Testing of Hypotheses Questions and Answers

Answer: C

The true statement is the power function is decreasing on \((-\infty, 5)\).

For a normal mean with known variance, the likelihood ratio test of \(H_{0}: \mu = 5\) against \(\mu \neq 5\) rejects when \(|\sqrt{n}(\bar{X} - 5)| \gt 1.96\). Its power function is

\[ \begin{aligned} \beta(\mu) &= \Phi(-1.96 - \delta) + \Phi(-1.96 + \delta), \\ \delta &= \sqrt{n}(\mu - 5) \end{aligned} \]

This is symmetric about \(\mu = 5\), takes its minimum value \(0.05\) there, decreases on \((-\infty, 5)\) and increases on \((5, \infty)\), tending to 1 far from 5 on either side.

So \(\mu = 5\) is a local minimum, not a maximum; the power increases, not decreases, on \((5, \infty)\); and it decreases, not increases, on \((0, 5)\).

Answer: A

The true statement is \(0.20 \lt \frac{3}{4}\alpha + \frac{1}{4}\beta \lt 0.25\).

The sum of independent chi-square variables is chi-square with the degrees of freedom added. Under \(H_{0}\), \(\sum X_{i} \sim \chi_{4}^{2}\); under \(H_{1}\), \(\sum X_{i} \sim \chi_{8}^{2}\). From the values given:

\[ \begin{aligned} \alpha &= P(\chi_{4}^{2} \gt 6) = 0.1991 \\ 1 - \beta &= P(\chi_{8}^{2} \gt 6) = 0.6472 \\ \beta &= 0.3528 \end{aligned} \]

\[ \begin{aligned} &\tfrac{3}{4}(0.1991) + \tfrac{1}{4}(0.3528) \\ &\approx 0.1493 + 0.0882 = 0.2375 \end{aligned} \]

which lies between 0.20 and 0.25. The other statements fail: \(\alpha = 0.1991\) is just below 0.20, \(\beta = 0.3528\) is above 0.30, and the power, 0.6472, exceeds \(\frac{1}{2}\). The values for 1 and 2 degrees of freedom are distractors, since the test uses the sum of four observations.

Answer: C

The statement that is NOT true is for all \(\theta \gt 0\), \(\beta(\theta) \lt 1\).

The power function is \(\beta(\theta) = P_{\theta}\left(X \lt \frac{3}{4}\right)\):

\[ \beta(\theta) = \begin{cases} 1, & 0 \lt \theta \leq \frac{3}{4}, \\ \frac{3}{4\theta}, & \theta \gt \frac{3}{4}. \end{cases} \]

For \(\theta \leq \frac{3}{4}\) every observation is below \(\frac{3}{4}\), so the power equals 1 and the statement is false.

The other three are true. The size is \(\sup_{\theta \geq 1}\frac{3}{4\theta} = \frac{3}{4}\), attained at \(\theta = 1\). Over \(\theta \lt 1\), \(\beta\) decreases towards \(\frac{3}{4}\) as \(\theta \to 1\) without reaching it, so the infimum is \(\frac{3}{4}\). And \(\beta(\theta) = 1\) for all small \(\theta\), so the limit as \(\theta \to 0\) is 1.

Answer: C

The test is the t-test.

The t-test was devised by W. S. Gosset, writing as 'Student', in 1908 to draw inferences about means from small samples, usually of fewer than 30, when the population standard deviation is unknown. It uses the t-distribution, which allows for the extra uncertainty of small samples.

The chi-square test is for frequencies and associations between categorical variables, ANOVA compares the means of three or more groups, and the mean is a descriptive statistic, not a hypothesis test.

Answer: A

The degrees of freedom are 2.

For a chi-square test of independence on an r × c contingency table:

\[ \begin{aligned} df &= (r - 1)(c - 1) \\ &= (3 - 1)(2 - 1) = 2 \end{aligned} \]

Once the row and column totals are fixed, only two cell values can be chosen freely; the rest follow from the totals. 5 is the result of subtracting 1 from the total number of cells (6), the rule for a goodness-of-fit test, not a contingency table. 3 is the number of rows, and 6 is the number of cells.

Answer: C

There exists a unique randomized most powerful test of size 0.05.

The likelihood ratio is

\[ \begin{aligned} \frac{f_{1}(x)}{f_{0}(x)} &= \frac{(2/3)^{x}(1/3)^{1 - x}}{(1/3)^{x}(2/3)^{1 - x}} \\ &= 2^{2x - 1} \end{aligned} \]

which is 2 at x = 1 and 1/2 at x = 0. By the Neyman-Pearson lemma the most powerful test rejects for the larger ratio, that is on X = 1.

