Trading skills of institutional investors. The trading skills of institutional stock investors were quantified and analyzed in The Journal of Finance (April 2011). The study focused on “round-trip” trades, i.e., trades in which the same stock was both bought and sold in the same quarter. Consider a random sample of 200 round-trip trades made by institutional investors. Suppose the sample mean rate of return is 2.95% and the sample standard deviation is 8.82%. If the true mean rate of return of round-trip trades is positive, then the population of institutional investors is considered to have performed successfully.

a. Specify the null and alternative hypotheses for determining whether the population of institutional investors performed successfully.

b. Find the rejection region for the test using\(\alpha = 0.05\).

c. Interpret the value of\(\alpha \)in the words of the problem.

d. A Minitab printout of the analysis is shown below. Locate the test statistic and p-value on the printout. (Note: For large samples, z ≈ t.)

e. Give the appropriate conclusion in the words of the problem.

Short Answer

Expert verified

a.\({H_0}:\mu = 0\)and\({H_a}:\mu > 0\)are the null and the alternative hypothesis.

b. The rejection region is \({z_c} > 1.645\).

c. The probability of the true mean rate of return of round-trip trades is positive.

d.From the MINITAB output, the test statistic is 4.73, and the p-value is 0.000.

e. It can be concluded that the population of institutional investors did not perform successfully at the significance level \(\alpha = 0.05\).

Step by step solution

01

Given information

The sample size is 200, the sample mean is 2.95, and the sample standard deviation is 8.82.

Also, MINITAB output is as follows

02

Setting up the null and alternative hypothesis

a.

Null hypothesis:

\({H_0}:\mu = 0\)

That is, the population of institutional investors performed successfully.

Alternative hypothesis:

\({H_a}:\mu > 0\)

That is, the population of institutional investors did not perform successfully.

03

Finding the rejection region

b.

Here, the test is the right tail, and the significance level is 0.05. The critical value \({z_{0.05}}\) is obtained from the standard normal table.

Thus, the required \({z_{0.05}}\) critical value is 1.645.

The rejection region for the right tail test is\({z_c} > {z_a}\)

Hence, the rejection region is \({z_c} > 1.645\).

04

Interpreting the value of \(\alpha \) \(\)

c.

The probability of Type I error is denoted as\(\alpha \). In other words, it is the probability of the error committed to rejecting a null hypothesis\(\left( {{H_0}} \right)\)when it is true.

In the study, the interpretation \(\alpha \) is that the probability of the population of institutional investors did not perform successfully. That is, the probability of the true mean rate of return of round-trip trades is positive.

05

Finding the test statistic and the p-value

d.

From the MINITAB output, the test statistic is 4.73, and the p-value is 0.000.

06

Conclusion

e.

Here, the p-value is 0.000, which is less than the value \(\alpha = 0.05\).

If the p-value <\(\alpha \), then the null hypothesis is rejected.

Hence, Reject the null hypothesis\({H_0}\)

Thus, it can be concluded that the population of institutional investors did not perform successfully at the significance level\(\alpha = 0.05\).

That is, the true mean rate of return of round-trip trades is positive. \(\)

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

Consider the test \({H_0}:\mu = 70\) versus \({H_a}:\mu \ne 70\) using a large sample of size n = 400. Assume\(\sigma = 20\).

a. Describe the sampling distribution of\(\bar x\).

b. Find the value of the test statistic if\(\bar x = 72.5\).

c. Refer to part b. Find the p-value of the test.

d. Find the rejection region of the test for\(\alpha = 0.01\).

e. Refer to parts c and d. Use the p-value approach to

make the appropriate conclusion.

f. Repeat part e, but use the rejection region approach.

g. Do the conclusions, parts e and f, agree?

Refer to Exercise 7.99.

a. Find b for each of the following values of the population mean: 74, 72, 70, 68, and 66.

b. Plot each value of b you obtained in part a against its associated population mean. Show b on the vertical axis and m on the horizontal axis. Draw a curve through the five points on your graph.

c. Use your graph of part b to find the approximate probability that the hypothesis test will lead to a Type II error when m = 73.

d. Convert each of the b values you calculated in part a to the power of the test at the specified value of m. Plot the power on the vertical axis against m on the horizontal axis. Compare the graph of part b with the power curve of this part.

e. Examine the graphs of parts b and d. Explain what they reveal about the relationships among the distance between the true mean m and the null hypothesized mean m0, the value of b, and the power.

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

Suppose the sample in Exercise 7.64 has produced \(\hat p = .83\) and we wish to test \({H_0}:P = 0.9\) against the alternative \({H_a}:p < .9\)

a. Calculate the value of the z-statistic for this test.

A border protection avatar. The National Center for Border Security and Protection has developed the "Embodied Avatar"—a kiosk with a computer-animated border guard that uses artificial intelligence to scan passports, check fingerprints, read eye pupils, and asks questions of travellers crossing the U.S. border. (National Defense Magazine, February 2014.) Based on field tests, the avatar's developer claims that the avatar can detect deceitful speech correctly 75% of the time.

a. Identify the parameter of interest.

b. Give the null and alternative hypotheses for testing the claim made by the avatar's developer.

c. Describe a Type I error in the words of the problem.

d. Describe a Type II error in the words of the problem

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