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Chapter 9: Q 9.22 (page 358)

Data on salaries in the public school system are published annually in Ranking of the States and Estimates of School Statistics by the National Education Association. The mean annual salary of (public) classroom teachers is $55.4thousand. A hypothesis test is to be performed to decide whether the mean annual salary of classroom teachers in Ohio is greater than the national mean.

Short Answer

Expert verified

The Null Hypothesis is H0:渭=$55.4and the Alternative Hypothesis is H0:渭>$55.4.

The hypothesis test is right-tailed.

Step by step solution

01

Step 1. Given Information.

The objective is to perform a hypothesis test to decide whether the mean annual salary of classroom teachers in Ohio is greater than the national mean.

02

Step 2. Null Hypothesis.

The Null Hypothesis is:

H0:渭=$55.4

The mean annual salary of the classroom teachers in Ohio is not greater than the national mean.

03

Step 3. Alternative Hypothesis.

The Alternative Hypothesis is:

H0:渭>$55.4

The mean annual salary of the classroom teachers in Ohio is greater than the national mean.

04

Step 4. Checking the hypothesis.

The hypothesis test is right-tailed.

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Most popular questions from this chapter

Determine the critical value(s) for a one-mean z-test at the 1 % significance level if the test is

a. right tailed.

b. left tailed.

c. two tailed.

The following graph portrays the decision criterion for a onemean z-test, using the critical-value approach to hypothesis testing. The curve in the graph is the normal curve for the test statistic under the assumption that the null hypothesis is true.

Determine the

a. rejection region.

b. nonrejection region.

c. critical value(s).

d. significance level.

e. Draw a graph that depicts the answers that you obtained in parts (a)-(d).

f. Classify the hypothesis test as two tailed, left tailed, or right tailed.

9.95 Two-Tailed Hypothesis Tests and CIs. As we mentioned on page 378 , the following relationship holds between hypothesis tests and confidence intervals for one-mean z-procedures: For a two-tailed hypothesis test at the significance level 伪, the null hypothesis H0:渭=渭0 will be rejected in favor of the alternative hypothesis Ha:渭鈮�渭0if and only if 渭0 lies outside the (1-伪)-level confidence interval for 渭. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean z-interval (Procedure 8.1 on page 322 ) and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise 9.84
b. Exercise 9.87

Ankle Brachial Index. The ankle brachial index (ABI) compares the blood pressure of a patient's arm to the blood pressure of the patient's leg. The ABI can be an indicator of different diseases, including arterial diseases. A healthy (or normal) ABI is 0.9 or greater. In a study by M. McDermott et al. titled "Sex Differences in Peripheral Arterial Disease: Leg Symptoms and Physical Functioning" (Journal of the American Geriatrics Society, Vol. 51, No. 2, Pp. 222-228), the researchers obtained the ABI of 187 women with peripheral arterial disease. The results were a mean ABI of 0.64 with a standard deviation of 0.15 At the 1 % significance level, do the data provide sufficient evidence to conclude that, on average, women with peripheral arterial disease have an unhealthy ABI?

Betting the Spreads. College basketball, and particularly the NCAA basketball tournament, is a popular venue for gambling, from novices in office betting pools to high rollers. To encourage uniforn betting across teams, Las Vegas oddsmakers assign a point spread to each game. The point spread is the oddsmakers" prediction for th number of points by which the favored team will win. If you bet of the favorite, you win the bet provided the favorite wins by more than the point spread; otherwise, you lose the bet. Is the point spread a good measure of the relative ability of the two teams? H. Stern and B. Mock addressed this question in the paper "College Basketball Upsets: Will a 16-Seed Ever Beat a 1-Seed?" (Chance, Vol. 11(1), pp. 27-31). They obtained the difference between the actual margin of victory and the point spread, called the point-spread error, for 2109 college basketball games. The mean point-spread error was found to be 鈭�0.2 point with a standard deviation of10.9 points. For a particular game, a point-spread error of 0 indicates that the point spread was a perfect estimate of the two teams' relative abilities.
(a) If, on average, the oddsmakers are estimating correctly, what is the (population) mean point-spread error?
(b) Use the data to decide, at the 5% significance level, whether the (population) mean point-spread error differs from 0 .
c) Interpret your answer in part (b).
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