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Suppose $\theta >0$ and a positive estimator $T_n$ satisfies $\sqrt{n}(T_n-\theta )/\theta \stackrel{d}{\to }N(0,1)$. Choose the standard normal quantile $c={\Phi }^{-1}(1-\alpha /2)$. For $n>c^2$, an asymptotic confidence interval of level $1-\alpha$ isProof. Convergence in distribution gives $\mathbb{P}(-c<\sqrt{n}(T_n-\theta )/\theta <c)\to 1-\alpha$. Multiplication by the positive $\theta$ and rearrangement makes this event equivalent to $\theta (1-c/\sqrt{n})<T_n<\theta (1+c/\sqrt{n})$. Both factors are positive when $n>c^2$, so division yields exactly the displayed interval. Thus its coverage probability tends to $1-\alpha$. For $95\%$ coverage, $c\approx 1.96$. This inversion retains the unknown scale $\theta$ in the normalized error, rather than replacing it by the observed estimator before inversion.

The Cornish-Fisher expansion inverts an Edgeworth expansion to approximate quantiles. For a standardized iid sum, let ${\gamma }_1$ be the skewness of one summand and $z_{\alpha }$ the standard normal quantile. The first distribution function correction is $-{\gamma }_1(z^2-1)\varphi (z)/(6\sqrt{n})$. Substituting $y_{\alpha }=z_{\alpha }+c/\sqrt{n}$ and expanding $\Phi$ gives $c={\gamma }_1(z_{\alpha }^2-1)/6$. This is a fixed-quantile asymptotic, with $\alpha$ fixed in $(0,1)$.

For $X\sim N(m,v)$ with $v>0$ and $0<\epsilon <1$, $P(X\ge C)\le \epsilon$ is equivalent to the displayed chance constraint, because the standard normal distribution function is continuous and strictly increasing. The coefficient is a standard normal quantile, which can be negative. If $v=0$, equality is an atom and the exact condition becomes $C>m$ instead.

Because the real error function is strictly increasing from $-1$ to $1$, its real inverse function maps $(-1,1)$ onto the real line. For $0<q<1$, the standard normal quantile is thereforeMore generally, the $q$-quantile of a normal distribution with expected value $\mu$ and positive standard deviation $\sigma$ is $\mu +\sigma \sqrt{2}{\mathrm{erf}}^{-1}(2q-1)$. This reverses the area calculation: instead of asking how much probability lies below an endpoint, a quantile asks which endpoint encloses a specified probability. The domain restriction matters; $q=0$ and $q=1$ correspond to infinite endpoints, rather than finite real values of the inverse error function.

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 $\mu =0$.

For a one-sided normal-mean test of size $\alpha$ with known standard deviation $\sigma$, a positive alternative ${\delta }^{\ast }$ and target statistical power $1-\beta >\alpha$, the required sample size is the ceiling of ${\sigma }^2(z_{1-\alpha }+z_{1-\beta }{)}^2/({\delta }^{\ast }{)}^2$. This follows by shifting the standard normal quantile in the alternative rejection probability.

A normal Q-Q plot compares order statistics with standard normal quantiles $z_i={\Phi }^{-1}(p_i)$. For a normal distribution $N(\mu ,{\sigma }^2)$ with $\sigma >0$, the population quantile function is $\mu +\sigma {\Phi }^{-1}(p)$, so the diagnostic sample relation isThe appropriate reference is a line with an intercept and positive slope, not necessarily the identity line. A line can be fitted through two central sample quantiles; R's qqline uses the quantile levels $p=0.25$ and $p=0.75$ by default. Neither that line nor its fitting rule is part of the definition of a normal Q-Q plot. ^[ref]

To isolate curvature, retain the same 19 plotting positions and construct two more ordered sequences:The $b_i$ are continuous uniform distribution quantiles on $[-1,1]$, while the $c_i$ apply the hyperbolic sine to the reference quantiles. Both examples are symmetric about zero. At ranks $1,10,19$, $b_i$ is approximately $-0.947368,0,0.947368$ and $c_i$ is approximately $-1.812618,0,1.812618$. Again, these are constructed values rather than random draws or real data.

Read these shapes relative to their central slope. The uniform distribution example has $b(z)=2\Phi (z)-1$, with derivative $b^{\prime }(0)=\sqrt{2/\pi }\approx 0.797885$; at the upper endpoint the tangent predicts $1.546246$, while $b_{19}=0.947368$. Its lower end lies above the tangent and its upper end below it: the outer quantiles are compressed. The hyperbolic sine example has $c^{\prime }(0)=0.7$; its tangent predicts $1.356552$ at the upper endpoint, while $c_{19}=1.812618$. Its lower end lies below the tangent and its upper end above it: the outer quantiles are stretched. Unequal departures at the two ends can suggest skewness. The direction of these descriptions assumes that reference quantiles are horizontal and observed values vertical. The graphs illustrate finite-range shape differences; a sample pattern cannot establish an asymptotic tail probability law or membership in a heavy-tailed distribution class.

A nearly straight normal Q-Q plot is a diagnostic, not a proof of a normal distribution or a formal hypothesis test. Random statistical samples fluctuate, particularly at extreme order statistics, and apparent curvature depends on sample size, plotting positions and the reference line. A calibrated hypothesis test or uncertainty band requires its own assumptions and must account for parameters estimated from the same observations. Sorting also discards the original observation order: rearranging a time series leaves its Q-Q plot unchanged, so the plot cannot establish independence.

For a normal linear model, R's diagnostic normal Q-Q plot uses standardized regression residuals, also called internally studentized residuals. ^[ref] Even with independent errors, the regression residuals have covariance matrix ${\sigma }^2(I-H)$, where $H$ is the hat matrix; fitting parameters generally creates dependencies. Dividing by a shared estimated standard deviation creates further dependence, and the distribution of an internally studentized Gaussian residual is not exactly the standard normal distribution. The normal Q-Q plot therefore helps assess error shape alongside other diagnostics rather than certifying independent normal random variables.

Assume equal allocation of $n$ independent offenders to each arm and a two-sided statistical test of significance level $\alpha$. Write $p_0=0.60$ for P30 and $p_1=0.56$ for initial community-service assignment, with $\Delta =p_0-p_1=0.04$ and $\bar{p}=(p_0+p_1)/2=0.58$. The two event counts have independent binomial distributions. Under the stated alternative, the difference of sample proportions is approximately a normal distribution with mean $\Delta$ and varianceThe pooled two-proportion test has approximate upper rejection threshold $z_{1-\alpha /2}\sqrt{2\bar{p}(1-\bar{p})/n}$ at the planning rates, where $z_q$ is the standard normal quantile. To obtain statistical power $0.80$ against a positive difference, place the alternative mean $z_{0.80}$ alternative-standard-errors above that threshold:Solving yields the usual sample size for comparing two proportions:per arm, rounded upwards, with total sample size $2n$. This targets the dominant rejection tail under the planning alternative; an exact power calculation can include the very small opposite-tail probability or use the binomial distributions directly.

For illustration, at $\alpha =0.05$, $z_{0.975}\simeq 1.960$ and $z_{0.80}\simeq 0.842$, giving $n\simeq 2388.8$. ThusFor a prespecified one-sided test, replace $z_{1-\alpha /2}$ by $z_{1-\alpha }$. The calculation assumes complete endpoint ascertainment, independent individual allocation and that the two initial-policy rates are appropriate; losses or clustered allocation would require corresponding design adjustments.

Set $Y_i=X_i^2$, $\overline{Y}=n^{-1}{\sum }_i{Y}_i$ and ${\tau }^2={\mathrm{Var}}_F(Y_1)={\mathbb{E}}_F{X}_1^4-{\theta }^2$. The fourth-moment assumption makes ${\tau }^2$ finite, so the variance of a sample mean is $v={\tau }^2/n$. Assume $n\ge 2$, as required for leave-one-out estimation.

The Jackknife variance estimator recomputes the original statistic on each sample with one observation removed. HereBy the definition of the Jackknife variance estimator,For completeness, expanding the centered sum givesConsequently $\mathbb{E}{\hat{v}}_{\mathrm{JACK}}={\tau }^2/n=v$. The jackknife variance estimator is unbiased for this statistic. This calculation uses only the finite fourth moment, not normality of the original observations.

Given the observed data, put $y_i=x_i^2$ and $\overline{y}=n^{-1}{\sum }_i{y}_i$. The empirical distribution assigns mass $1/n$ to each observed $x_i$, including repeated values with their multiplicities. A bootstrap sample consists of $n$ independent draws from this distribution, with replacement. The nonparametric bootstrap variance estimate is the conditional varianceIndependence of the bootstrap draws gives the exact conditional bootstrap variance of a sample mean:Thus its finite-sample expectation is $(n-1)v/n$, in contrast to the unbiased jackknife estimate. This is the jackknife and bootstrap variance of a sample second moment scaling difference.

To approximate it by Monte Carlo simulation, independently generate $B$ bootstrap samples. For replicate $b$, choose $n$ independent uniform indices $I_{b1},\dots ,I_{bn}\in \{1,\dots ,n\}$ and compute $T_b^{\ast }=n^{-1}{\sum }_j{x}_{I_{bj}}^2$. With ${\overline{T}}^{\ast }=B^{-1}{\sum }_b{T}_b^{\ast }$, useFor $B>1$ this is conditionally unbiased for the bootstrap variance and converges to it as $B\to \infty$. Uniform indices can be obtained by $I=\lceil nU\rceil$ for uniforms on $(0,1]$.

Let $G_{\ast }(t)=P_{\ast }(T^{\ast }\le t)$ be the conditional bootstrap distribution of the statistic. A nominal level $1-\alpha$ percentile bootstrap confidence interval isReuse the bootstrap replicates, sort them as $T_{(1)}^{\ast }\le \cdots \le T_{(B)}^{\ast }$, and take the order statistics with ranks $\lceil B\alpha /2\rceil$ and $\lceil B(1-\alpha /2)\rceil$ as empirical quantiles. These are quantiles of the statistic itself, rather than reflected quantiles of its centered error. Increasing $B$ reduces simulation error; it does not turn an approximate confidence interval into an exact finite-sample one.

An analytic alternative follows from the central limit theorem. If ${\tau }^2>0$, then $\sqrt{n}({\hat{\theta }}^n-\theta )\Rightarrow N(0,{\tau }^2)$, and the sample variance $s_Y^2=(n-1{)}^{-1}{\sum }_i(x_i^2-\overline{y}{)}^2$ converges to ${\tau }^2$ by the strong law of large numbers. The Slutsky theorem therefore gives the large-sample normal approximation intervalwhere $z_{1-\alpha /2}$ is the standard normal quantile. Its lower endpoint may be truncated to zero because the parameter is nonnegative. If ${\tau }^2=0$, $X_i^2=\theta$ almost surely and the statistic identifies the parameter exactly; the interval degenerates to that value. The nondegenerate normal interval is asymptotic, not an exact Student interval under arbitrary $F$.

Put $\hat{\Sigma }=X^{T}X/n$. The Debiased Lasso correction isFor the Nodewise Lasso, let $r_j=X_j-X_{-j}{\hat{\gamma }}^{(j)}$. Its Karush-Kuhn-Tucker conditions areso $z_{jk}=\mathrm{sgn}({\hat{\gamma }}_k^{(j)})$ at a nonzero coefficient and $|z_{jk}|\le 1$ at a zero coefficient. Multiplying by ${\hat{\gamma }}^{(j)}$ gives the nodewise-Lasso residual identityWe require nonzero design columns, a necessary qualification missing from the unrestricted printed formulation. Because ${\lambda }_j>0$, ${\hat{\tau }}_j^2=0$ would force both $r_j=0$ and ${\hat{\gamma }}^{(j)}=0$, and hence $X_j=0$. Thus nonzero columns suffice for ${\hat{\tau }}_j^2>0$ even when $p>n$ or the full Gram matrix is singular. A zero column instead gives ${\hat{\tau }}_j^2=0$ and its coefficient is unidentifiable from $Y$; the proposed normalization and confidence interval do not exist there.

