Pivotal Quantities & Exact CIs - Statistics Graduate Level

1. Which is the exact equal-tailed Garwood $95\%$ confidence interval for the event rate $\lambda$?
   Answer: $[0.0619,\ 0.8768]$ When you see a Poisson rate estimation problem, think immediately about the Garwood exact confidence interval, which inverts the Poisson CDF using chi-square quantiles. The key formulas are: $\lambda_L = \frac{\chi^2_{\alpha/2,\; 2x}}{2E}, \qquad \lambda_U = \frac{\chi^2_{1-\alpha/2,\; 2(x+1)}}{2E}$ With $x = 3$ and $E = 10$, the degrees of freedom are $2x = 6$ for the lower bound and $2(x+1) = 8$ for the upper bound. Lower bound: $\lambda_L = \dfrac{\chi^2_{0.025,\,6}}{2(10)} = \dfrac{1.237}{20} = 0.0619$ Upper bound: $\lambda_U = \dfrac{\chi^2_{0.975,\,8}}{2(10)} = \dfrac{17.535}{20} = 0.8768$ This gives the interval $[0.0619,\ 0.8768]$, confirming B is correct. Now for the distractors. A gets the lower bound right but uses $\chi^2_{0.975,6} = 14.449$ for the upper bound — a degrees-of-freedom error, failing to increment $x$ to $x+1$ for the upper tail. C uses $\chi^2_{0.025,8} = 2.180$ for the lower bound, incorrectly applying 8 degrees of freedom (meant for the upper bound) to the lower limit. D simply reads the raw chi-square values (1.237 and 17.535) without dividing by $2E$, omitting the exposure scaling entirely. A helpful mnemonic: **lower uses $2x$, upper uses $2(x+1)$**, and both are divided by $2E$. Always check that you've incremented the degrees of freedom asymmetrically — this asymmetry is the most common source of error on Garwood interval problems.
2. Which interval is the exact conditional $95\%$ confidence interval for $c^{\mathsf T}\beta$ under the stated model?
   Answer: $[1.128,\ 3.672]$ When building a confidence interval for a contrast $c^\mathsf{T}\beta$ in a Gaussian linear model, you need three ingredients: the point estimate, the standard error of that estimate, and the correct critical value. The key decision is which critical value applies — and that hinges on degrees of freedom. In this model, the residual degrees of freedom are $n - p = 20 - 4 = 16$. Because $s$ is estimated (not known), the pivot statistic follows a $t$-distribution with 16 degrees of freedom, making $t_{0.975,16} = 2.120$ the correct critical value. The standard error of the fitted contrast is $\text{SE} = s\sqrt{c^\mathsf{T}(X^\mathsf{T}X)^{-1}c} = 1.5\sqrt{0.16} = 1.5 \times 0.4 = 0.6$. The 95% CI is then $2.4 \pm 2.120 \times 0.6 = 2.4 \pm 1.272$, giving $[1.128,\ 3.672]$, which is answer A. Answer B uses $t_{0.975,19} = 2.093$, which corresponds to $n - 1 = 19$ degrees of freedom — a mistake you'd make if you forgot to subtract the number of parameters $p$, not just 1. Answer C uses the normal critical value $z_{0.975} = 1.960$, appropriate only when $\sigma^2$ is known, which it isn't here since $s$ is estimated from data. Answer D appears to use an incorrect standard error entirely (likely $\sqrt{0.16} = 0.4$ without multiplying by $s$), shrinking the interval incorrectly. The key study tip: always match your critical value to the correct degrees of freedom $n - p$, and remember that any time $\sigma$ is estimated, you must use a $t$-distribution, never $z$.
