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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

Cooling method for gas turbines. During periods of high electricity demand, especially during the hot summer months, the power output from a gas turbine engine can drop dramatically. One way to counter this drop in power is by cooling the inlet air to the gas turbine. An increasingly popular cooling method uses high-pressure inlet fogging. The performance of a sample of 67 gas turbines augmented with high-pressure inlet fogging was investigated in the Journal of Engineering for Gas Turbines and Power (January 2005). One performance measure is heat rate (kilojoules per kilowatt per hour). Heat rates for the 67 gas turbines are listed in the table below. Suppose that standard gas turbines have heat rates with a standard deviation of 1,500 kJ/kWh. Is there sufficient evidence to indicate that the heat rates of the augmented gas turbine engine are more variable than the heat rates of the standard gas turbine engine? Test using a = .05.

A simple random sample of 25 observations was selected from a normal population. The mean and standard deviation of this sample are 20 and 5, respectively.

a. Test H0:渭=22against Ha:渭鈮�22at the 10% significance level.

b. Test H0:渭鈮�22against H0:渭鈮�22at the 1% significance level.

A random sample of n observations is selected from a normal population to test the null hypothesis that 碌=10.Specify the rejection region for each of the following combinations of \(Ha,\alpha ,\) and n:

a.\(Ha:\)碌\( \ne 10;\alpha = .05.;n = 14\)

b.\(Ha:\)碌\( > 10;\alpha = .01;n = 24\)\(\)

c.\(Ha:\)碌\( > 10;\alpha = .10;n = 9\)

d.\(Ha:\)碌 <\(10:\alpha = .01;n = 12\)

e.\(Ha:\)碌\( \ne 10;\alpha = .10;n = 20\)

f. \(Ha:\)碌<\(10;\alpha = .05;n = 4\)

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.

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.
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