For each $j$, define $c_j\in {\mathbb{R}}^p$ by $(c_j{)}_j=1$ and $(c_j{)}_{-j}=-{\hat{\gamma }}^{(j)}$. Set row $j$ of $\hat{\Theta }$ to $c_j^T/{\hat{\tau }}_j^2$. Since $X{c}_j=r_j$,This row construction is the nodewise normalization used by van de Geer and coauthors. The matrix $\hat{\Theta }$ need not be symmetric or be an actual inverse.

Substitute the linear model into the correction. The exact decomposition isIn the fixed-design normal linear model, or with $\epsilon \mid X\sim N_n(0,{\sigma }^{2}I)$ and nodewise estimates and tuning chosen from $X$ alone, a linear transformation of a multivariate normal distribution gives $W\mid X\sim N_p(0,{\sigma }^2\hat{\Omega })$. Marginal normality of $\epsilon$ without this conditional assumption would not suffice for a response-dependent random design. No independence between $W$ and $\Delta$ is asserted. The diagonal cancellation and the Holder inequality give the debiased-Lasso remainder bound

For a coordinate variance, ${\hat{\Theta }}_j{X}^T=r_j^T/{\hat{\tau }}_j^2$, whenceWriting $z_{1-\alpha /2}$ for the corresponding standard normal quantile, an approximate confidence interval isEquivalently its half-width is $z_{1-\alpha /2}\sigma \Vert X_j-X_{-j}{\hat{\gamma }}^{(j)}{\Vert }_2/(n{\hat{\tau }}_j^2)$. We assume $0<\alpha <1$ and $\sigma >0$. For asymptotic coverage, the standardized remainder ${\Delta }_j/(\sigma \sqrt{{\hat{\Omega }}_{jj}})$ must have convergence in probability to zero and the coordinate variance must be nondegenerate. Exact conditional normality of $W$ alone does not give finite-sample exact coverage for $\hat{b}$.

One sufficient random-design regime is the Gaussian identity design for debiased Lasso: take independent rows $X_i\sim N_p(0,I_p)$, independently of $\epsilon$, fixed $\sigma >0$, $p=p_n\to \infty$, $s=s_n\ge 1$, andChoose $\lambda =C\sigma \sqrt{\log p/n}$ and all ${\lambda }_j=D\sqrt{\log p/n}$ for sufficiently large fixed $C,D$. Here is why this regime has the requested remainder bound. Uniformly over columns, their squared norms divided by $n$ tend to one, while all off-diagonal sample cross-correlations have magnitude at most a constant times $\sqrt{\log p/n}$ with probability tending to one. Taking $D$ larger makes the zero vector satisfy all nodewise Karush-Kuhn-Tucker conditions strictly, so ${\hat{\gamma }}^{(j)}=0$ for every $j$ on that event, and ${\hat{\tau }}_j^2=\Vert X_j{\Vert }_2^2/n\ge 1/2$ eventually. This also gives ${\hat{\Omega }}_{jj}=1/{\hat{\tau }}_j^2\to 1$ uniformly.

In this regime, the entrywise entrywise concentration of a Gaussian Gram matrix and Lasso cone condition argument in Gaussian identity design for debiased Lasso gives a Compatibility condition for the Lasso bounded away from zero with probability tending to one. Together with the noise-score event and the Basic inequality for the Lasso, this yields, for a fixed $C_1$,For example, the cone inequality and compatibility bound give $\Vert \hat{\beta }-{\beta }^0{\Vert }_1\le 12\lambda s/{\varphi }^2$ on their joint event. Therefore, with a fixed $A\ge 2D{C}_1$,The stated sparsity condition makes this remainder vanish and supplies the preceding confidence-interval requirement. Both dimensions may grow: $p=n^2$ and $s=\lfloor n^{1/4}\rfloor$ satisfy these assumptions. The concentration and Gaussian-design compatibility results are sufficient standard conditions here; the question does not require their proofs. The displayed rate is a high-probability fixed-constant bound, rather than only a claim of boundedness in probability.

Write $y_{WP}$ and $y_{AP}$ for the estimated log odds ratios relative to the common control. Indirect treatment comparison uses the consistency relation ${\delta }_{WA}={\delta }_{WP}-{\delta }_{AP}$. For independent trials, subtraction adds their variances, givingThe substantive assumption is transitivity in network meta-analysis: the study populations, outcome definitions and follow-up are sufficiently comparable, particularly in their distributions of effect modifiers, that the two control comparisons estimate effects applicable to the same target population. The common control's name alone does not justify that assumption. This also assumes comparable marginal odds ratios; these need not equal covariate-adjusted odds ratios even without confounding.

Using the supplied rounded variances and the given standard normal quantile, the approximate confidence interval isOn the odds ratio scale this is $e^{-1.26}\simeq 0.284$, with approximate limits $(0.104,0.771)$. The estimated odds of recurrent stroke under treatment $W$ are about 28% of those under $A$, a reduction of about 72% in odds. The interval excludes equal odds, so this comparison provides evidence of lower odds under $W$ if the indirect treatment comparison assumptions hold. These are odds, not a 72% reduction in probability; a confidence interval describes the long-run coverage of the procedure, not a posterior probability for this particular interval.

One standard construction uses the Debiased Lasso. Starting from the Square-root Lasso estimate $\hat{\beta }$, estimate a vector ${\hat{\Theta }}_j$ that approximately inverts the $j$th column of the empirical Gram matrix $\hat{\Sigma }=X^{T}X/n$, for example by a Nodewise Lasso. Defineand estimate $\sigma$ by $\hat{\sigma }=\Vert Y-X\hat{\beta }{\Vert }_2/\sqrt{n}$. The approximate two-sided level-$\alpha$ test rejects $H_j^{\beta }$ whenwhere $z_{1-\alpha /2}$ is a standard normal quantile.

Sufficient high-dimensional conditions include a Compatibility condition for the Lasso bounded away from zero, $\gamma \asymp \sqrt{\log p/n}$, $\log p=o(n)$, andtogether with the corresponding sparsity and consistency conditions for the nodewise inverse-Gram estimate. Under these assumptions the Debiased-Lasso asymptotic normality makes the rejection probability under $H_j^{\beta }$ tend to $\alpha$.

Apply the central limit theorem to the independent and identically distributed random variables $X_i^2$, whose mean and variance are $\theta$ and ${\theta }^2$. It gives the approximate standard normal distributionThe supplied standard normal quantile $c=1.96$ leaves probability $0.025$ in each tail. HenceSince $\theta >0$, this event is equivalent toFor $n>1.96^2$, equivalently integer $n\ge 4$, both denominators below are positive. Inverting the inequalities gives the approximate $95\%$ confidence intervalThe approximation is justified for large $n$; positivity at $n\ge 4$ by itself does not guarantee accurate coverage. This construction is a confidence interval from a relative normal pivot: the unknown scale remains inside the pivotal ratio until the inequalities are inverted. Its meaning is that the random interval covers the fixed statistical parameter $\theta$ with probability approximately $0.95$ under repeated sampling.

The central limit theorem givesWrite $z=z_{1-\alpha /2}$ for the indicated standard normal quantile. Inverting the approximate pivot inequality and substituting $Y/n=1/\hat{\theta }$ gives the approximate confidence intervalIts coverage is approximately $1-\alpha$ for large $n$. The intersection respects the positive rate parameter; asymptotically the lower endpoint is positive. This is an asymptotic interval, not an exact finite-sample confidence interval.

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.

Condition on the total of 400 complaints. Assume each complaint independently has the same probability $p$ of being received on Monday; in particular, repeated complaints from one incident must not create material dependence. The Monday count $S$ then has a binomial distribution with parameters $(400,p)$. Test $H_0:p=0.2$ against $H_1:p>0.2$, rejecting for large $S$ as suggested by the monotone likelihood ratio.