3. Which expression gives the exact equal-tailed $95\%$ confidence interval for $\theta$?
   Answer: $\left[\frac{8}{(0.975)^{1/5}},\ \frac{8}{(0.025)^{1/5}}\right]$ When constructing confidence intervals via a pivot, your core task is to find a function of the data and parameter whose distribution is fully known, then algebraically invert probability statements to isolate the parameter. Here, the pivot is $U = M/\theta$, where $M = \max(X_1,\ldots,X_5)$. Since each $X_i \sim \text{Uniform}[0,\theta]$, the CDF of $M$ is $F_M(m) = (m/\theta)^5$, so $U = M/\theta$ follows a $\text{Beta}(5,1)$ distribution on $[0,1]$ with CDF $F_U(u) = u^5$. For an equal-tailed 95% interval, you need $P(u_L \leq U \leq u_U) = 0.95$ with 2.5% in each tail. Solving $u_L^5 = 0.025$ and $u_U^5 = 0.975$ gives $u_L = (0.025)^{1/5}$ and $u_U = (0.975)^{1/5}$. Now invert: $u_L \leq M/\theta \leq u_U$ becomes $M/u_U \leq \theta \leq M/u_L$. Substituting $M = 8$ yields $\left[\frac{8}{(0.975)^{1/5}},\ \frac{8}{(0.025)^{1/5}}\right]$, confirming A is correct. B is wrong because it never inverts the pivot — it places $\theta$ on the numerator side, which is backwards and produces values less than 8 (impossible, since $\theta \geq M$). C uses an asymmetric cutoff $(0.05)^{1/5}$ on one side instead of $(0.025)^{1/5}$, violating the equal-tailed requirement. D reflects a one-sided 95% lower bound using $0.95^{1/5}$, not a two-sided interval. The key strategy: always track the direction of inequality when inverting a pivot. Because $\theta$ appears in the denominator of $U$, dividing flips the inequalities, putting the larger quantile in the denominator of the lower bound.
4. Which interval is the Clopper–Pearson interval, and not merely an approximation or a one-sided interval?
   Answer: $\left[0,\ 1-(0.025)^{1/20}\right]$ When you see a Clopper–Pearson interval question, your anchor is the exact beta-distribution inversion formula. For $x$ successes in $n$ trials, the equal-tailed two-sided interval at confidence level $1-\alpha$ sets each tail to $\alpha/2$. The upper bound solves $P(X = 0 \mid p_U) = (\alpha/2)$, i.e., $(1-p_U)^n = \alpha/2$, giving $p_U = 1 - (\alpha/2)^{1/n}$. The lower bound is 0 since no failures-from-above constrain it. With $n=20$, $x=0$, and $\alpha = 0.05$, you need $\alpha/2 = 0.025$. Plugging in: $p_U = 1-(0.025)^{1/20}$. The interval is $\left[0,\ 1-(0.025)^{1/20}\right]$, confirming C is correct. Each wrong answer embodies a specific error. A uses the exponent $1/21$ instead of $1/20$—this would arise from an off-by-one mistake, perhaps confusing $n+1$ with $n$ in a related formula (like a Bayesian credible interval with a uniform prior). B uses the full $\alpha = 0.05$ rather than the half-tail $\alpha/2 = 0.025$—a classic error of forgetting to split $\alpha$ for a two-sided interval. D gives $3/20$, which is the well-known "rule of three" approximation ($3/n$) used for quick estimation when $x=0$; it is convenient but not the exact Clopper–Pearson bound. Your study tip: always write out $\alpha/2$ explicitly before computing Clopper–Pearson bounds. The most common trap on exam questions like this is using the full $\alpha$, which would be appropriate only for a one-sided interval.
5. Which interval is the exact equal-tailed $95\%$ confidence interval for the rate $\lambda$?
   Answer: $[0.288,\ 1.202]$ When you see exponential lifetimes and need a confidence interval for the rate parameter $\lambda$, the key pivot is recognizing that if $X_i \sim \text{Exp}(\lambda)$, then $2\lambda \sum X_i \sim \chi^2_{2n}$. With $n=8$, the pivot is $2\lambda \sum_{i=1}^8 X_i \sim \chi^2_{16}$, which gives you an exact interval. Inverting the probability statement $P\!\left(\chi^2_{0.025,16} \leq 2\lambda\sum X_i \leq \chi^2_{0.975,16}\right) = 0.95$ yields: $\lambda \in \left[\frac{\chi^2_{0.025,16}}{2\sum X_i},\ \frac{\chi^2_{0.975,16}}{2\sum X_i}\right]$ Plugging in $\sum X_i = 12$, $\chi^2_{0.025,16} = 6.908$, and $\chi^2_{0.975,16} = 28.845$: $\text{Lower: } \frac{6.908}{24} \approx 0.288, \quad \text{Upper: } \frac{28.845}{24} \approx 1.202$ This confirms answer A, $[0.288,\ 1.202]$. Answer B is wrong because it divides by $\sum X_i = 12$ instead of $2\sum X_i = 24$, forgetting the factor of 2 in the pivot. Answer C divides by $2n = 16$ rather than $2\sum X_i = 24$, confusing the degrees of freedom with the observed sum. Answer D appears to use incorrect quantile values or a different (non-exact) approximation method entirely. A useful memory anchor: the factor of 2 appears because the exponential is a special case of the gamma, and the chi-squared pivot requires doubling both $\lambda$ and the sufficient statistic. Always write out the full pivot $2\lambda\sum X_i \sim \chi^2_{2n}$ before plugging in numbers — skipping that step is exactly how you land on distractors B or C.
6. Which statement about the resulting set of retained values is correct?
   Answer: Under the constant-effect model, it is an exact randomization-based $95\%$ confidence set for $\tau$, with possible conservatism because only finitely many treatment assignments exist and attainable test sizes may fall below $0.05$. When you see a question about randomization-based inference, anchor yourself to this principle: validity comes entirely from the random assignment mechanism, not from distributional assumptions about outcomes. The procedure described is Fisher's method of inverting randomization tests to construct confidence sets. For each candidate value $\tau_0$, you impute the missing potential outcomes under the sharp null $H_0: \tau = \tau_0$, then test whether the observed assignment looks unusual relative to all possible randomizations. The set of non-rejected values forms a valid confidence set. B is correct because under the constant-effect model, the sharp null is exactly equivalent to testing that specific $\tau$, giving exact $1 - \alpha$ coverage in repeated randomizations. However, because only finitely many treatment assignments exist, the discrete distribution of the test statistic may prevent the rejection region from attaining exactly level $0.05$—the true size may be slightly below $0.05$, making the procedure potentially conservative. A is tempting but wrong: with heterogeneous effects, imputing potential outcomes using a single constant $\tau_0$ is misspecified. The sharp null no longer cleanly corresponds to a statement about the average treatment effect, so inversion doesn't yield a valid confidence interval for the ATE under heterogeneity. C is wrong because randomization inference requires no distributional assumptions—normality is irrelevant. The validity is entirely design-based. D is wrong because randomization inference explicitly conditions on the observed units and their fixed potential outcomes; the only randomness is the assignment mechanism itself, which is perfectly known and controlled by the experimenter. Study tip: Remember that randomization tests are nonparametric and design-based—their validity never depends on outcome distributions, only on the assignment mechanism.
7. Which interval is the exact equal-tailed $95\%$ confidence interval for $\rho$?
   Answer: $[0.499,\ 6.084]$ When comparing two population variances from independent normal samples, you build a confidence interval for $\rho = \sigma_1^2/\sigma_2^2$ using the pivot $F = \frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2} \sim F_{n_1-1,\, n_2-1}$. Rearranging this pivot gives the exact equal-tailed 95% CI: $\left[\frac{S_1^2}{S_2^2}\cdot\frac{1}{F_{0.975;\,9,14}},\quad \frac{S_1^2}{S_2^2}\cdot\frac{1}{F_{0.025;\,9,14}}\right]$ First compute the observed ratio: $S_1^2/S_2^2 = 8/5 = 1.6$. Then: $\text{Lower} = \frac{1.6}{3.209} \approx 0.499, \qquad \text{Upper} = \frac{1.6}{0.263} \approx 6.084$ This confirms D is correct. Choice A divides by the wrong quantiles entirely — the bounds $[0.164, 2.004]$ are too narrow and don't correspond to any standard manipulation of the given values. Choice B yields $[0.421, 5.134]$, which comes from mistakenly using $S_2^2/S_1^2 = 0.625$ as the ratio (swapping which sample is "1"), then dividing correctly — a labeling error. Choice C produces $[0.706, 2.467]$, consistent with incorrectly using the square roots of the variances (i.e., treating $S_1/S_2$ as the ratio) rather than $S_1^2/S_2^2$. The key study tip: always divide the sample variance ratio by the larger F-quantile to get the lower bound and by the smaller F-quantile for the upper bound. Because $F_{0.025} < 1 < F_{0.975}$, dividing by a small number produces the wider (upper) end — which is why the CI is asymmetric around the point estimate.
8. Which is the exact equal-tailed $95\%$ confidence interval for the population variance $\sigma^2$?
   Answer: $[4.52,\ 25.94]$ Whenever you see a confidence interval for a population variance from a normal distribution, your go-to pivot statistic is $\frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1}$. An equal-tailed 95% interval traps the middle 95% of the chi-square distribution, leaving 2.5% in each tail, so you use the critical values $\chi^2_{0.025,11}$ and $\chi^2_{0.975,11}$. The confidence interval formula for $\sigma^2$ is: $\left[\frac{(n-1)S^2}{\chi^2_{0.975,11}},\ \frac{(n-1)S^2}{\chi^2_{0.025,11}}\right]$ Notice the larger chi-square value goes in the denominator of the lower bound (since you're dividing, bigger denominator gives smaller result). Plugging in $n-1=11$ and $S^2=9$: $\text{Lower bound: } \frac{11 \times 9}{21.920} = \frac{99}{21.920} \approx 4.52$ $\text{Upper bound: } \frac{11 \times 9}{3.816} = \frac{99}{3.816} \approx 25.94$ This confirms **answer A: $[4.52,\ 25.94]$**. Choice B ($[4.93, 28.30]$) results from using $n=12$ instead of $n-1=11$ in the numerator — a classic off-by-one error. Choice C ($[3.12, 17.93]$) comes from mistakenly using $S=3$ (the standard deviation) rather than $S^2=9$. Choice D ($[5.03, 21.64]$) flips the chi-square critical values — placing $\chi^2_{0.025}$ in the lower bound's denominator, which reverses the interval endpoints. Your study tip: always remember you divide by chi-square values, so the smaller critical value produces the larger bound. Sketch the chi-square distribution and label the tails to avoid swapping the critical values under pressure.
