Skip to content
Part BCSIR NET June 2025a-heavier-tail-parameter-pushes-the-sample-down-so-the-ump-test-rejects-small-values

A heavier tail parameter pushes the sample down so the ump test rejects small values

Let X1,X2,…,XnX_{1}, X_{2}, \dots, X_{n} be a random sample from a continuous distribution with the common probability density function f(x|θ)=2θθ/\theta) = 2^\theta \theta / xθ+1x^{\theta+1} if x > 2, and 0 otherwise, where θ(>0)\theta (> 0) is an unknown parameter. Suppose P(Y>χm,β2)P(Y > \chi^{2}_{m,\beta}) =β= \beta, where Y ~ χm2\chi^{2}_m. For testing H0:θ=1H_{0} : \theta = 1 against H1:θ>1,aH_{1} : \theta > 1, a uniformly most powerful test of size α,0<α<1\alpha, 0 < \alpha < 1, will reject H0H_{0} if

  1. A.∑i\sum_{i}₌1nln⁡Xi>(1/2)χ2n,α2_{1}^{n} \ln X_{i} > (1/2)\chi^{2}_{2n,\alpha} + n ln 2
  2. B.∑i\sum_{i}₌1nln⁡Xi<(1/2)χ2n,1−α2_{1}^{n} \ln X_{i} < (1/2)\chi^{2}_{2n,1-\alpha} + n ln 2✓ Correct
  3. C.∑i\sum_{i}₌1nln⁡Xi>χn,α2_{1}^{n} \ln X_{i} > \chi^{2}_{n,\alpha} + n ln 2
  4. D.∑i\sum_{i}₌1nln⁡Xi<χn,1−α2_{1}^{n} \ln X_{i} < \chi^{2}_{n,1-\alpha} + n ln 2

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests what the inference means.

See pricing

50 are analysed free — try those first.

The trap it tests

What the inference means

p-values, size, power and error types say specific things. This was not one of them.

Drill statements like this

Related counterexample: A UMP test exists for every testing problem

More on this topic

The chapter behind this: Neyman–Pearson and UMP tests — free to read

From Hypothesis Testing › Neyman–Pearson lemma and UMP tests

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram