Chapter 10, Problem 115SE

True or False.

a If the p-value for a test is .036, the null hypothesis can be rejected at the α = .05 level of significance.

b In a formal test of hypothesis, α is the probability that the null hypothesis is incorrect.

c If the p-value is very small for a test to compare two population means, the difference between the means must be large.

d Power (θ*) is the probability that the null hypothesis is rejected when θ = θ*.

e Power (θ) is always computed by assuming that the null hypothesis is true.

f If .01 < p-value < .025, the null hypothesis can always be rejected at the α = .02 level of significance.

g Suppose that a test is a uniformly most powerful α-level test regarding the value of a parameter θ. If θ_a is a value in the alternative hypothesis, β(θ_a) might be smaller for some other α-level test.

h When developing a likelihood ratio test, it is possible that $L({\hat{\Omega }}_0)>L(\hat{\Omega })$.

i −2 ln(λ) is always positive.

To determine

Check whether the given statement as true or false.

Answer to Problem 115SE

True.

Explanation of Solution

Decision rule:

• If the p-value is less than the level of significance (α), reject the null hypothesis.
• Otherwise, fail to reject the null hypothesis.

In this context, the p-value of 0.036 is less than the level of significance of 0.05. Hence, the null hypothesis is rejected at $\alpha =0.05$. Therefore, the given statement is true.

To determine

Pinpoint the given statement as true or false.

Answer to Problem 115SE

False.

Explanation of Solution

Level of significance$\alpha$:

Type I error is defined as the probability of rejecting the null hypothesis, when it is actually true.

In a test of hypothesis, $\alpha$ is the probability that the null hypothesis is true. Therefore, the given statement is false.

To determine

State whether the given statement as true or false.

Answer to Problem 115SE

False.

Explanation of Solution

In a test of hypothesis, in order to compare the two populations, the pooled variance can also be small that leads to the greater value of the test statistic as pooled variance. Therefore, the given statement is false.

To determine

Define whether the given statement as true or false.

Answer to Problem 115SE

True.

Explanation of Solution

In a test of hypothesis, power is the probability that the null hypothesis is rejected when $\theta =\theta *$. Therefore, the given statement is true.

To determine

Delineate whether the given statement as true or false.

Answer to Problem 115SE

False.

Explanation of Solution

Normally, the power of test is computed by assuming that the null hypothesis is false and computed for the specific values of $H_a$.

To determine

Find whether the given statement as true or false.

Answer to Problem 115SE

False.

Explanation of Solution

Based on the decision rule provided in Part (a), when $0.01<p\text{-value}<0.025$ can be either greater or smaller than the level of significance of 0.02. The null hypothesis is not rejected for the given scenario. Therefore, the given statement is false.

To determine

Discover whether the given statement as true or false.

Answer to Problem 115SE

False.

Explanation of Solution

In this context, the given statement is false. This is because the uniformly most powerful test may have the highest power against all other α-level tests, and for some values of ${\theta }_a$ in $H_a$. In such case, the power of test is equal to $1-\beta \left({\theta }_a\right)$.

To determine

Determine whether the given statement as true or false.

Answer to Problem 115SE

False.

Explanation of Solution

Likelihood-ratio test:

A likelihood-ratio test of $H_0:\Theta \in {\Omega }_0\text{versus }{H}_a:\Theta \in {\widehat{\Omega }}_a$ denote $\lambda$ as the test statistic and the rejection region as $\lambda \le k$.

The test statistic $\lambda$ is denoted as follows:

$\begin{array}{c}\lambda =\frac{L\left({\widehat{\Omega }}_0\right)}{L\left(\widehat{\Omega }\right)}\\ =\frac{\underset{\Theta \in {\widehat{\Omega }}_0}{\max }L\left({\widehat{\Omega }}_0\right)}{\underset{\Theta \in \widehat{\Omega }}{\max }L\left(\widehat{\Omega }\right)}\end{array}$

In this scenario, the provided statement is false, it is because $L\left(\widehat{\Omega }\right)=\max \left\{L\left({\widehat{\Omega }}_0\right),L\left({\widehat{\Omega }}_1\right)\right\}$.

To determine

Determine the given statement as true or false.

Answer to Problem 115SE

True.

Explanation of Solution

Based on Theorem 10.2, let $Y_1,Y_2,\mathrm{...}{Y}_n$ have a joint likelihood function $L\left(\Theta \right)$. Denote $r_0$ as the number of free parameter that specifies $H_0:\Theta \in {\Omega }_0$ and denote r as the number of free parameter that specifies $\Theta \in \Omega$. Then, for the large sample of size n, $-2\ln \left(\lambda \right)$ (always positive) has approximately a ${\chi }^2$ distribution with $r_0-r$ df.

Based on the theorem, the provided statement is True.