Under the null hypothesis, ${\mathbb{E}}_{0}S=80$ and ${\mathrm{Var}}_0(S)=400(0.2)(0.8)=64$. Both expected successes and expected failures are large, so the central limit theorem gives a suitable normal approximation:Using the paper's supplied rounded table, the upper-tail p-value is approximatelyA more accurate evaluation of the standard normal distribution cumulative distribution function gives about 0.0062. For example, choose significance level 0.05 before examining the data; the supplied upper 5% standard normal quantile is approximately 1.6, and $2.5>1.6$. Reject $H_0$ at 5%, and also at 1%, using this approximation. The observed binomial proportion is $100/400=0.25$, providing evidence that the Monday probability exceeds 0.2 under the sampling assumptions.

If applying a continuity correction, approximate ${\mathbb{P}}_0(S\ge 100)$ by $1-\Phi ((99.5-80)/8)=1-\Phi (2.4375)\approx 0.0074$; this gives the same 5% and 1% conclusions. The uncorrected value 0.005 is approximate, so a decision exactly at the 0.5% boundary is not warranted by the supplied rounded table.

Sort the observations into order statistics $x_{(1)}\le \cdots \le x_{(n)}$. Choose plotting positions $p_i=(i-1/2)/n$ and calculate the standard normal quantiles $z_i={\Phi }^{-1}(p_i)$. The normal Q-Q plot consists of the pairs $(z_i,x_{(i)})$.

If the normal distribution model is appropriate, the points should lie approximately on a straight line with positive slope, because its population quantile function isThe intercept is approximately $\mu$ and the slope approximately $\sigma$; neither needs to be known in advance. The reference need not be the identity line. One justification is the probability integral transform: for an independent normal statistical sample, $\Phi ((X_{(i)}-\mu )/\sigma )$ are uniform order statistics, whose typical values are near their plotting positions. The equality is approximate for a finite sample, particularly at the extremes. Systematic curvature, unusually heavy tails, or isolated extreme departures argue against the proposed distributional shape. A straight marginal Q-Q plot does not itself establish independence of observations.

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.

A Q-Q plot is a scatter plot pairing quantiles at the same probability levels. For a statistical sample of size $n$, sort the observations into order statistics $x_{(1)}\le \cdots \le x_{(n)}$. Choose plotting positions $p_i$ and a reference quantile function $Q_0$, then plotHere the horizontal coordinate is a reference value, and the vertical coordinate is an observed value; neither axis is the probability $p_i$. The displayed plotting positions are our chosen convention, placing every level strictly between zero and one. They are not a claim that the $i$th order statistic equals that population quantile. For two statistical samples, use sample quantiles at matching levels, with a stated interpolation convention when their sizes differ. ^[ref]

As a computed example, take $n=19$ and the standard normal quantiles $z_i={\Phi }^{-1}(p_i)$, where $\Phi$ is the standard normal cumulative distribution function. Construct the ordered values $a_i=10+2z_i$. At ranks $1,10,19$, the plotting positions are $1/38,1/2,37/38$, the reference quantiles are approximately $-1.937932,0,1.937932$, and the constructed values are approximately $6.124137,10,13.875863$. Every point lies on $a=10+2z$ by construction. These 19 values are deliberately chosen, not random draws or real measurements.

The usefulness of a Q-Q plot is that it turns a comparison of probability distributions into geometry: if a reference random variable $Z$ is transformed into $X=\mu +\sigma Z$ with $\sigma >0$, its quantile function satisfies $Q_X(p)=\mu +\sigma Q_Z(p)$. This follows because a positive scaling preserves ordering. Thus the location-scale family produces a straight population Q-Q plot, and an approximate sample line suggests a compatible shape over the observed range. Its intercept describes location and its slope describes scale. The identity line requires the same location, scale and units on both axes; even then its displayed angle is 45 degrees only with equal axis scales. Curvature is often more informative than a histogram because it shows which quantiles depart from the reference. ^[ref]