Testing of Hypotheses Questions and Answers

Answer: A

The true statement is that for simple against simple, the likelihood ratio principle leads to the most powerful test.

The Neyman-Pearson lemma says that the most powerful test of a given size for \(H_{0} : \theta = \theta_{0}\) against \(H_{1} : \theta = \theta_{1}\) rejects when the likelihood ratio \(L(\theta_{1})/L(\theta_{0})\) exceeds a constant.

The others fail. The two error probabilities \(\alpha\) and \(\beta\) need not add up to 1. Randomized tests are used to attain an exact size (level) for discrete distributions, not a desired power. And UMP tests against composite alternatives often do not exist, for example against two-sided alternatives in the normal mean problem.

Answer: C

The MVUE of \(\sigma^{2}\) is (1/n) Σ X_i².

The mean is known to be 0, so \(\sum X_{i}^{2}\) is a complete sufficient statistic for \(\sigma^{2}\), and

\[ \begin{aligned} E\left(\frac{1}{n}\sum X_{i}^{2}\right) &= \frac{1}{n} \cdot n\sigma^{2} \\ &= \sigma^{2} \end{aligned} \]

An unbiased function of a complete sufficient statistic is the MVUE (Lehmann-Scheffe).

Dividing by n − 1 makes it biased when the mean is known. The estimators using \(\bar{X}\) waste the known mean: \(\frac{1}{n-1}\sum(X_{i} - \bar{X})^{2}\) is unbiased but has larger variance, and the version with n is biased.

Answer: C

She should test H₀ : p = 0.80 against H₁ : p > 0.80.

The claim she wants evidence for, that more than 80% enjoyed the class, goes in the alternative hypothesis, and the null states the boundary value of no effect, p = 0.80. Rejecting the null then supports the claim.

\(p \lt 0.80\) tests the opposite direction, and \(p \neq 0.80\) is two-sided although only one direction interests her. Putting the claim \(p \gt 0.80\) in the null, with an equality alternative, reverses the roles of the hypotheses.

Answer: C

The expected frequency is n_i. n_.j / n.

Under independence, \(P(A_{i} \cap B_{j}) = P(A_{i})P(B_{j})\), estimated by \(\frac{n_{i.}}{n} \cdot \frac{n_{.j}}{n}\). Multiplying by the total n gives the expected count

\[ \begin{aligned} e_{ij} &= n \cdot \frac{n_{i.}}{n} \cdot \frac{n_{.j}}{n} \\ &= \frac{n_{i.}\, n_{.j}}{n} \end{aligned} \]

The product \(n_{i.}n_{.j}\) without dividing is far too large, dividing by \(n^{2}\) gives the estimated probability rather than the expected count, and \(n_{ij}/n\) is the observed proportion, not an expected frequency.

Answer: B

A UMP test is unbiased.

Compare the UMP level-\(\alpha\) test \(\phi\) with the trivial test \(\phi_{0} \equiv \alpha\), which has power exactly \(\alpha\) everywhere. Since \(\phi\) is most powerful among all level-\(\alpha\) tests, its power at every alternative is at least that of \(\phi_{0}\), namely \(\alpha\). Power at least the size under every alternative is the definition of an unbiased test.

That rules out 'not unbiased' and 'power less than its size'. \(\phi \equiv 1\) always rejects, so its size is 1, not \(\alpha\), and it is not a level-\(\alpha\) test.

Answer: B

A UMP level \(\alpha\) test exists for \(P_{1}\) but not for \(P_{2}\).

The normal family has monotone likelihood ratio in \(\overline{X}\). For the one-sided alternative \(\mu \gt \mu_{0}\), the Neyman-Pearson test 'reject when \(\overline{X} \gt \mu_{0} + z_{\alpha}/\sqrt{n}\)' is the same for every \(\mu \gt \mu_{0}\), so it is UMP (Karlin-Rubin).

For the two-sided alternative, the most powerful test against \(\mu \gt \mu_{0}\) rejects for large \(\overline{X}\) and the most powerful test against \(\mu \lt \mu_{0}\) rejects for small \(\overline{X}\). No single test is best on both sides, so no UMP test exists for \(P_{2}\); the usual two-sided test is only UMP among unbiased tests.

Answer: A

Such a test is an unbiased test.

A test is unbiased when its power, the probability of rejecting \(H_{0}\) when \(H_{0}\) is false, is at least its size, the probability of rejecting \(H_{0}\) when it is true. A test whose rejection probability is higher under the alternative than under the null meets that definition.

A biased test is the opposite, with power below the size for some alternative. A consistent test is one whose power tends to 1 as the sample size grows, and a uniformly most powerful test has the greatest power among all tests of its size; neither is defined by the comparison in the question.

Answer: B

The true statement is that Karl Pearson's statistic \(\sum \frac{(o_{i} - e_{i})^{2}}{e_{i}}\) is asymptotically \(\chi^{2}_{k-1}\).

Under a fully specified \(H_{0}\), the k cell counts are tied by the single constraint \(\sum o_{i} = n\), so the statistic has \(k - 1\) degrees of freedom.

The likelihood ratio statistic \(-2\log\Lambda\) is also asymptotically \(\chi^{2}_{k-1}\), with degrees of freedom equal to the number of free parameters, not \(n - k\). Using \(\chi^{2}_{k}\) forgets the constraint. The version with \(o_{i}\) in the denominator (Neyman's statistic) is also asymptotically \(\chi^{2}_{k-1}\), so option D's \(\chi^{2}_{k}\) is wrong too.