald confidence interval isIts coverage probability approaches 0.95 for each fixed $\theta >0$ as $n\to \infty$. The intersection enforces positivity and does not change whether the true positive statistical parameter is covered. The value 1.64 would give the wrong central two-sided coverage probability.

The estimated standard error is ${\hat{\theta }}^{3/2}/\sqrt{n}$. Using the central 95% standard normal quantiles $\pm 1.96$, a Wald confidence interval isConsistency of $\hat{\theta }$ and Slutsky's theorem justify replacing the unknown scale ${\theta }^{3/2}$ by its estimate. The coverage probability is asymptotic, not exact at finite $n$. Intersecting with the positive parameter space removes impossible values without changing whether the true positive parameter is covered.

Given the observed data, a $100(1-\alpha )\%$ credible interval $[a,b]$ for the scalar parameter satisfiesThe probability refers to the posterior distribution, proportional to $\pi (\theta ){\prod }_{i}f(x_i\mid \theta )$ on the parameter space. It conditions on the observations; it is not the repeated-sampling coverage probability definition of a confidence interval.

Fix $N>1$ real population values $y_1,\dots ,y_N$. Define the finite population mean and two normalizations of the finite population variance byThe values are fixed; only the selected subset $A$ is random. For its sample mean ${\bar{y}}_A=n^{-1}{\sum }_{i\in A}{y}_i$, the exact design expectation and variance areThe sample mean is an unbiased estimator of the finite population mean. At $n=N$, every unit is observed and ${\bar{y}}_A=\bar{Y}$ exactly, so the variance is zero. The $N=1$ case is deterministic and is handled directly, since $S^2$ is undefined then.^[ref]

The finite population correction depends on the baseline. With the $N-1$ normalization it is $1-n/N$ multiplying $S^2/n$. With the $N$ normalization it is $(N-n)/(N-1)$ multiplying ${\sigma }_N^2/n$, the exact variance for independent uniform sampling with replacement. The standard error uses the square root of the relevant variance factor. Mixing the two normalizations gives a wrong answer even at $n=1$, when replacement cannot matter. For the six-value example, $\bar{Y}=7$, ${\sum }_i(y_i-7{)}^2=70$, $S^2=14$ and ${\sigma }_N^2=35/3$, reproducing $14/3$ without replacement and $35/6$ with replacement.

A short proof exposes the statistical dependence. Let $X_r$ be the value at position $r$ in uniform sequential sampling without replacement, so each $X_r$ has expected value $\bar{Y}$ and variance ${\sigma }_N^2$. Write $d_i=y_i-\bar{Y}$. Every ordered pair of distinct labels is equally likely, and ${\sum }_i{d}_i=0$, hence for $r\ne s$,Linearity of expectation proves unbiasedness without requiring independence. The variance of a sum includes all $n(n-1)$ ordered covariance terms, givingFor the original population, each pair of distinct draw positions has covariance $-7/3$. Drawing a high value leaves a lower mean among the remaining values, balancing subsequent draws. Equal marginal probability distributions therefore coexist with negative covariance.

When $2\le n\le N$, the sample variance $s_A^2=(n-1{)}^{-1}{\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2$ is an unbiased estimator of $S^2$. To check this independently, use ${\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2={\sum }_{i\in A}(y_i-\bar{Y}{)}^2-n({\bar{y}}_A-\bar{Y}{)}^2$: its expected value is $n{\sigma }_N^2-n\mathrm{Var}({\bar{y}}_A)=(n-1)S^2$. Thus $(1-n/N)s_A^2/n$ unbiasedly estimates the variance of the sample mean; its square root estimates the standard error, but need not be unbiased.^[ref]

The diagram compares variance with the expected fraction of distinct population labels observed. Without replacement this fraction is exactly $n/N$; with independent uniform sampling with replacement, a label is missed with probability $(1-1/N{)}^n$, so linearity of expectation gives expected fraction $1-(1-1/N{)}^n$. At six draws, replacement covers only about $66.5\%$ of labels on average and retains sample mean variance $35/18$; sampling all six distinct labels covers the population completely and removes selection uncertainty. This population coverage is distinct from confidence interval coverage probability.

The correction matters when the sampling fraction is appreciable. It calibrates uncertainty about this fixed population under the specified selection mechanism. It does not correct unequal selection weights, missing observations or measurement errors; observing every label also does not eliminate uncertainty about a larger unobserved population. Moreover, an exact variance alone does not guarantee the coverage probability of a proposed confidence interval: a small discrete sampling distribution can make a normal approximation inaccurate.

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.

Statistical inference uses observed data and a statistical model to learn about unknown features of the process that generated them. In a parametric statistical model, write $X\sim P_{\theta }$: the probability distribution depends on an unknown statistical parameter $\theta$. An estimand specifies the target, such as a mean or a difference of means. A statistic is a function of the observations; an estimator is a statistic used to estimate the target. Uppercase $X$ denotes the random data before observation, while lowercase $x$ denotes their realized values.

Three questions lead to different outputs. Statistical estimation asks which target value the observations suggest. A confidence interval asks for a range produced by a procedure with calibrated coverage probability. A hypothesis test asks whether the observations depart sufficiently from a specified null hypothesis to enter its rejection region. These outputs complement one another: an estimated effect size describes magnitude, while a standard error describes the sampling distribution of an estimator. A p-value describes a tail probability under the null hypothesis.

The appeal of statistical inference is that it turns variable observations into reproducible uncertainty statements. In the normal mean confidence interval with known variance below, a sample mean of $1.2$ has standard error $0.5$: both the estimated shift and its precision matter. The calibration depends on the statistical model, including independence and the specified probability distribution. Increasing sample size improves precision under those assumptions; it does not repair a systematically misleading sampling procedure. Bayesian inference provides another framework, using a prior distribution and a likelihood function to obtain a posterior distribution.