Question

8.117 Trading skills of institutional investors. The Journal of Finance (Apr. 2011) published the results of an analysis of trading skills of institutional investors. 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 ran- dom 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 perfor- mance of institutional investors is a small variance in the rates of return of round-trip trades, say, a standard devia tion of less than 10%. a. Specify the null and alternative hypotheses for deter- mining whether the population of institutional inves- tors performs consistently. b. Find the rejection region for the test using a = .05. c. Interpret the value of a 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. Test and Cl for One Variance Method Null hypothesis Sigma - 10 Alternative hypothesis Sigma < 10 The chi-square method is only for the normal distribution. The Bonett method cannot be calculated with summarized data. Statistics N St Dev Variance 2008.82 77. Tests Method Chi-Square Test Statistic DF 154.81 199 P-value 0.009

Compute the test statistic and state the test result.

Compute a p-value

Answer 1

A chi-square test can be used to test if the variance of a population is equal to a specified value. Here we are testing if the population of institutional investors performs consistently i.e if there is small variance in the rates of returns in the round trip trades also it is specified by 10%

a) So, null hypothesis, H0 : $\sigma$=10 vs alternative hypothesis, H1 :$\sigma$$<$10

b) rejection region :

Reject the null hypothesis that the variance is a specified value, σ₀², if

T<χ2$\alpha ,N-1$

for a lower one-tailed alternative

where N is the sample size , $\alpha$=0.05

Test statistic is :

T=(N−1)(s/$\sigma 0$)2

where N is the sample size and s is the sample standard deviation=8.82, $\sigma 0$=specified value =10

c) the alpha level is the probability of rejecting the null hypothesis when the null hypothesis is true. Translation: It's the probability of making a wrong decision.

d) in the print out of MINITAB, the test statistic value is T=154.81 and p-value is 0.009

e) Conclusion : As the p-value =0.009<0.05=alpha, we reject H0 and conclude there is small variance (<10) in the rates of returns in the round trip trades.

f) Assumptions : it requires the assumption that the samples are distributed normally. Moreover, it also assumed that the observations are independent and identically distributed.

COMPUTATION OF TEST STATISTICS AND P-VALUE :

T=(200-1)(8.82/10)^2=154.80

p-value = P ( χ² < 154.80)=0.009