Statistics Graduate Level Quiz: Neyman Pearson Lemma

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Neyman Pearson LemmaQuestion 1 of 10

A single observation has distribution $N(0,1)$ under $H_0$ and distribution $N(1,4)$ under $H_1$ . For a fixed significance level $0<\alpha<1$ , which form must a Neyman–Pearson most powerful test have, up to possible randomization on its boundary?

Reject when

\lvert X+1/3\rvert>d_\alpha

for a constant

d_\alpha

chosen to attain size

\alpha

Reject when

X>d_\alpha

because the mean under

H_1

is larger than the mean under

H_0

Reject when

\lvert X-1\rvert>d_\alpha

because deviations from the alternative mean favor the larger variance.

Reject when

X<d_\alpha

because sufficiently negative observations are more probable under the larger-variance alternative.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Neyman Pearson Lemma

Practice Neyman Pearson Lemma 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 Neyman Pearson Lemma, 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

Reject when $\lvert X+1/3\rvert>d_\alpha$ for a constant $d_\alpha$ chosen to attain size $\alpha$ . (correct answer)
Reject when $X>d_\alpha$ because the mean under $H_1$ is larger than the mean under $H_0$ .
Reject when $\lvert X-1\rvert>d_\alpha$ because deviations from the alternative mean favor the larger variance.
Reject when $X<d_\alpha$ because sufficiently negative observations are more probable under the larger-variance alternative.

Explanation: Whenever you see a Neyman–Pearson most powerful test question, your first move should always be to write out the likelihood ratio and simplify — don't let intuition about means or variances substitute for algebra. Here,

f_0(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}

and

f_1(x) = \frac{1}{\sqrt{2\pi}\cdot 2}e^{-(x-1)^2/8}

. The Neyman–Pearson lemma says to reject when the likelihood ratio

\Lambda(x) = f_1(x)/f_0(x)

exceeds a threshold. Taking the log and keeping only terms involving

x

\log \Lambda(x) \propto -\frac{(x-1)^2}{8} + \frac{x^2}{2} = \frac{3x^2 + 2x - 1}{8} = \frac{(3x-1)(x+1)}{8}.

Rejecting when

\log \Lambda(x) > k

means rejecting when

3x^2 + 2x - 1 > c

, i.e.,

3(x + 1/3)^2 > c'

, which simplifies to

|x + 1/3| > d_\alpha

. This is exactly answer A — a two-sided rejection region centered at

-1/3

, not at either distribution's mean. Answer B is wrong because simply comparing means ignores the variance difference; the larger variance under

H_1

creates probability mass in both tails. Answer D captures the intuition that large negative values favor the alternative, but it's only a one-sided rule — it misses the upper tail, where the large variance also inflates

f_1

. Answer C uses the wrong center (1 instead of

-1/3

), reflecting a misunderstanding of where the likelihood ratio tips in favor of

H_1

. The key study tip: always derive the rejection region from the likelihood ratio algebra. Intuition about means or variances alone will lead you astray when the two hypotheses differ in both parameters simultaneously.

Question 2

Let $X_1,\ldots,X_n$ be independent $N(\mu,\sigma^2)$ observations, where $\sigma^2$ is unknown. A researcher claims that the Neyman–Pearson lemma directly proves that the usual one-sided Student test is most powerful for testing $H_0:\mu=0$ against $H_1:\mu=\delta$ , where $\delta>0$ is fixed. Which assessment is most accurate?

The claim is correct because fixing $0$ and $\delta$ makes both hypotheses simple despite the unknown variance.
The claim is incorrect only when $n$ is small; asymptotic normality makes the hypotheses simple for sufficiently large $n$ .
The claim is correct because replacing $\sigma$ by the sample standard deviation leaves the likelihood ratio unchanged.
The claim is incorrect because the unknown variance makes each hypothesis composite, so additional invariance or conditioning arguments are required. (correct answer)

Explanation: Whenever you see the Neyman–Pearson (NP) lemma invoked, your first instinct should be to check whether both hypotheses are truly simple — that is, whether each hypothesis specifies a single, fully determined probability distribution. The NP lemma guarantees a most powerful test only in that setting. Here, even though

\mu

is fixed under both

H_0

and

H_1

, the variance

\sigma^2

is unknown and unspecified. This means each hypothesis actually represents an entire family of distributions indexed by

