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Chapter 7: Q36 E (page 410)

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 鈥渞ound-trip鈥� 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 鈮� 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. \(\)

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Most popular questions from this chapter

Americans鈥� favorite sport. The Harris Poll (December 2013) conducted an online survey of American adults to determine their favorite sport. Your friend believes professional (National Football League [NFL]) football鈥攚ith revenue of about $13 billion per year鈥攊s the favorite sport for 40% of American adults. Specify the null and alternative hypotheses for testing this belief. Be sure to identify the parameter of interest.

Jury trial outcomes. Sometimes, the outcome of a jury trial defies the 鈥渃ommon sense鈥� expectations of the general public (e.g., the 1995 O. J. Simpson verdict and the 2011 Casey Anthony verdict). Such a verdict is more acceptable if we understand that the jury trial of an accused murderer is analogous to the statistical hypothesis-testing process. The null hypothesis in a jury trial is that the accused is innocent. (The status-quo hypothesis in the U.S. system of justice is innocence, which is assumed to be true until proven beyond a reasonable doubt.) The alternative hypothesis is guilt, which is accepted only when sufficient evidence exists to establish its truth. If the vote of the jury is unanimous in favor of guilt, the null hypothesis of innocence is rejected, and the court concludes that the accused murderer is guilty. Any vote other than a unanimous one for guilt results in a 鈥渘ot guilty鈥� verdict. The court never accepts the null hypothesis; that is, the court never declares the accused 鈥渋nnocent.鈥� A 鈥渘ot guilty鈥� verdict (as in the Casey Anthony case) implies that the court could not find the defendant guilty beyond a reasonable doubt

a. Define Type I and Type II errors in a murder trial.

b. Which of the two errors is the more serious? Explain.

c. The court does not, in general, know the values of 伪 and 尾 ; but ideally, both should be small. One of these probabilities is assumed to be smaller than the other in a jury trial. Which one, and why?

d. The court system relies on the belief that the value of is made very small by requiring a unanimous vote before guilt is concluded. Explain why this is so.

e. For a jury prejudiced against a guilty verdict as the trial begins, will the value of伪 increase or decrease? Explain.

f. For a jury prejudiced against a guilty verdict as the trial begins, will the value of 尾 increase or decrease? Explain

For each of the following rejection regions, sketch the sampling distribution for z and indicate the location of the rejection region.

a. \({H_0}:\mu \le {\mu _0}\) and \({H_a}:\mu > {\mu _0};\alpha = 0.1\)

b. \({H_0}:\mu \le {\mu _0}\) and \({H_a}:\mu > {\mu _0};\alpha = 0.05\)

c. \({H_0}:\mu \ge {\mu _0}\) and \({H_a}:\mu < {\mu _0};\alpha = 0.01\)

d. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.05\)

e. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.1\)

f. \({H_0}:\mu = {\mu _0}\) and \({H_a}:\mu \ne {\mu _0};\alpha = 0.01\)

g. For each rejection region specified in parts a鈥揻, state the probability notation in z and its respective Type I error value.

If a hypothesis test were conducted using 伪= 0.05, for which of the following p-values would the null hypothesis be rejected?

a. .06

b. .10

c. .01

d. .001

e. .251

f. .042

Producer's and consumer's risk. In quality-control applications of hypothesis testing, the null and alternative hypotheses are frequently specified as\({H_0}\)The production process is performing satisfactorily. \({H_a}\): The process is performing in an unsatisfactory manner. Accordingly, \(\alpha \) is sometimes referred to as the producer's risk, while \(\beta \)is called the consumer's risk (Stevenson, Operations Management, 2014). An injection molder produces plastic golf tees. The process is designed to produce tees with a mean weight of .250 ounce. To investigate whether the injection molder is operating satisfactorily 40 tees were randomly sampled from the last hour's production. Their weights (in ounces) are listed in the following table.

a. Write \({H_0}\) and \({H_a}\) in terms of the true mean weight of the golf tees, \(\mu \).

b. Access the data and find \(\overline x \)and s.

c. Calculate the test statistic.

d. Find the p-value for the test.

e. Locate the rejection region for the test using\({H_a} = 0.01\).

f. Do the data provide sufficient evidence to conclude that the process is not operating satisfactorily?

g. In the context of this problem, explain why it makes sense to call \(\alpha \)the producer's risk and \(\beta \)the consumer's risk.
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