Statistics Graduate Level Quiz: Monte Carlo Estimation

10 questions · exam conditions

0:00

Monte Carlo EstimationQuestion 1 of 10

A target distribution assigns probabilities $p(0)=0.8$ and $p(1)=0.2$ . The integrand satisfies $h(0)=1$ and $h(1)=9$ . An importance-sampling estimate of $E_p[h(X)]$ is formed by averaging $h(X)p(X)/q(X)$ over independent draws from a proposal $q$ .

Which proposal gives the smallest variance among the choices, and what is that variance per replication?

q=(0.8,0.2)

, with variance

10.24

because sampling from the target avoids weight variability

q=(0.5,0.5)

, with variance

1

because it oversamples the larger integrand value

q=(4/13,9/13)

, with variance

0

because the weighted output is constant

q=(1/13,12/13)

, with variance

0.64

because it places most probability on the rare state

← Back to quizzes

Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Monte Carlo Estimation

Practice Monte Carlo Estimation in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Monte Carlo Estimation, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

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.

All questions

Question 1

Which proposal gives the smallest variance among the choices, and what is that variance per replication?

$q=(0.8,0.2)$ , with variance $10.24$ because sampling from the target avoids weight variability
$q=(0.5,0.5)$ , with variance $1$ because it oversamples the larger integrand value
$q=(4/13,9/13)$ , with variance $0$ because the weighted output is constant (correct answer)
$q=(1/13,12/13)$ , with variance $0.64$ because it places most probability on the rare state

Explanation: Importance sampling works by rewriting

E_p[h(X)] = E_q\left[\frac{h(X)p(X)}{q(X)}\right]

, so the estimator's variance depends entirely on how much the importance weights

w(x) = h(x)p(x)/q(x)

vary across draws. The holy grail is a proposal that makes those weights constant, because a constant random variable has zero variance. The optimal proposal is proportional to

|h(x)|p(x)

. Here,

h(0)p(0) = 1 \times 0.8 = 0.8

and

h(1)p(1) = 9 \times 0.2 = 1.8

, giving a normalizing constant of

0.8 + 1.8 = 2.6 = 13/5

. This yields

q^*(0) = 0.8/2.6 = 4/13

and

q^*(1) = 1.8/2.6 = 9/13

. Under this proposal, every weight equals

h(x)p(x)/q^*(x) = 2.6

regardless of which

x

is drawn — the estimator is constant, so its variance is exactly 0. This confirms C is correct. Choice A uses the target itself as the proposal, so weights are all 1 and the estimator simply averages

h(X)

. The variance is

E_p[h^2] - (E_p[h])^2 = (0.8 \cdot 1 + 0.2 \cdot 81) - (0.8 + 1.8)^2 = 17 - 6.76 = 10.24

. This is valid but far from optimal. Choice B uses uniform weights; you can verify the weighted outputs are

1.6

and

3.6

with mean

2.6

and variance

1

, better than A but not zero. Choice D inverts the sensible direction — piling mass on the rare, high-value state without the precise optimal ratio still leaves weight variability, yielding variance

0.64

. Your takeaway: whenever you see an importance-sampling variance question, immediately check whether any proposal makes

h(x)p(x)/q(x)

constant. That proposal,

q \propto h(x)p(x)

, always achieves zero variance and is the theoretical benchmark you should recognize on sight.

Question 2

Two stochastic systems are compared using common random numbers. Under a shared random input, their outputs have variances $9$ and $4$ and correlation $0.75$ . The target is the difference between their expected outputs, estimated by the average of paired output differences over $n$ independent shared inputs.

What is the variance of this paired estimator, and how does it compare with using independent random inputs for the two systems?

$4/n$ , which is smaller than the independent-input variance by a factor of $13/4$ (correct answer)
$9/n$ , which is smaller than the independent-input variance by a factor of $13/9$
$13/n$ , which is unchanged because common random numbers preserve both marginal variances
$22/n$ , which is larger because positive correlation adds variability to a difference

Explanation: When comparing two stochastic systems, the key question is how correlation between paired outputs affects the variance of their difference. Recall that for any two random variables,

\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y) - 2\text{Cov}(X,Y)

, where

\text{Cov}(X,Y) = \rho\sqrt{\text{Var}(X)\cdot\text{Var}(Y)}

. Here,

\text{Var}(X) = 9

\text{Var}(Y) = 4

, and

\rho = 0.75

. So

\text{Cov}(X,Y) = 0.75\sqrt{9 \cdot 4} = 0.75 \times 6 = 4.5

. The variance of a single paired difference is

9 + 4 - 2(4.5) = 13 - 9 = 4

. Averaging over

n

independent pairs gives variance

4/n

, confirming A is correct. Under independent inputs, the systems are uncorrelated, so the variance would be

(9 + 4)/n = 13/n

. The ratio is

13/4

, meaning the paired estimator is smaller by exactly that factor — the whole point of common random numbers (CRN) as a variance reduction technique. Choice B uses

9/n

, which would follow if you mistakenly dropped

\text{Var}(Y)

from the calculation. Choice C gives

13/n

— this is the independent-input variance, reflecting a failure to apply the covariance term at all; it's the benchmark you're comparing against, not the CRN result. Choice D claims variance

22/n

, which would arise from adding the covariance term instead of subtracting — a sign error that would only apply if you were computing

\text{Var}(X + Y)

. The key study tip: CRN works precisely because positive correlation reduces the variance of a difference. Whenever you see paired simulation experiments with positive correlation, expect the paired variance to be strictly less than the sum of marginals — and always subtract

2\text{Cov}

for differences.

Question 3

A target density is known only up to a normalizing constant: $p(x)=\widetilde p(x)/Z$ . Independent draws are obtained from a proposal density $q$ , and unnormalized weights are $w(x)=\widetilde p(x)/q(x)$ . The self-normalized importance-sampling estimator is $\widehat\mu=\sum_{i=1}^n w_i h(X_i)/\sum_{i=1}^n w_i$ .

Assuming the proposal covers the target support and the needed first moments are finite, which statement is generally correct?

$\widehat\mu$ is exactly unbiased for every $n$ because the unknown constant cancels from the ratio
$\widehat\mu$ is generally biased for finite $n$ but consistent for $E_p[h(X)]$ as $n$ increases (correct answer)
$\widehat\mu$ is inconsistent unless the weights are rescaled to have sample mean exactly $1$
$\widehat\mu$ is unbiased only when $q=p$ , but otherwise converges to an expectation under $q$

Explanation: Whenever you see a question involving ratio estimators in importance sampling, your instinct should be to think carefully about the distinction between exact unbiasedness and asymptotic unbiasedness (consistency). The self-normalized estimator

\widehat\mu = \frac{\sum_i w_i h(X_i)}{\sum_i w_i}

is a ratio of two random quantities. Even though the unknown normalizing constant

Z

cancels algebraically, the ratio of two unbiased estimators is not generally unbiased — this is a classic result in statistics. Formally,

E\left[\frac{A}{B}\right] \neq \frac{E[A]}{E[B]}

unless

A

and

B

are independent or the ratio is deterministic. However, by the Law of Large Numbers, both

\frac{1}{n}\sum_i w_i h(X_i) \xrightarrow{p} E_q[w(X)h(X)]

and

\frac{1}{n}\sum_i w_i \xrightarrow{p} E_q[w(X)] = Z

, so their ratio converges in probability to

E_p[h(X)]

, confirming B is correct. A is wrong because cancellation of

Z

does not eliminate the finite-sample bias introduced by taking the expectation of a ratio. The bias is

O(1/n)

and vanishes only asymptotically. C is wrong on two counts: rescaling weights to have mean 1 is algebraically equivalent to

\widehat\mu

already written, and the estimator is consistent without any such additional requirement. D is wrong because

\widehat\mu

converges to

E_p[h(X)]

regardless of the choice of proposal

q

(provided

q

covers the support of

p

), not to an expectation under

q

. Study tip: On graduate-level exams, always flag ratio estimators — they are virtually never exactly unbiased, but they are often consistent. The phrase "bias-variance tradeoff" and "

O(1/n)

bias" are your signals that self-normalization is the topic at hand.

Question 4

A stationary Markov chain is used to estimate $E_\pi[h(X)]$ . Under the stationary distribution, $\operatorname{Var}_\pi(h)=4$ , and the lag- $k$ autocorrelation of $h(X_t)$ is $\rho_k=0.5^k$ . The usual large-sample Markov chain variance approximation is assumed.

Approximately how many post-burn-in iterations are needed for the Monte Carlo standard error of the sample mean to be at most $0.02$ ?

