Question

Trading skills of institutional investors. Refer to The Journal of Finance (April 2011) analysis of trading skills of institutional investors, Exercise 7.36 (p. 410). Recall that the study focused on “round-trip” trades, i.e., trades in which the same stock was both bought and sold in the same quarter. In a random sample of 200 round-trip trades made by institutional investors, the sample standard deviation of the rates of return was 8.82%. One property of a consistent performance of institutional investors is a small variance in the rates of return of round-trip trades, say, a standard deviation of less than 10%.

a. Specify the null and alternative hypotheses for determining whether the population of institutional investors performs consistently.

b. Find the rejection region for the test using$\mathit{\alpha }\mathbf{=}\mathbf{.}\mathbf{05}$

c. Interpret the value of in the words of the problem.

d. A Minitab printout of the analysis is shown (next column). Locate the test statistic and$\mathit{p}\mathbf{}\mathbf{-}\mathbf{}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{e}$ on the printout.

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

f. What assumptions about the data are required for the inference to be valid?

Short answer

a. $H_0:{\sigma }^2=100,H_a:{\sigma }^2<100$

b. ${\chi }^2<167.361$

c. The chance of concluding that $\sigma <0.01$when actually $\sigma =10$is 0.05.

d. ${\chi }^2=154.71,p-value=0.009$

e. Reject $H_0$

f. The rates of return should follow a normal distribution.

Step-by-step solution

Specifying the null and the alternative hypothesis

The null hypothesis is stated as follows:

$H_0:{\sigma }^2=100$(The population of institutional investors does not perform consistently).

The alternate hypothesis is stated as follows:

$H_a:{\sigma }^2<100$(The population of institutional investors performs consistently).

Finding the rejection region for the test

Using $\alpha =0.05$and $(n-1)=199$, ${\chi }^2$value of rejection is found in table IV of Appendix D. However, since we have to determine the lower tail value, which has $\alpha =0.05$to its left, we used the column ${\chi }_{0.95}^2$in table IV.

Therefore the rejection region is given by ${\chi }^2<167.361$.

Interpreting the value of  α

The value of $\alpha$means that the chance of concluding that the alternative hypothesis is true when actually the null hypothesis is true is 0.05.

In the words of the problem it means, chance of concluding that $\sigma <10$when actually $\sigma =10$is 0.05.

Locating the test statistic and the p - value from the Minitab printout.

From the Minitab printout, we find,

The test statistic is ${\chi }^2=154.71$and the $p-value=0.009$.

Giving the conclusion in the words of the problem.

Since the test statistic gives us a value of ${\chi }^2=154.71$which does not fall in the rejection region, therefore we will reject the null hypothesis $H_0:{\sigma }^2=100$which says the population of institutional investors does not perform consistently.

Therefore, the alternative hypothesis $H_a:{\sigma }^2<100$,which specifies that the population of institutional investors performs consistently, is accepted.

Stating the assumption required for the inference to be valid.

The assumption required is that the rates of return must follow a normal distribution for the inference to be valid.