But rejecting whenever X = 1 has size \(P_{0}(X = 1) = 1/3\), far above 0.05. The only way to reach size exactly 0.05 is to reject with probability \(\gamma\) when X = 1 and never when X = 0:

\[ \gamma \cdot \tfrac{1}{3} = 0.05 \Rightarrow \gamma = 0.15 \]

That test is randomized, and the lemma makes it unique, since the rejection probability is fixed on both sample points. No non-randomized test has size 0.05 (the only non-randomized sizes are 0, 1/3, 2/3 and 1). Rejecting on X = 0 is the least powerful choice, not the most powerful.

Answer: B

The UMP test rejects \(H_{0}\) if \(X_{(1)} \gt C\).

The density is \(f(x; \theta) = e^{-(x - \theta)}\) for \(x \gt \theta\), so the likelihood is

\[ L(\theta) = e^{-\sum x_{i} + n\theta} I(x_{(1)} \gt \theta) \]

It depends on \(\theta\) only through the smallest observation \(X_{(1)}\), which is sufficient. For \(\theta_{1} \gt \theta_{0}\), the ratio \(L(\theta_{1})/L(\theta_{0})\) is 0 when \(x_{(1)} \le \theta_{1}\) and the constant \(e^{n(\theta_{1} - \theta_{0})}\) otherwise, a non-decreasing function of \(x_{(1)}\). So the family has monotone likelihood ratio in \(X_{(1)}\), and the UMP test for the one-sided alternative rejects for large \(X_{(1)}\). C comes from \(n(X_{(1)} - \theta_{0})\) being exponential with mean 1 under \(H_{0}\).

\(\bar{X}\) and \(X_{(n)}\) are not sufficient for the location, and \(\sum(X_{i} - X_{(1)})\) is ancillary: its distribution does not depend on \(\theta\) at all, so it cannot detect a shift.

Answer: C

The technique is the chi-square test as a test of goodness-of-fit.

The claim is about the whole shape of the distribution of daily customer counts, not about a single parameter. The goodness-of-fit test compares the observed frequencies of each count with the frequencies expected under the fitted Binomial distribution:

\[ \chi^{2} = \sum \frac{(O_{i} - E_{i})^{2}}{E_{i}} \]

with degrees of freedom equal to the number of classes minus 1, minus the number of parameters estimated (one, if p is estimated from the data). A large value rejects the Binomial model.

A test of a mean, or of a difference of means, checks one or two averages and cannot tell a Binomial from any other distribution with the same mean. Correlation analysis measures association between two variables, which is not the question here.

Answer: B

The critical value is \(\mu_{0} + 3.290/\sqrt{n}\).

Under \(H_{0}\), \(\bar{X} \sim N(\mu_{0}, \sigma^{2}/n)\) with \(\sigma = 2\). The alternative is \(\mu_{1} \gt \mu_{0}\), so the best critical region is the upper tail, and its size must be 0.05:

\[ \begin{aligned} P_{0}(\bar{X} \gt K) &= 0.05 \\ \frac{K - \mu_{0}}{2/\sqrt{n}} &= 1.645 \\ K &= \mu_{0} + \frac{3.290}{\sqrt{n}} \end{aligned} \]

\(\mu_{0} + 1.645/\sqrt{n}\) forgets that \(\sigma = 2\), not 1. The values with 1.96 and 3.92 use the two-sided 5% point, which is wrong for a one-sided alternative.

Answer: B

The value of \(\chi^{2}\) is 0.794.

For a \(2 \times 2\) table with cells a = 10, b = 20, c = 30, d = 40 and N = 100, the shortcut formula is

\[ \chi^{2} = \frac{N(ad - bc)^{2}}{(a+b)(c+d)(a+c)(b+d)} \]

Here \(ad - bc = 400 - 600 = -200\), the row totals are 30 and 70, and the column totals are 40 and 60:

\[ \begin{aligned} \chi^{2} &= \frac{100 \times 40000}{30 \times 70 \times 40 \times 60} \\ &= \frac{4000000}{5040000} \approx 0.794 \end{aligned} \]

The same value comes from \(\sum (O - E)^{2}/E\) with expected counts 12, 18, 28 and 42. The other values do not agree with either formula. With 1 degree of freedom, 0.794 is far below the 5% point 3.84, so independence is not rejected.