Question

Trading skills of institutional investors. Managers of stock portfolios make decisions as to what stocks to buy and sell in a given quarter. The trading skills of these institutional 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 trips 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-trips is positive, then the population of institutional investors is considered to have preformed successfully.

a) Specify the null and alternative hypotheses for determining whether the population of institutional investors preformed 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) Give the appropriate conclusion in the words of the problem.

Answer 1

Solution:

Given: a random sample of 200 round trips made by institutional investors is selected.

Thus sample size = n = 200

Sample mean = = 2.95%

Sample standard deviation = s = 8.82%

If the true mean rate of return of round-trips is positive, then the population of institutional investors is considered to have preformed successfully.

We have to test if the population of institutional investors preformed successfully or not.

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

Vs

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

Since sample size = n = 200 is large we use approximate Normal distribution, thus find z critical value.

Since this is right tailed test, look in z table for an Area = 1 - 0.05 = 0.95

and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_critical = 1.645

Thus the rejection region for the test is :

Reject H0, when z test statistic value > Z_critical = 1.645

Part c) Interpret the value of alpha in the words of the problem.

the value of alpha is

the value of alpha is the probability of making type I error.

That means we assume there 5% chance of making type I error.

that is: there is 5% chance we reject null hypothesis H0 when H0 is true.

that is: there is 5% chance that we conclude that the population of institutional investors is considered to have preformed successfully in fact they do not preformed successfully.

Part d) Give the appropriate conclusion in the words of the problem.

Find z test statistic value:

Since population standard deviation is unknown, we use its point estimate sample standard deviation s.

Since z test statistic value = z = 4.73 > z critical value = 1.645 , we reject H0 and thus we conclude that: the population of institutional investors is considered to have preformed successfully