$10{,}000$ iterations, treating the correlated draws as independent
$20{,}000$ iterations, using only the one-sided sum of autocorrelations
$40{,}000$ iterations, replacing autocorrelations by their absolute doubled sum
$30{,}000$ iterations, using the full integrated autocorrelation time (correct answer)

Explanation: When estimating

E_\pi[h(X)]

with a correlated Markov chain, the variance of the sample mean is not simply

\text{Var}_\pi(h)/n

. Instead, you must account for autocorrelation through the integrated autocorrelation time

\tau

, giving:

\text{Var}(\bar{h}_n) \approx \frac{\text{Var}_\pi(h)}{n} \cdot \tau, \quad \text{where} \quad \tau = 1 + 2\sum_{k=1}^{\infty} \rho_k

With

\rho_k = 0.5^k

, the one-sided sum is

\sum_{k=1}^{\infty} 0.5^k = 1

, so

\tau = 1 + 2(1) = 3

. Setting the Monte Carlo standard error (MCSE)

\leq 0.02

\frac{\tau \cdot \text{Var}_\pi(h)}{n} \leq (0.02)^2 \implies n \geq \frac{3 \times 4}{0.0004} = 30{,}000

This confirms D is correct. A is wrong because it ignores autocorrelation entirely, using

n \geq 4/0.0004 = 10{,}000

. This severely underestimates the required sample size — a classic trap when students forget that correlated draws carry less information than independent ones. B uses only the one-sided sum (

\tau = 1 + 1 = 2

), yielding

n = 20{,}000

. This forgets that the spectral variance formula requires doubling the lagged autocorrelation sum, so it underestimates

\tau

by half. C applies some inflated doubling incorrectly, arriving at

\tau = 4

and

n = 40{,}000

. There's no justification for replacing

\tau = 3

with

4

— this overcorrects without theoretical basis. Study tip: Always memorize

\tau = 1 + 2\sum_{k=1}^{\infty}\rho_k

for geometric autocorrelations — the geometric series simplifies cleanly, making this a common exam calculation. Never treat MCMC draws as independent when autocorrelation is given.

Question 5

For each independent simulation replication, an output $Y$ and a control variable $C$ are recorded. It is known that $E[C]=\mu_C$ , $\operatorname{Var}(Y)=25$ , $\operatorname{Var}(C)=4$ , and $\operatorname{Corr}(Y,C)=0.6$ . Consider the estimator $\overline{Y}-\beta(\overline{C}-\mu_C)$ .

Which coefficient is variance minimizing, and what is the resulting variance per replication before division by the number of replications?

$\beta=0.75$ and variance $19$ , using the correlation divided by the control standard deviation
$\beta=1.50$ and variance $16$ , using covariance divided by the control variance (correct answer)
$\beta=-1.50$ and variance $34$ , because the control term is subtracted in the estimator
$\beta=3.00$ and variance $25$ , using the output standard deviation divided by the control standard deviation

Explanation: Whenever you encounter a control variate problem, your instinct should be to minimize the variance of the adjusted estimator

\overline{Y} - \beta(\overline{C} - \mu_C)

. Since replications are independent, the variance of this estimator (per replication, before dividing by

n

) is:

\operatorname{Var}(Y - \beta C) = \operatorname{Var}(Y) - 2\beta\operatorname{Cov}(Y,C) + \beta^2\operatorname{Var}(C)

To minimize, take the derivative with respect to

\beta

and set it to zero:

-2\operatorname{Cov}(Y,C) + 2\beta\operatorname{Var}(C) = 0 \implies \beta^* = \frac{\operatorname{Cov}(Y,C)}{\operatorname{Var}(C)}

Now plug in the given values. Since

\operatorname{Corr}(Y,C) = 0.6

\operatorname{Var}(Y)=25

\operatorname{Var}(C)=4

, we get

\operatorname{Cov}(Y,C) = 0.6\sqrt{25}\sqrt{4} = 0.6 \times 5 \times 2 = 6

. Therefore

\beta^* = 6/4 = 1.50

. The minimized variance is:

25 - \frac{[\operatorname{Cov}(Y,C)]^2}{\operatorname{Var}(C)} = 25 - \frac{36}{4} = 25 - 9 = 16

This confirms B is correct. Choice A misuses the formula by dividing correlation by the control standard deviation (

0.6/2 = 0.3 \times 5 = 0.75

) — that's not the covariance-to-variance ratio. Choice C incorrectly flips the sign of

\beta

; the subtraction in the estimator is already absorbed into the optimization setup. Choice D uses a ratio of standard deviations (

5/2 = 2.50

... or

\sqrt{25}/\sqrt{4}

), which is related to regression slope only if correlation were 1. Study tip: Always memorize

\beta^* = \operatorname{Cov}(Y,C)/\operatorname{Var}(C)

and the resulting variance reduction formula

\operatorname{Var}(Y)(1 - \rho^2)

— a quick sanity check here gives

25(1-0.36)=16

✓.

Question 6

A population is divided into two strata with weights $W_1=0.8$ and $W_2=0.2$ . The within-stratum standard deviations are $\sigma_1=1$ and $\sigma_2=4$ . Independent simulation observations cost $c_1=1$ unit in stratum $1$ and $c_2=4$ units in stratum $2$ . A stratified mean estimator is used under the budget constraint $n_1+4n_2=600$ .

Ignoring integer-rounding issues in the optimization, which allocation minimizes the estimator's variance?

$n_1=200$ and $n_2=100$ , balancing weighted variability against square-root cost (correct answer)
$n_1=120$ and $n_2=120$ , applying ordinary Neyman allocation without cost adjustment
$n_1=300$ and $n_2=75$ , allocating observations in proportion to the stratum weights
$n_1=400$ and $n_2=50$ , allocating according to weighted stratum variances

Explanation: When you see a stratified sampling question with both unequal variances and unequal costs, your first instinct should be optimal (cost-adjusted Neyman) allocation, not plain Neyman allocation. The optimal rule says allocate sample sizes proportional to

W_h \sigma_h / \sqrt{c_h}

, then scale to meet the budget. Compute the allocation weights:

Stratum 1: $W_1\sigma_1/\sqrt{c_1} = 0.8 \times 1 / \sqrt{1} = 0.8$
Stratum 2: $W_2\sigma_2/\sqrt{c_2} = 0.2 \times 4 / \sqrt{4} = 0.4$

The ratio is

n_1 : n_2 = 0.8 : 0.4 = 2:1

. Now apply the budget constraint

n_1 + 4n_2 = 600

with

n_1 = 2n_2

: substituting gives

2n_2 + 4n_2 = 6n_2 = 600

, so

n_2 = 100

and

n_1 = 200

. This confirms answer A is correct. Answer B (

n_1=120, n_2=120

) uses plain Neyman allocation — proportional to

W_h\sigma_h

— but ignores the cost differential entirely. This is a classic trap: Neyman allocation only minimizes variance under equal costs. Answer C (

n_1=300, n_2=75

) allocates purely by stratum weight (0.8 and 0.2), ignoring both within-stratum variance and cost. This is proportional allocation, which is even cruder than Neyman. Answer D (

n_1=400, n_2=50

) doesn't correspond to any standard optimality criterion — it overloads the cheaper stratum without a principled justification. Study tip: Always check whether costs are equal before applying Neyman allocation. If

c_h

differs across strata, you must include the

1/\sqrt{c_h}

factor — forgetting it is the single most common error on optimal allocation problems.

Question 7

To estimate $E[e^U]$ for $U\sim\operatorname{Uniform}(0,1)$ , an analyst generates one uniform draw $U$ and reports the antithetic-pair average $\{e^U+e^{1-U}\}/2$ . Another analyst uses the average of $e^{U_1}$ and $e^{U_2}$ for two independent uniform draws.

Which statement correctly compares the two estimators?

The antithetic estimator is biased because $U$ and $1-U$ are deterministically dependent
The estimators have equal variance because both average exactly two uniform marginal draws
The antithetic estimator has smaller variance because the two transformed outputs have negative covariance (correct answer)
The antithetic estimator has zero variance because the two uniform inputs sum to $1$

Explanation: Variance reduction via antithetic variates is a core Monte Carlo technique worth mastering deeply. When you see a question comparing two estimators that both "use two uniform draws," immediately ask: are those draws independent or negatively correlated? That distinction drives everything. The antithetic estimator is

\hat{\mu}_A = \frac{e^U + e^{1-U}}{2}

. Since

1-U \sim \text{Uniform}(0,1)

, both terms have the correct marginal distribution, so

E[\hat{\mu}_A] = E[e^U] = e - 1

—the estimator is unbiased. Its variance is

\frac{1}{4}\left(\text{Var}(e^U) + \text{Var}(e^{1-U}) + 2\text{Cov}(e^U, e^{1-U})\right)

. The key insight: because

e^x

is convex and monotone, when

U

is large,

1-U

is small, making

e^U

and

e^{1-U}

move in opposite directions. This produces a negative covariance, shrinking the variance below that of the independent-pair estimator

\hat{\mu}_I = \frac{e^{U_1}+e^{U_2}}{2}

, whose covariance term is zero. Answer C correctly identifies this mechanism. A is wrong because dependence does not imply bias—both

U

and

1-U

are marginally uniform, preserving unbiasedness. B is wrong because equal marginal distributions do not guarantee equal variances; the covariance structure differs critically between the two estimators. D is wrong because the inputs summing to 1 does not make the outputs constant—

e^U + e^{1-U}

still varies with

U

(it equals

2\cosh(U - 0.5)

), so variance is positive, just reduced. Study tip: For antithetic variates, always verify two things—same marginals (ensures unbiasedness) and negative covariance (ensures variance reduction). Monotone transformations of

U

and

1-U

always achieve the latter.

Question 8

Let $X$ and $Y$ be independent exponential random variables with rate $1$ . A crude Monte Carlo estimator of $P(X+Y>2)$ averages indicators of the event. A conditional Monte Carlo estimator instead conditions on the simulated value of $X$ .

Which quantity should be averaged over independent draws of $X$ to obtain the conditional estimator, and what variance guarantee does it satisfy?

$e^{-(2-X)}\,I(X<2)$ , with variance strictly smaller than the crude estimator for every integrand
$I(X\geq 2)+e^{-(2-X)}\,I(X<2)$ , with variance no larger than the crude estimator by the law of total variance (correct answer)
$I(X\geq 2)+e^{-(2+X)}\,I(X<2)$ , with variance equal to the crude estimator because iterated expectation preserves the mean
$e^{-2}\,I(X<2)+I(X\geq 2)$ , with variance no larger because it replaces the random variable $Y$ by its unconditional mean

Explanation: Conditional Monte Carlo is a variance reduction technique based on the law of total variance:

\text{Var}(Z) = \text{Var}(E[Z|X]) + E[\text{Var}(Z|X)]

, which guarantees

\text{Var}(E[Z|X]) \leq \text{Var}(Z)

always. The strategy is to replace a noisy indicator with its conditional expectation given observed information. For

Z = I(X+Y>2)

with

Y \sim \text{Exp}(1)

independent of

X

, compute

E[Z \mid X]= P(Y > 2-X \mid X)

. When

X \geq 2

, the event is certain:

P(Y > 2-X) = 1

. When

X < 2

, you need

P(Y > 2-X) = e^{-(2-X)}

by the exponential survival function. This gives the conditional estimator

I(X \geq 2) + e^{-(2-X)}I(X < 2)

, averaged over draws of

X

— exactly answer B. The law of total variance then guarantees variance no larger than the crude estimator. Choice A omits the

I(X \geq 2)

term entirely, so it's a biased estimator — it assigns zero when

X \geq 2

even though the event is guaranteed. Its variance claim of "strictly smaller for every integrand" is also too strong; the law of total variance only guarantees non-strict inequality. Choice C uses

e^{-(2+X)}

instead of

e^{-(2-X)}

, which is simply the wrong survival probability — a sign error inside the exponent. The variance claim about iterated expectation is also a non sequitur. Choice D replaces

e^{-(2-X)}

with the constant

e^{-2}

, as if substituting

X=0

, which is neither the conditional expectation nor an unbiased quantity. Your study tip: whenever you condition on a simulated variable, always write out

E[I(\text{event}) \mid \text{observed}]

explicitly — identify each case (certain, impossible, or in-between) before writing the formula.

Question 9

The rare-event probability $P(Z>5)$ is to be estimated for $Z\sim N(0,1)$ . To generate more exceedances, independent observations $X_i$ are sampled from the proposal distribution $N(5,1)$ .

Which sample average is an unbiased importance-sampling estimator of the rare-event probability?

