Testing of Hypotheses Questions and Answers

Answer: D

The degrees of freedom are \((r - 1)(k - 1)\).

The data form an \(r \times k\) contingency table. Each of the \(r\) rows has \(k - 1\) free probabilities, \(r(k - 1)\) parameters in all. Under \(H_{0}\) the rows share one probability vector with \(k - 1\) free parameters, estimated from the data. The degrees of freedom are the difference:

\[ r(k - 1) - (k - 1) = (r - 1)(k - 1) \]

This is the same as the chi-square test of homogeneity or independence. \(r(k - 1)\) forgets the estimated common probabilities, and \(rk - 1\) treats the whole table as one multinomial.

Answer: C

The UMP test rejects \(H_{0}\) if \(\bar{X} \gt c\).

The Poisson family has a monotone likelihood ratio in \(\sum X_{i}\) (equivalently \(\bar{X}\)): for \(\theta_{1} \gt \theta_{0}\),

\[ \begin{aligned} &\frac{L(\theta_{1})}{L(\theta_{0})} \\ &= e^{-n(\theta_{1} - \theta_{0})}\left(\frac{\theta_{1}}{\theta_{0}}\right)^{\sum x_{i}} \end{aligned} \]

increases with \(\sum x_{i}\). By the Neyman-Pearson lemma and the Karlin-Rubin theorem, the UMP test of \(\theta = 1\) against \(\theta \gt 1\) rejects for large \(\bar{X}\) (with randomisation if needed, since the distribution is discrete).

Rejecting for small \(\bar{X}\) suits \(\theta \lt 1\), and two-sided or interval regions are not UMP for a one-sided alternative.

Answer: A

For the t-test, the distributions of both X and Y must be normal with equal variances.

The two-sample t statistic \((\bar{X} - \bar{Y}) / (s_{p}\sqrt{1/n_{1} + 1/n_{2}})\) has an exact t distribution with \(n_{1} + n_{2} - 2\) degrees of freedom only when both populations are normal and share a common variance, so that the pooled estimate \(s_{p}^{2}\) is valid.

Normality alone (B) is not enough, because with unequal variances the statistic is not exactly t (the Behrens-Fisher problem). Equal variances alone (D) is not enough without normality, and C drops the normality assumption, which the exact test needs.

Answer: D

The degrees of freedom are 6.

For a chi-square goodness of fit test, df = (number of classes used) - 1 - (number of parameters estimated). Combining two classes leaves 9 classes, and fitting a normal distribution estimates two parameters, \(\mu\) and \(\sigma^{2}\):

\[ \text{df} = 9 - 1 - 2 = 6 \]

9 is the number of classes, 8 forgets the estimated parameters, and 7 subtracts only one of them.

Answer: C

(i) and (ii) both are correct.

For this one-sided test the rejection region is \(\bar{X} \gt \mu_{0} + z_{\alpha}/\sqrt{n}\). A larger \(\alpha\) gives a smaller critical value \(z_{\alpha}\), so the rejection regions are nested: the region at \(\alpha = 0.05\) lies inside the region at \(\alpha = 0.10\), and the region at \(\alpha = 0.01\) lies inside the one at 0.05.

(i) A sample rejected at 0.05 falls in the larger 0.10 region, so it is rejected there too. (ii) A sample accepted at 0.05 is outside the 0.05 region, hence outside the smaller 0.01 region, so it is accepted at 0.01. Equivalently, H₀ is rejected at level \(\alpha\) exactly when the p-value is at most \(\alpha\).

Answer: A

The size of the test is 0.

The size is the probability of rejecting \(H_{0}\) when it is true. Under \(H_{0}\), X is uniform on (0, 1), so it can never exceed 1:

\[ \alpha = P_{\theta = 1}(X \gt 1) = 0 \]

(The paper prints the alternative as '\(H_{1} = \theta = 2\)', meaning \(H_{1} : \theta = 2\).) The test still has power: under \(\theta = 2\), \(P(X \gt 1) = 1/2\). The values 0.01, 0.05 and 0.10 are conventional significance levels, not the size of this particular critical region.

Answer: B

The hypotheses are \(H_{0} : \mu = 8.5\), \(H_{a} : \mu \lt 8.5\).

The claim being challenged is the advertised mean loss of 8.5 lb, so it goes into the null hypothesis. The experiment is looking for evidence that patients lose less than advertised, so the alternative is one-sided to the left.

\(H_{a} : \mu \gt 8.5\) would test whether patients lose more than claimed, which is not the question. A two-sided alternative wastes power on the uninteresting direction. And \(H_{0} : \mu \gt 8.5\) against \(H_{a} : \mu \le 8.5\) puts the equality in the alternative, which is not how a null is set up: the null must contain the boundary value so the test statistic has a known distribution under it.

Answer: B

A test based on the sample mean is not consistent.

For a Cauchy sample the sample mean \(\overline{X}_{n}\) has exactly the same \(C(\theta, 1)\) distribution as a single observation, whatever \(n\) is: the characteristic function of the mean is \((e^{i\theta t/n - |t|/n})^{n} = e^{i\theta t - |t|}\). So its distribution never concentrates, and a test based on it cannot have power tending to 1 against a fixed \(\theta \neq 0\).

Sample quantiles behave well for the Cauchy. The sample median converges in probability to \(\theta\), and the first and third quartiles converge to \(\theta - 1\) and \(\theta + 1\), so tests built on any of them are consistent. That is why the median is the standard location estimator for the Cauchy.

Answer: C

The appropriate test is the paired t-test, if normality can be assumed.

Each person receives both creams, one on each arm, so the observations come in matched pairs. The two responses from the same person share that person's skin type and other characteristics, so they are dependent. The right analysis works with the within-person differences \(d_{i} = A_{i} - B_{i}\) and tests whether their mean is zero, which is the paired t-test when the differences are approximately normal.

The two-sample t-test, the Mann-Whitney test and the two-sample Kolmogorov-Smirnov test all assume two independent samples. Using them here ignores the pairing, which both violates their assumptions and throws away the large reduction in variance that the design provides. (The nonparametric counterpart for paired data would be the Wilcoxon signed-rank test.)

Answer: D

The size is 42/64.

The size of a test is the probability of the critical region when \(H_{0}\) is true, here \(P = \frac{1}{2}\) with n = 6:

\[ \begin{aligned} P(X \le 3) &= \frac{\binom{6}{0} + \binom{6}{1} + \binom{6}{2} + \binom{6}{3}}{2^{6}} \\ &= \frac{1 + 6 + 15 + 20}{64} = \frac{42}{64} \end{aligned} \]

22/64 stops at \(X \le 2\), 35/64 and 15/64 do not correspond to the region \(X \le 3\) under \(P = \frac{1}{2}\). The alternative \(P = \frac{1}{4}\) would be used for the power, not the size.