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Part CCSIR NET December 2025the-critical-value-sits-at-1-342-not-a-round-number-1-3-is-just-under-it-not-over

The critical value sits at 1 342 not a round number 1 3 is just under it not over

Let X1,X2,,X15X_{1}, X_{2}, \dots, X_{15} be a random sample from an Exponential distribution with the probability density function f(x∣σ)=(1/σ)exp(x/σ)\sigma) = (1/\sigma)\exp(-x/\sigma) if x>0, 0 elsewhere, where the unknown parameter σ\sigma is positive. Let Xˉ=(1/15)115Xi{\bar{X}} = (1/15)\sum_{1}^{15}X_{i}. Suppose that φ\varphi denotes the likelihood ratio test for testing H0:σ1H_{0}: \sigma\le1 against H1:σ>1H_{1}: \sigma>1 at level α=0.1\alpha=0.1. It is given that χ152,0.1=22.307,χ152,0.9=8.547,χ302,0.1=40.256,χ302,0.9=20.599\chi^{2}_{15},_{0}._{1}=22.307, \chi^{2}_{15},_{0}._{9}=8.547, \chi^{2}_{30},_{0}._{1}=40.256, \chi^{2}_{30},_{0}._{9}=20.599, where P(W>χn2,a)=αP(W>\chi^{2}_{n},_{a})=\alpha and W~χn2\chi^{2}_{n}. Then which of the following statements are true?

  1. A.If the observed value of X̄ is 0.6, then φ\varphi does not reject H0H_{0}
  2. B.If the observed value of X̄ is 1.6, then φ\varphi rejects H0H_{0}
  3. C.If the observed value of X̄ is 1.3, then φ\varphi rejects H0H_{0}
  4. D.If the observed value of X̄ is 1.2, then φ\varphi does not reject H0H_{0}

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.

See pricing

50 are analysed free — try those first.

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 TestingLikelihood ratio and standard tests

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

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