Question:Accounting and Machiavellianism. Refer to the Behavioral Research in Accounting (January 2008) study of Machiavellian traits in accountants, Exercise 6.19 (p. 341). A Mach rating score was determined for each in a random sample of 122 purchasing managers, with the following results: = 99.6 s = 12.6. Recall that a director of purchasing at a major firm claims that the true mean Mach rating score of all purchasing managers is 85.

a. Suppose you want to test the director’s claim. Specify the null and alternative hypotheses for the test.

b. Give the rejection region for the test using α = 0.10.

c. Find the value of the test statistic.

d. Use the result, part c, to make the appropriate conclusion.

Short Answer

Expert verified
  1. The null and alternative hypotheses are H0 : µ = 85against Ha : µ ≠ 85.
  2. Reject the null hypothesis if z < -1.64 or z > 1.64.
  3. The value of the test statistic is z = 12.7986.
  4. There is not enough evidence to support the director’s claim that all purchasing managers' true mean Mach rating score is 85.

Step by step solution

01

Given Information

  • The claim of the director of purchasing at a major firm is that the true Mach score of all purchasing managers is 85.
  • The significance level for the hypothesis testing problem is α = 0.10.
  • A Mach rating score for a random sample 122 purchasing managers is provided.
  • The summary statistics are:x̄ = 99.6 s = 12.6.
02

Setting the null and alternative hypothesis

(a)

The researchers are interested in the central value of the population, commonly known as the parameters. The claim about the population parameters is validated by using hypothesis testing.

The null and alternative hypotheses for the director’s claim are:

The null and alternative hypotheses are H0 : µ = 85against Ha : µ ≠ 85.

03

Obtaining the rejection region

(b)

The rejection region for the two-tailed test lies at both the tail of the test statistic distribution.

The z-critical value at a 10% level of significance for a two-tailed test is:

The critical values are obtained by using the z-table.

Therefore, reject the null hypothesis if z < -1.64 or z > 1.64.

04

Computing the test statistic

(c)

In a hypothesis testing problem, when a sufficiently large sample is drawn from a population z-test is used.

The test statistic is

The value of the test statistic is z = 12.7986.

05

Conclusion

(d)

Referring to part c., the test statistic is z = 12.7986. The rejection region is z < -1.65 or z > 1.65.

The value of the test statistic is checked; if it lies in the rejection region, the null hypothesis is rejected.

The test statistic 12.7986 lies in the rejection region; hence the null hypothesis is rejected.

Therefore, there is not enough evidence to support the director’s claim that all purchasing managers' true mean Mach rating score is 85.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample of five measurements, randomly selected from a normally distributed population, resulted in the following summary statistics: \(\bar x = 4.8\), \(s = 1.3\) \(\) .

a. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis, µ<6. Use\(\alpha = .05.\)

b. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis, µ\( \ne 6\). Use\(\alpha = .05.\)

c. Find the observed significance level for each test.

Refer to Exercise 6.44 (p. 356), in which 50 consumers taste-tested a new snack food. Their responses (where 0 = do not like; 1 = like; 2 = indifferent) are reproduced below

  1. Test \({H_0}:p = .5\) against \({H_0}:p > .5\), where p is the proportion of customers who do not like the snack food. Use \(\alpha = 0.10\).
    1 0 0 1 2 0 1 1 0 0 0 1 0 2 0 2 2 0 0 1 1 0 0 0 0 1 0 2 0 0 0 1 0 0 1 0 0 1 0 1 0 2 0 0 1 1 0 0 0 1

Calories in school lunches. A University of Florida economist conducted a study of Virginia elementary school lunch menus. During the state-mandated testing period, school lunches averaged 863 calories (National Bureau of Economic Research, November 2002). The economist claims that after the testing period ends, the average caloric content of Virginia school lunches drops significantly. Set up the null and alternative hypotheses to test the economist’s claim.

Revenue for a full-service funeral. According to the National Funeral Directors Association (NFDA), the nation's 19,000 funeral homes collected an average of \(7,180 per full-service funeral in 2014 (www.nfda.org). A random sample of 36 funeral homes reported revenue data for the current year. Among other measures, each reported its average fee for a full-service funeral. These data (in thousands of dollars) are shown in the following table.

a. What are the appropriate null and alternative hypotheses to test whether the average full-service fee of U. S. funeral homes this year is less than \)7,180?

b. Conduct the test at\(\alpha = 0.05\). Do the sample data provide sufficient evidence to conclude that the average fee this year is lower than in 2014?

c. In conducting the test, was it necessary to assume that the population of average full-service fees was normally distributed? Justify your answer

Minimizing tractor skidding distance. Refer to the Journal of Forest Engineering (July 1999) study of minimizing tractor skidding distances along a new road in a European forest, Exercise 6.37 (p. 350). The skidding distances (in meters) were measured at 20 randomly selected road sites. The data are repeated below. Recall that a logger working on the road claims the mean skidding distance is at least 425 meters. Is there sufficient evidence to refute this claim? Use \(\alpha = .10\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free