9. What is the exact coverage probability of the distribution-free interval $[X_{(2)},X_{(9)}]$ for $m$?
   Answer: $\frac{1002}{1024}\approx0.9785$ When working with distribution-free confidence intervals for a median, the key framework is sign statistics and order statistics. For any continuous distribution with median $m$, each observation independently satisfies $P(X_i < m) = P(X_i > m) = 0.5$. The interval $[X_{(j)}, X_{(k)}]$ covers $m$ if and only if fewer than $j$ observations fall below $m$ AND fewer than $n-k+1$ observations fall above $m$. For $[X_{(2)}, X_{(9)}]$ with $n=10$, the interval fails to cover $m$ when either: (1) fewer than 2 observations exceed $m$ (i.e., 0 or 1 fall above), or (2) fewer than 2 observations fall below $m$ (i.e., 0 or 1 fall below). Let $Y \sim \text{Binomial}(10, 0.5)$ count observations below $m$. Then: $P(\text{miss}) = P(Y \leq 1) + P(Y \geq 9) = 2 \cdot P(Y \leq 1)$ by symmetry. Computing: $P(Y=0) = \frac{1}{1024}$, $P(Y=1) = \frac{10}{1024}$, so $P(Y \leq 1) = \frac{11}{1024}$. Thus $P(\text{miss}) = \frac{22}{1024}$, giving coverage $= \frac{1002}{1024} \approx 0.9785$, confirming D. Choice A ($912/1024$) arises from using $[X_{(1)}, X_{(10)}]$-style logic incorrectly. Choice B ($1012/1024$) likely miscounts boundary terms, forgetting to include both $Y=0$ and $Y=1$ cases. Choice C ($990/1024$) corresponds to a different interval, perhaps $[X_{(3)}, X_{(8)}]$, where failure counts extend to $Y \leq 2$. Study tip: Always compute coverage as $1 - P(\text{miss})$, where "miss" means too few observations on either side of $m$. Symmetry of the Binomial(n, 0.5) cuts your work in half.
10. Which statement correctly describes the pivotal basis and exactness of interval estimation for $\mu_1-\mu_2$?
    Answer: Under equal variances, the pooled statistic has a $t_{18}$ distribution, giving an exact interval; if variances differ, Welch's statistic has a distribution that depends on nuisance variance components, so Welch–Satterthwaite intervals are approximate. When comparing two normal population means with unknown but assumed equal variances, the pivotal quantity is the pooled two-sample $t$-statistic. Its degrees of freedom come from adding the two sample degrees of freedom: $(n_1 - 1) + (n_2 - 1) = 8 + 10 = 18$. This gives an exact $t_{18}$ distribution under the equal-variance assumption, producing exact confidence intervals for $\mu_1 - \mu_2$. When that equal-variance assumption breaks down, you can no longer use $S_p^2$ validly. Welch's approach estimates the two variances separately, forming a statistic whose distribution is no longer exactly Student's $t$ — it depends on the unknown ratio $\sigma_1^2/\sigma_2^2$, which are nuisance parameters. The Welch–Satterthwaite approximation replaces those nuisance components with estimates to produce an approximate degrees-of-freedom value, yielding intervals that are approximate, not exact. This is precisely what C describes. A is wrong on two counts: $S_p^2$ does not make the statistic standard normal (finite samples require $t$, not $Z$), and switching to unequal variances does far more than change interval width — it invalidates the pivot itself. B correctly identifies the equal-variance case but then claims Welch's statistic is exactly $t$-distributed with data-dependent degrees of freedom. That claim is false — the Satterthwaite degrees of freedom are an approximation, not an exact result. D gives the wrong degrees of freedom entirely ($t_{16}$ instead of $t_{18}$) and invents a nonsensical "label-swapping" remedy for unequal variances. Study tip: Always compute degrees of freedom as $(n_1-1)+(n_2-1)$ for the pooled $t$-test, and remember that "Welch = approximate" is a fundamental distinction examiners love to test.

Start this quiz