\sigma^2 \in (0,\infty)

, making both hypotheses composite. The NP lemma simply does not apply directly. To establish that the one-sided

t

-test is most powerful, you need additional arguments — specifically, you appeal to invariance (the problem is invariant under scale transformations) and a conditional argument (conditioning on a sufficient statistic for

\sigma^2

, namely

S^2

), which together show the

t

-test is uniformly most powerful among scale-invariant tests. D is therefore correct: the unknown variance creates composite hypotheses, and extra structure beyond the NP lemma is needed. A is wrong because fixing

\mu

under both hypotheses does not make them simple —

\sigma^2

remains free, so infinitely many distributions are compatible with each hypothesis. B is wrong because asymptotic normality approximates the distribution of

\bar{X}

, but

\sigma^2

remains unknown even asymptotically; this does not convert composite hypotheses into simple ones. C is wrong because substituting

S

for

\sigma

fundamentally changes the likelihood ratio statistic — it is no longer a pure likelihood ratio, and the NP lemma's conclusion does not transfer automatically. A good study rule: before citing the NP lemma, always verify that both hypotheses are fully specified (simple). Any free nuisance parameter immediately makes the lemma inapplicable.

Question 3

For simple hypotheses with densities $f_0$ and $f_1$ , let $L=f_1/f_0$ . A level- $\alpha$ Neyman–Pearson test $\phi^*$ rejects when $L>k$ . Assume $P_0(L=k)=0$ . Another test $\phi$ has exact size $\alpha$ and differs from $\phi^*$ on a set having positive $P_0$ probability. What follows?

The two tests have equal power because they have the same exact type I error probability.
The test $\phi$ has strictly smaller power because no positive-probability likelihood-ratio boundary can account for the difference. (correct answer)
The test $\phi$ can have greater power if it rejects more frequently where $f_0$ is relatively large.
Only weak power ordering follows unless $P_1(L=k)=0$ is separately assumed under the alternative.

Explanation: Whenever you see a question invoking the Neyman–Pearson Lemma, anchor your thinking to its optimality guarantee: among all level-

\alpha

tests, the likelihood-ratio test

\phi^*

uniquely maximizes power under

P_1

. The key result is the Neyman–Pearson inequality. For any competing test

\phi

with

E_0[\phi] \leq \alpha = E_0[\phi^*]

, we have:

E_1[\phi^*] - E_1[\phi] \geq k\bigl(E_0[\phi^*] - E_0[\phi]\bigr)

This follows because

(\phi^* - \phi)(f_1 - kf_0) \geq 0

pointwise —

\phi^*

rejects exactly when

f_1 > kf_0

. Since both tests have equal size (

E_0[\phi^*] = E_0[\phi] = \alpha

), the right side is zero, giving

E_1[\phi^*] \geq E_1[\phi]

. Now, since

\phi

differs from

\phi^*

on a positive-

P_0

set and

P_0(L=k)=0

, that difference must occur where

L \neq k

, meaning the pointwise inequality

(\phi^* - \phi)(f_1 - kf_0) \geq 0

is strict on a positive-measure set. With

k > 0

, this forces

E_1[\phi^*] > E_1[\phi]

strictly, confirming B. A is wrong because equal size does not imply equal power — that confuses a necessary condition with a sufficient one. C is the critical trap: if

\phi

rejects more where

f_0

is large (i.e., where

L

is small), it wastes its size budget in a region that contributes little under

P_1

, strictly reducing power. D is a red herring; the assumption

P_0(L=k)=0

already ensures no boundary mass matters under

P_0

, and the strict inequality follows without any separate condition on $$P_1$. Study tip: Always remember that the NP Lemma gives a strict power ranking when the competing test genuinely differs from the LRT — "same size" never rescues a suboptimal rejection region.

Question 4

Independent Bernoulli trials are observed until the first success or until three trials have been observed. Thus, the possible stopped paths are $1,01,001,000$ . For testing $H_0:p=0.2$ against $H_1:p=0.5$ , which ordering lists the paths from largest to smallest likelihood ratio?

$1,01,000,001$ , because three failures are stronger evidence than a delayed success.
$001,01,1,000$ , because longer paths contain more information in favor of the alternative.
$1,01,001,000$ , because each additional observed failure decreases the likelihood ratio. (correct answer)
$000,001,01,1$ , because the alternative assigns greater probability to every stopped path.

Explanation: When comparing two simple hypotheses, the likelihood ratio (LR) is your primary tool for ordering evidence. For each stopped path, compute

\Lambda = \frac{L(p=0.5)}{L(p=0.2)}

and rank accordingly. Each path has probability

p^s(1-p)^f

, where

s \in \{0,1\}

is the number of successes and

f

is the number of failures observed. So the likelihood ratio for a path with one success and

f

failures is

\frac{(0.5)(0.5)^f}{(0.2)(0.8)^f} = \frac{0.5}{0.2}\left(\frac{0.5}{0.8}\right)^f = 2.5 \cdot (0.625)^f

. For the all-failure path

000

\frac{(0.5)^3}{(0.8)^3} = \frac{0.125}{0.512} \approx 0.244

. Now compute each:

Path $1$ ( $f=0$ ): $2.5 \cdot 1 = 2.5$
Path $01$ ( $f=1$ ): $2.5 \cdot 0.625 = 1.5625$
Path $001$ ( $f=2$ ): $2.5 \cdot 0.391 \approx 0.977$
Path $000$ (no success): $\approx 0.244$

This gives the decreasing order

1, 01, 001, 000

, confirming C. Each additional failure before a success multiplies the LR by

0.625 < 1

, steadily reducing it, and the pure-failure path has the smallest LR of all. Choice A incorrectly claims

000

beats

001

, but three consecutive failures actually favor

H_0

(LR < 1). Choice B inverts the ordering entirely with faulty intuition about "information." Choice D incorrectly asserts the alternative dominates on every path —

000

and

001

actually favor

H_0

. Study tip: Always compute LRs explicitly rather than relying on intuition about "more data = more evidence for

H_1

." The direction of evidence depends critically on what was observed, not just how much.

Question 5

A single observation is used to test $H_0:X\sim\operatorname{Uniform}(0,1)$ against $H_1:X\sim\operatorname{Uniform}(0,2)$ . Which description gives a Neyman–Pearson test of exact level $\alpha$ and correctly states its power?

Reject on any interval of length $\alpha$ inside $[0,1]$ and never reject when $X>1$ , because the alternative also has positive density there; the power equals $\alpha/2$ .
Reject only when $X>1$ , since this region already satisfies the level constraint; the power equals $1/2$ for all values of $\alpha$ .
Reject whenever $X>1$ and also reject on a subset of $[0,1]$ of null probability $\alpha$ ; the power equals $(1+\alpha)/2$ . (correct answer)
Reject whenever $X>1-\alpha$ , using the upper tail of the null distribution to set the critical value; the power equals $\alpha/2$ .

Explanation: When applying the Neyman–Pearson Lemma, your goal is to find the most powerful test at a fixed level

\alpha

by comparing likelihood ratios. Here, the likelihood ratio is

f_1(x)/f_0(x)

, where

f_0(x) = 1

[0,1]

and

f_1(x) = 1/2

[0,2]

. This ratio equals

1/2

for

x \in [0,1]

and is undefined (infinity, practically) for

x \in (1,2]

since

f_0 = 0

there. The NP test rejects where the likelihood ratio is largest — so you must always reject when

X > 1

, because

f_0 = 0

there and the ratio is effectively infinite. That region alone has null probability zero, so it contributes nothing to your level. To reach exact level

\alpha

, you supplement by also rejecting on some subset of

[0,1]

with null probability exactly

\alpha

. The total power is then

P_1(X > 1) + P_1(\text{subset of } [0,1]) = \frac{1}{2} + \frac{\alpha}{2} = \frac{1+\alpha}{2}

, confirming C is correct. Choice A is wrong because it ignores the region

X > 1

, which has infinite likelihood ratio and must be included in any most-powerful test. Choice B gets the rejection region right for the

X > 1

part but fails to add the supplementary region needed for exact level

\alpha

, and incorrectly claims the power is always

1/2

regardless of

\alpha

. Choice D uses only a tail interval of the null distribution without accounting for the non-overlap region

(1,2]

, yielding power

\alpha/2

— far from optimal. A key study tip: always check where

f_0 = 0

but

f_1 > 0

. Those points have infinite likelihood ratio and belong in every NP rejection region before you even touch the level constraint.

Question 6

Suppose $X_1,\ldots,X_n$ are independent $\operatorname{Poisson}(\lambda)$ variables. A test rejects $H_0:\lambda=\lambda_0$ for sufficiently large values of $T=\sum_{i=1}^n X_i$ , with boundary randomization if needed to obtain level $\alpha$ . Which statement best explains why this test is uniformly most powerful against $H_1:\lambda>\lambda_0$ ?

The Neyman–Pearson lemma directly applies because the alternative $\lambda>\lambda_0$ is itself a simple hypothesis.
For every fixed $\lambda_1>\lambda_0$ , the likelihood ratio is increasing in the same statistic $T$ , so the same level- $\alpha$ test is most powerful. (correct answer)
The statistic $T$ is sufficient for $\lambda$ , and sufficiency alone implies that every test based on $T$ is uniformly most powerful.
The Poisson family has increasing variance, so large values of $T$ must yield a uniformly most powerful test without using likelihood ratios.

Explanation: When you see a question about uniformly most powerful (UMP) tests with a one-sided composite alternative, your first instinct should be the Karlin-Rubin theorem, which extends the Neyman–Pearson lemma beyond simple hypotheses. Here's the core logic: For Poisson

(\lambda)

, the likelihood ratio for any simple alternative

\lambda_1 > \lambda_0

\frac{L(\lambda_1)}{L(\lambda_0)} = e^{-n(\lambda_1-\lambda_0)}\left(\frac{\lambda_1}{\lambda_0}\right)^T,

which is strictly increasing in

T = \sum X_i

. This means the most powerful level-

\alpha

test against any fixed

\lambda_1 > \lambda_0

always rejects for large

T

, with the same critical region regardless of which

\lambda_1

you choose. Because a single test simultaneously achieves maximum power against every point in

H_1

, it is uniformly most powerful. That's exactly what B captures. A is wrong because

H_1: \lambda > \lambda_0

is a composite (not simple) hypothesis — the Neyman–Pearson lemma alone only handles simple-vs-simple comparisons and cannot directly deliver UMP results. C is wrong because sufficiency is necessary but not sufficient for UMP status. Many sufficient statistics exist for families that admit no UMP test (e.g., two-sided alternatives). Sufficiency alone proves nothing about optimality. D is wrong and frankly a red herring — increasing variance is a property of the Poisson family but has no logical connection to the UMP criterion. Power optimality comes from likelihood ratio structure, not moment properties. Study tip: Whenever a question asks why a one-sided test is UMP, look for the monotone likelihood ratio (MLR) property — it's the engine behind Karlin-Rubin and the correct framing for these problems.

Question 7

Let $X_1,\ldots,X_5$ be independent normal variables with known mean $0$ . Consider testing $H_0:\sigma^2=1$ against $H_1:\sigma^2=4$ . If $q_{5,1-\alpha}$ is the upper $\alpha$ critical value of a chi-square distribution with $5$ degrees of freedom, which test and power are prescribed by the Neyman–Pearson lemma?

Reject when $\sum X_i^2>q_{5,1-\alpha}$ ; the power is $1-F_{\chi^2_5}(q_{5,1-\alpha}/4)$ . (correct answer)
Reject when $\sum X_i^2<q_{5,\alpha}$ ; the power is $F_{\chi^2_5}(4q_{5,\alpha})$ .
Reject when $\left\lvert\sum X_i\right\rvert>q_{5,1-\alpha}$ ; the power is $1-F_{\chi^2_5}(q_{5,1-\alpha}/4)$ .
Reject when $\sum X_i^2>4q_{5,1-\alpha}$ ; the power is $1-F_{\chi^2_5}(q_{5,1-\alpha})$ .

Explanation: When you see a question about testing variance with normal data, your first instinct should be the Neyman–Pearson lemma: form the likelihood ratio and find what statistic it reduces to. Here, with

X_i \sim N(0, \sigma^2)

, the likelihood ratio is:

\frac{L(\sigma^2=4)}{L(\sigma^2=1)} = \left(\frac{1}{4}\right)^{5/2} \exp\!\left(-\frac{\sum X_i^2}{2}\cdot\left(\frac{1}{4}-1\right)\right) = C \cdot \exp\!\left(\frac{3\sum X_i^2}{8}\right)

This is increasing in

\sum X_i^2

, so the NP lemma says: reject when $\sum X_i^2$ is large. Under

H_0

\sum X_i^2 \sim \chi^2_5

, so the size-

\alpha

cutoff is

q_{5,1-\alpha}

. Under

H_1

\sum X_i^2/4 \sim \chi^2_5

, so power is

P(\sum X_i^2 > q_{5,1-\alpha} \mid \sigma^2=4) = 1 - F_{\chi^2_5}(q_{5,1-\alpha}/4)

. This is exactly answer A. B rejects for small values of

\sum X_i^2

, which would be appropriate if

H_1

specified a smaller variance — the inequality is backwards for

\sigma^2=4>1

. C uses

|\sum X_i|

instead of

\sum X_i^2

. Since the mean is known to be zero, the sufficient statistic for

\sigma^2

\sum X_i^2

, not the sample sum; this reflects a confusion between testing the mean and testing the variance. D scales the rejection threshold by 4 (using

4q_{5,1-\alpha}

), which would correspond to computing the critical value under

H_1

rather than

H_0

— that's not how size is controlled. Strategy tip: Always derive the NP rejection region from the likelihood ratio first, then separately compute the critical value under

H_0

(for size) and evaluate the probability under

H_1

(for power) — keep those two steps distinct.

Question 8

A sample space consists of outcomes $u,v,w$ . Their probabilities under $H_0$ are $0.5,0.3,0.2$ , and their likelihood ratios $f_1/f_0$ are $0.4,1,2.5$ , respectively. A proposed level- $0.30$ test rejects only when $v$ occurs. Which modification gives a most powerful level- $0.30$ test?

Reject with probability $3/5$ whenever $u$ occurs and never reject at $v$ or $w$ .
Reject whenever $w$ occurs and reject with probability $1/5$ whenever $u$ occurs.
Reject whenever $v$ occurs because it alone uses the full available type I error probability.
Reject whenever $w$ occurs and reject with probability $1/3$ whenever $v$ occurs. (correct answer)

Explanation: When you see a question invoking the Neyman-Pearson lemma, your instinct should be: rank outcomes by their likelihood ratio

\Lambda = f_1/f_0

, then reject starting from the highest ratio downward until you've exhausted the level

\alpha

. Here the likelihood ratios are:

w: 2.5

v: 1.0

u: 0.4

. The most powerful level-0.30 test must prioritize rejecting

w

first (highest

\Lambda

), then move to

v

if budget remains, and use

u

last (lowest

\Lambda

). Rejecting

w

always uses

P_0(w) = 0.2

of your size budget, leaving

0.10

remaining. Since

P_0(v) = 0.3 > 0.10

, you can't fully reject at

v

, so you randomize: reject

v

with probability

\gamma

where

0.3\gamma = 0.10

, giving

\gamma = 1/3

. This is exactly answer D — reject whenever

w

occurs and reject

v

with probability

1/3

. Answer A fails because it rejects at

u

, which has the lowest likelihood ratio (0.4 < 1). Using your type I error budget on the least powerful outcome is the opposite of what the lemma prescribes. Answer B also rejects

u

(with probability

1/5

) alongside

w

, wasting power on the weakest outcome instead of

v

. Answer C is the original flawed test — rejecting only

v

ignores

w

, which has a strictly higher likelihood ratio and should be rejected first. Study tip: Always sort outcomes by

\Lambda

descending before building your rejection region. Fill your

\alpha

budget greedily from the top — randomize only at the threshold outcome where the budget runs out.

Question 9

An observation takes values $a,b,c$ . Under $H_0$ , their probabilities are $0.2,0.3,0.5$ , respectively; under $H_1$ , their probabilities are $0.4,0.6,0$ . Let $\phi_j$ denote the probability of rejection when outcome $j$ occurs. Which statement correctly characterizes the level- $0.25$ most powerful tests?

The unique test has $\phi_a=1$ , $\phi_b=1/6$ , and $\phi_c=0$ , with power $0.50$ .
Any test satisfying $0.2\phi_a+0.3\phi_b=0.25$ and $\phi_c=0$ is most powerful, with power $0.50$ . (correct answer)
Any test satisfying $0.4\phi_a+0.6\phi_b=0.25$ and $\phi_c=0$ is most powerful, with power $0.25$ .
The unique test has $\phi_a=0$ , $\phi_b=0$ , and $\phi_c=1/2$ , with power $0$ .

