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Part CCSIR NET June 2025the-likelihood-ratio-is-unimodal-so-small-values-occur-at-both-ends

The likelihood ratio is unimodal so small values occur at both ends

Let X be a random sample of size one from the probability density function f(x|θ)=\theta) = θe−θ(x−1)\theta{}e^{-\theta(x-1)} if x > 1, and 0 otherwise, where θ(>0)\theta (> 0) is the unknown parameter. Suppose we want to test the null hypothesis H0:θ=1H_{0} : \theta = 1 against the alternative hypothesis H1:θ≠1H_{1} : \theta \ne 1, based on the observed value x of X. Then, which of the following statements are true?

  1. A.The likelihood function is maximized at θ=1/(x−1)\theta = 1/(x - 1)✓ Correct
  2. B.The maximum value of the likelihood function is e−1/(xe^{-1}/(x − 1)✓ Correct
  3. C.The likelihood ratio test for testing H0H_{0} against H1H_{1} rejects H0H_{0} if (x − 1)e−x1)e^{-x} < k, for some k > 0✓ Correct
  4. D.The likelihood ratio test for testing H0H_{0} against H1H_{1} rejects H0H_{0} if x > c, for some c > 1

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: Any interval of the form [X̄ − (S/√n)t, ∞) with a 90% quantile t is a 90% confidence interval

More on this topic

The chapter behind this: Standard tests and confidence intervals — free to read

From Hypothesis Testing › Likelihood ratio and standard tests

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