Statistics Graduate Level Quiz: Pivotal Quantities And Exact Cis

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Pivotal Quantities And Exact CisQuestion 1 of 10

In a Gaussian linear model with fixed full-rank design matrix, there are $n=20$ observations and $p=4$ regression coefficients, including the intercept. For a contrast $c^{\mathsf T}\beta$ , the fitted contrast is $c^{\mathsf T}\hat\beta=2.4$ , the residual standard deviation is $s=1.5$ , and $c^{\mathsf T}(X^{\mathsf T}X)^{-1}c=0.16$ . Also, $t_{0.975,16}=2.120$ , $t_{0.975,19}=2.093$ , and $z_{0.975}=1.960$ .

Which interval is the exact conditional $95\%$ confidence interval for $c^{\mathsf T}\beta$ under the stated model?

[1.128,\ 3.672]

[1.224,\ 3.576]

[1.144,\ 3.656]

[1.552,\ 3.248]

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Pivotal Quantities And Exact Cis

Practice Pivotal Quantities And Exact Cis in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Pivotal Quantities And Exact Cis, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Which interval is the exact conditional $95\%$ confidence interval for $c^{\mathsf T}\beta$ under the stated model?

$[1.128,\ 3.672]$ (correct answer)
$[1.224,\ 3.576]$
$[1.144,\ 3.656]$
$[1.552,\ 3.248]$

Explanation: 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

Question 2

Let $X_1,\ldots,X_5$ be independent observations from the uniform distribution on $[0,\theta]$ , and suppose the observed sample maximum is $M=8$ . An equal-tailed interval is to be constructed using the pivot $U=M/\theta$ .

Which expression gives the exact equal-tailed $95\%$ confidence interval for $\theta$ ?

$\left[\frac{8}{(0.975)^{1/5}},\ \frac{8}{(0.025)^{1/5}}\right]$ (correct answer)
$\left[8(0.025)^{1/5},\ 8(0.975)^{1/5}\right]$
$\left[\frac{8}{(0.975)^{1/5}},\ \frac{8}{(0.05)^{1/5}}\right]$
$\left[\frac{8}{(0.95)^{1/5}},\ \infty\right)$

Explanation: 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

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.

Question 3

Two independent samples are drawn from normal populations. Under the working model, both populations have the same unknown variance. The sample sizes are $n_1=9$ and $n_2=11$ , and the pooled variance estimator is denoted by $S_p^2$ . Subsequent subject-matter analysis raises concern that the population variances may actually differ.

Which statement correctly describes the pivotal basis and exactness of interval estimation for $\mu_1-\mu_2$ ?

Under equal variances, the pooled statistic follows a standard normal distribution because $S_p^2$ is consistent; if variances differ, estimating two separate variances changes only the numerical width of the resulting interval.
Under equal variances, the pooled statistic has a $t_{20}$ distribution, giving an exact interval; if variances differ, Welch's statistic remains exactly distributed as Student's $t$ with a data-dependent degrees-of-freedom adjustment.
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. (correct answer)
Under equal variances, the pooled statistic has a $t_{16}$ distribution, giving an exact interval; if variances differ, exchanging the sample labels restores the symmetry required for exact pivotality.

Explanation: 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.

Question 4

Let $X_1,…,X_{10}$ be independent observations from an unknown continuous distribution having a unique median $m$ . Denote the ordered observations by $X_{(1)}<\cdots<X_{(10)}$ .

What is the exact coverage probability of the distribution-free interval $[X_{(2)},X_{(9)}]$ for $m$ ?

$\frac{912}{1024}\approx0.8906$
$\frac{1012}{1024}\approx0.9883$
$\frac{990}{1024}\approx0.9668$
$\frac{1002}{1024}\approx0.9785$ (correct answer)

Explanation: 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.

Question 5

In a completely randomized experiment, a fixed number of units is assigned to treatment. Investigators posit the constant additive-effect model $Y_i(1)=Y_i(0)+\tau$ . For every candidate value $\tau_0$ , they subtract $\tau_0$ from treated outcomes, conduct a level- $0.05$ randomization test of the resulting sharp null hypothesis, and retain all candidate values not rejected.

Which statement about the resulting set of retained values is correct?

Even with heterogeneous treatment effects, it is an exact $95\%$ confidence interval for the finite-population average treatment effect, because random assignment ensures that the missing potential outcomes can be validly imputed.
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 $0.05$ . (correct answer)
Its exact coverage requires normally distributed potential outcomes, although the randomized design removes the need for equal outcome variances across treatment and control groups.
Its coverage is defined only over repeated superpopulation sampling, because conditioning on the observed experimental units and their potential outcomes eliminates any remaining source of randomization uncertainty.

Explanation: 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.

Question 6

Independent lifetimes $X_1,\ldots,X_8$ follow an exponential distribution with rate parameter $\lambda$ , so that $f(x)=\lambda e^{-\lambda x}$ for $x>0$ . The observed total lifetime is $\sum_{i=1}^8X_i=12$ . Relevant quantiles are $\chi^2_{0.025,16}=6.908$ and $\chi^2_{0.975,16}=28.845$ .

Which interval is the exact equal-tailed $95\%$ confidence interval for the rate $\lambda$ ?

$[0.288,\ 1.202]$ (correct answer)
$[0.832,\ 3.474]$
$[0.576,\ 2.404]$
$[0.235,\ 1.088]$

Explanation: 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.

Question 7

In $n=20$ independent Bernoulli trials, no successes are observed. A two-sided equal-tailed Clopper–Pearson confidence interval with confidence coefficient $95\%$ is required for the success probability $p$ .

Which interval is the Clopper–Pearson interval, and not merely an approximation or a one-sided interval?

$\left[0,\ 1-(0.025)^{1/21}\right]$
$\left[0,\ 1-(0.05)^{1/20}\right]$
$\left[0,\ 1-(0.025)^{1/20}\right]$ (correct answer)
$\left[0,\ \frac{3}{20}\right]$

Explanation: 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.

Question 8

Two independent normal samples have sizes $n_1=10$ and $n_2=15$ , with sample variances $S_1^2=8$ and $S_2^2=5$ . Let $\rho=\sigma_1^2/\sigma_2^2$ . Relevant quantiles are $F_{0.025;9,14}=0.263$ and $F_{0.975;9,14}=3.209$ .

Which interval is the exact equal-tailed $95\%$ confidence interval for $\rho$ ?

$[0.164,\ 2.004]$
$[0.421,\ 5.134]$
$[0.706,\ 2.467]$
$[0.499,\ 6.084]$ (correct answer)

Explanation: 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.

Question 9

During a total exposure of $E=10$ unit-years, an investigator observes $X=3$ events. Assume $X\sim\operatorname{Poisson}(E\lambda)$ . For the needed chi-square distributions, $\chi^2_{0.025,6}=1.237$ , $\chi^2_{0.975,6}=14.449$ , $\chi^2_{0.025,8}=2.180$ , and $\chi^2_{0.975,8}=17.535$ .

Which is the exact equal-tailed Garwood $95\%$ confidence interval for the event rate $\lambda$ ?

$[0.0619,\ 0.7225]$
$[0.0619,\ 0.8768]$ (correct answer)
$[0.1090,\ 0.8768]$
$[0.1237,\ 1.7535]$

Explanation: 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

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.

Question 10

A random sample of size $n=12$ is drawn from a normal population with unknown mean and variance. The unbiased sample variance is $S^2=9$ . For a chi-square random variable with $11$ degrees of freedom, $\chi^2_{0.025,11}=3.816$ and $\chi^2_{0.975,11}=21.920$ .

Which is the exact equal-tailed $95\%$ confidence interval for the population variance $\sigma^2$ ?

$[4.52,\ 25.94]$ (correct answer)
$[4.93,\ 28.30]$
$[3.12,\ 17.93]$
$[5.03,\ 21.64]$

Explanation: 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.