$n^{-1}\sum_{i=1}^n I(X_i>5)\exp(-5X_i+12.5)$ (correct answer)
$n^{-1}\sum_{i=1}^n I(X_i>5)\exp(5X_i-12.5)$
$n^{-1}\sum_{i=1}^n I(X_i>0)\exp(-5X_i+12.5)$
$n^{-1}\sum_{i=1}^n I(X_i>5)\exp(-5X_i-12.5)$

Explanation: Whenever you see an importance sampling problem, your first instinct should be to write out the likelihood ratio (the Radon-Nikodym derivative) between the target and proposal distributions. This ratio is the engine of the estimator. You want

P(Z > 5) = E_f[I(Z > 5)]

where

f

is the standard normal density. Importance sampling rewrites this as

E_g\!\left[I(X > 5)\frac{f(X)}{g(X)}\right]

, where

g

is the proposal density. Here

f = N(0,1)

and

g = N(5,1)

, so the likelihood ratio is:

\frac{f(x)}{g(x)} = \frac{\exp(-x^2/2)}{\exp(-(x-5)^2/2)} = \exp\!\left(-\frac{x^2}{2} + \frac{(x-5)^2}{2}\right) = \exp(-5x + 12.5)

The unbiased estimator is therefore

n^{-1}\sum_{i=1}^n I(X_i > 5)\exp(-5X_i + 12.5)

, which is answer A. B flips the sign of the exponent, giving

\exp(5X_i - 12.5)

. This would correspond to the ratio

g/f

rather than

f/g

— it's the reciprocal of the correct weight and produces a heavily biased, explosive estimator. C uses the wrong indicator

I(X_i > 0)

instead of

I(X_i > 5)

, so it estimates the wrong probability entirely —

P(Z > 0) = 0.5

, not the rare-event probability. D uses

\exp(-5X_i - 12.5)

, with the wrong sign on the constant 12.5; this comes from misexpanding

(x-5)^2

and loses the cross-term correction. Study tip: Always expand the exponent in

f(x)/g(x)

term-by-term and double-check the sign of every piece — sign errors in the weight are the most common trap in importance sampling questions.

Question 10

A pilot simulation based on $400$ independent replications gives a sample variance of $9$ for the output whose expectation is being estimated. The investigator treats this variance estimate as the planning value for a larger independent Monte Carlo run.

Using a normal approximation, what is the smallest total number of replications required for a two-sided $95\%$ Monte Carlo confidence interval to have half-width at most $0.10$ ?

$865$ replications, because the standard error itself must be at most $0.10$
$1{,}537$ replications, because the current interval width scales inversely with sample size
$3{,}458$ replications, because the half-width scales inversely with the square root of sample size (correct answer)
$13{,}830$ replications, because both endpoints contribute separately to the required precision

Explanation: Whenever you see a Monte Carlo confidence interval question, anchor yourself to the standard formula: the half-width of a two-sided 95% interval is

z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}

, where

z_{0.025} = 1.96

\sigma^2

is the planning variance, and

n

is the number of replications. Your goal is to solve for the smallest

n

such that this half-width is at most the target. Here, the pilot study gives

\hat{\sigma}^2 = 9

, so

\hat{\sigma} = 3

. Setting the half-width condition:

1.96 \cdot \frac{3}{\sqrt{n}} \leq 0.10

Solving:

\sqrt{n} \geq \frac{1.96 \times 3}{0.10} = 58.8

, so

n \geq 58.8^2 = 3{,}457.44

. Rounding up gives

n = 3{,}458

, confirming C is correct. The half-width shrinks at rate

1/\sqrt{n}

, not

1/n

, which is the critical insight. Choice A confuses the half-width with the standard error itself — it skips multiplying by

z_{0.025} = 1.96

and sets

\sigma/\sqrt{n} \leq 0.10

, yielding only

n \geq 900

. That ignores the confidence coefficient entirely. Choice B applies an incorrect

1/n

scaling relationship instead of

1/\sqrt{n}

, producing a nonsensical intermediate formula. Choice D doubles the correct answer as if both endpoints each independently require the full half-width precision, which is a misunderstanding — the half-width already accounts for both directions symmetrically. Your study tip: always write the half-width formula explicitly before solving. The most common trap on precision-planning problems is forgetting either the

z

-multiplier (leading to A) or using

n

instead of

\sqrt{n}

as the denominator (leading to B or D).