Graphical Analysis In Exercises 13 and 14, state whether each standardized test statistic X^2 allows you to reject the null hypothesis. Explain.


a. X^2=2.091

Finding Critical Values and Rejection Regions In Exercises 23–28, find the critical value(s) and rejection region(s) for the type of z-test with level of significance α. Include a graph with your answer.

Left-tailed test, α = 0.09

Finding Critical Values and Rejection Regions In Exercises 23–28, find the critical value(s) and rejection region(s) for the type of z-test with level of significance α. Include a graph with your answer.

Right-tailed test, α = 0.08

Finding Critical Values and Rejection Regions In Exercises 23–28, find the critical value(s) and rejection region(s) for the type of z-test with level of significance α. Include a graph with your answer.

Two-tailed test, α = 0.12

Graphical Analysis In Exercises 21 and 22, state whether each standardized test statistic z allows you to reject the null hypothesis. Explain your reasoning.

a. z = -1.301

b. z = 1.203

c. z = 1.280

d. z = 1.286

Graphical Analysis In Exercises 13 and 14, state whether each standardized test statistic X^2 allows you to reject the null hypothesis. Explain.

b. X^2=23.309

Getting at the Concept Explain why a level of significance of α=0 is not used.

Writing A null hypothesis is rejected with a level of significance of 0.10. Is it also rejected at a level of significance of 0.05? Explain.

Hypothesis Testing Using Rejection Regions In Exercises 23–30, (a) identify the claim and state H0 and Ha, (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic X^2, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the population is normally distributed.

Salaries The annual salaries (in dollars) of 12 randomly chosen nursing supervisors are shown in the table at the left. At α=0.10, is there enough evidence to reject the claim that the standard deviation of the annual salaries is $18,630?