ISS 2016 Statistics Paper-2 Solution: Question 60 (Power of a Test for Uniform Distribution)

Continuing our question-by-question walkthrough of the ISS Statistics Paper-2 previous year papers, today we pick up right where the series left off and work through 2016's Question 60. This one tests whether you really understand what "power of a test" means in concrete, computable terms, not just as a definition to memorize.

Quick Summary

• Topic: Power of a test, hypothesis testing with a Uniform distribution

• Reference: ISS 2016, Statistics Paper-2, Question 60

• Correct Answer: (d) 9/16

The Question As Asked

If the pdf of a random variable X is

f(x; θ) = 1 − θ², for 0 ≤ x ≤ 1/(1 − θ²)

and if we reject H0 : θ = 1/2 against H1 : θ = 3/4 whenever x ≥ 1, then what is the power of the test?

• (a) 15/16

• (b) 1/2

• (c) 7/16

• (d) 9/16

Step 1: Recognize What Distribution This Actually Is

The formula f(x; θ) = 1 − θ² looks unfamiliar at first glance because the θ² is hiding what's really going on. Notice that the density does not depend on x at all — it is a constant for every x in its range. A density that is constant over its support is, by definition, a Uniform distribution.

For a Uniform distribution on the interval [0, b], the density is always f(x) = 1/b over that interval. Compare that with our given density, 1 − θ², which is constant and equal to 1/[1/(1 − θ²)]. So here, b = 1/(1 − θ²), meaning:

X ~ Uniform(0, b), where b = 1/(1 − θ²)

You can double-check this interpretation by confirming the density integrates to 1 over its support, which it must for any valid pdf: (1 − θ²) × [1/(1 − θ²) − 0] = 1. It checks out.

Step 2: Work Out the Support Under the Alternative Hypothesis

The power of a test is defined as the probability of correctly rejecting H0 when H1 is actually true. So we need to work entirely under θ = 3/4 (the alternative), not under θ = 1/2.

Substitute θ = 3/4 into b = 1/(1 − θ²):

θ² = (3/4)² = 9/16

1 − θ² = 1 − 9/16 = 7/16

b = 1/(7/16) = 16/7

So, under H1, X is Uniform on the interval [0, 16/7], and its constant density over that interval is 7/16.

Step 3: Translate "Power" Into a Probability Statement

The rejection rule given in the question is: reject H0 whenever x ≥ 1. So the power of the test is:

Power = P(reject H0 | H1 is true) = P(X ≥ 1 | θ = 3/4)

Since X is Uniform(0, 16/7) under H1, and 16/7 ≈ 2.2857 is comfortably greater than 1, the event X ≥ 1 is a perfectly valid, non-trivial slice of the distribution's range.

Pro Tip: The most common slip on this type of question is computing the probability under H0 (θ = 1/2) instead of H1 (θ = 3/4) — that mistake gives you the

Step 4: Compute the Probability

For a Uniform(0, b) random variable, the probability that X falls below some point c (where 0 ≤ c ≤ b) is simply c/b, because the density is flat and probability is just "length × constant height."

P(X < 1) = (length of interval [0,1]) × (density) = 1 × (7/16) = 7/16

Since total probability must sum to 1:

P(X ≥ 1) = 1 − P(X < 1) = 1 − 7/16 = 9/16

So the power of the test is 9/16.

Step 5: Confirm Against the Official Answer Key

This matches option (d) 9/16, and it agrees with the official ISS 2016 Statistics Paper-2 answer key (Series A), which also records the answer to Question 60 as (d). Our independent derivation and the official key are consistent, so we can be confident in this result.

As a bonus sanity check, if you repeat the exact same steps using θ = 1/2 instead of θ = 3/4, you'd be computing the size (significance level) of this test rather than its power. Working through that gives b = 1/(1 − 1/4) = 4/3, and P(X ≥ 1) = 1 − (3/4) = 1/4 — a 25% significance level, which is a reasonable, exam-typical value and reinforces that our reading of the question's setup is correct.

Why This Question Matters

This question is a great example of why ISS Paper-2 rewards candidates who can see past unfamiliar notation. Many students panic the moment they see f(x; θ) = 1 − θ² because it doesn't immediately look like a distribution they've memorized a formula for. But the entire problem collapses into a two-line Uniform distribution calculation once you recognize that a constant density simply means "Uniform." Beyond the specific numbers, this question is really testing whether you understand the conceptual difference between the size of a test (computed under H₀) and the power of a test (computed under H₁) — a distinction that shows up repeatedly across ISS, IIT JAM, and GATE Statistics papers in slightly different disguises.

Frequently Asked Questions

What is the difference between the size of a test and the power of a test?

The size (or significance level, α) is the probability of rejecting H₀ when H₀ is actually true — it measures the Type I error rate. The power is the probability of correctly rejecting H₀ when H₁ is actually true. Both are computed using the same rejection region, but size uses the null parameter value while power uses the alternative parameter value.

How do I know when a given pdf is actually a Uniform distribution in disguise?

Check whether the density expression depends on x. If it does not — if it is a constant with respect to x over some range — then the random variable is uniformly distributed on that range, and the constant itself must equal 1 divided by the length of that range.

Why is the power calculated using θ = 3/4 and not θ = 1/2?

Because power is, by definition, the probability of rejection assuming the alternative hypothesis is the true state of the world. Since H₁ here specifies θ = 3/4, every probability calculation for power must use θ = 3/4, including the support of the distribution itself.

Can the power of a test ever be lower than its size?

In a well-constructed, sensible test it normally shouldn't be, since a useful test should be more likely to detect a true effect than to falsely flag a null effect. In this question, the size works out to 1/4 and the power to 9/16, and 9/16 is indeed comfortably larger than 1/4, which is the expected and reassuring direction.

Is this topic likely to reappear in future ISS Paper-2 papers?

Yes. Power functions, size versus power, Neyman-Pearson type rejection regions, and most powerful tests are core testing-of-hypotheses topics that appear almost every year in some form across ISS and related statistics exams, so it is worth mastering the mechanics shown here thoroughly rather than memorizing this one answer.

What should I do if my calculated support doesn't make sense, like getting a negative or undefined value?

Stop and recheck your substitution immediately — it usually means you plugged in a θ value that makes 1 − θ² zero or negative, which is a strong signal you've mixed up which hypothesis (null or alternative) you were supposed to be working under for that particular step.

If any step above wasn't clear, or you spotted something worth double-checking, drop a comment below — and if this helped untangle the size-versus-power distinction for you, consider sharing it with a fellow ISS aspirant who might be stuck on the same question.