Explanation: When you see a hypothesis testing problem like this, your first instinct should be the Neyman-Pearson lemma: reject when the likelihood ratio

\Lambda(x) = p_1(x)/p_0(x)

is large, and never reject where

\Lambda = 0

. Compute the likelihood ratios:

\Lambda(a) = 0.4/0.2 = 2

\Lambda(b) = 0.6/0.3 = 2

\Lambda(c) = 0/0.5 = 0

. Since

\Lambda(c) = 0

, outcome

c

should never cause rejection — so

\phi_c = 0

in any most powerful test. Outcomes

a

and

b

have equal likelihood ratios, which means the Neyman-Pearson critical region is not unique at this boundary. Any randomized test that sets

\phi_c = 0

and satisfies the level constraint

0.2\phi_a + 0.3\phi_b = 0.25

(keeping

\phi_a, \phi_b \in [0,1]

) is most powerful. The power of any such test is

0.4\phi_a + 0.6\phi_b

. Substituting

\phi_b = (0.25 - 0.2\phi_a)/0.3

and simplifying confirms the power is always

0.50

, regardless of which valid

(\phi_a, \phi_b)

pair you choose. This confirms B is correct. A is wrong because it identifies one valid solution (

\phi_a=1, \phi_b=1/6

) but falsely claims uniqueness — the equal likelihood ratios mean infinitely many tests achieve the same power. C confuses the power equation with the size constraint and states an incorrect power of

0.25

. D rejects only on

c

, where the likelihood ratio is zero — exactly backwards from what NP dictates. The key study tip: when two outcomes share the same likelihood ratio at the boundary, the MP test is non-unique — any mixture satisfying the size constraint is equally optimal.

Question 10

Let $X\sim\operatorname{Binomial}(2,p)$ . Consider testing $H_0:p=1/2$ against $H_1:p=3/4$ at exact level $\alpha=0.20$ . What is the Neyman–Pearson most powerful test, and what is its power?

Reject with probability $0.80$ when $X=2$ ; the resulting power is $0.45$ . (correct answer)
Reject whenever $X=2$ ; the resulting size is $0.25$ and the power is $0.5625$ .
Reject with probability $0.20$ when $X=2$ ; the resulting power is $0.1125$ .
Reject when $X=2$ and randomize when $X=1$ ; the resulting power is $0.60$ .

Explanation: When you encounter a Neyman–Pearson (NP) testing problem, your first move should always be to compute the likelihood ratio and identify the critical region — then check whether a randomized test is needed to hit the exact level. Here,

X \sim \text{Binomial}(2, p)

, so the possible values are

X \in \{0, 1, 2\}

. Under

H_0: p=1/2

, the probabilities are

P(X=0)=0.25

P(X=1)=0.50

P(X=2)=0.25

. Under

H_1: p=3/4

P(X=0)=0.0625

P(X=1)=0.375

P(X=2)=0.5625

. The likelihood ratio is largest at

X=2

, so the NP test rejects for large

X

. Since

P_0(X=2)=0.25 > 0.20

, rejecting only when

X=2

would give size 0.25, exceeding our budget. We cannot simply include

X=2

in the rejection region. Instead, we randomize: reject with probability

\gamma

when

X=2

such that

\gamma \cdot 0.25 = 0.20

, giving

\gamma = 0.80

. The power at

p=3/4

0.80 \times P_1(X=2) = 0.80 \times 0.5625 = 0.45

. That confirms A is correct. Choice B incorrectly rejects deterministically at

X=2

, producing size 0.25, which violates the

\alpha=0.20

constraint. Choice C randomizes with

\gamma=0.20

, which satisfies

\alpha

but is far too conservative — you've wasted power by under-rejecting. Choice D introduces

X=1

into the rejection region without justification; the likelihood ratio at

X=1

is smaller than at

X=2

, so NP theory says you'd never include

X=1

before fully exhausting

X=2

. Study tip: Whenever the exact size constraint can't be met by a deterministic test, always randomize only on the boundary point — compute

\gamma = (\alpha - P_0(\text{interior region})) / P_0(\text{boundary})

and apply it exclusively there.