Stocks are classified as conservative (C), moderately aggressive (M), or aggress
ID: 3339257 • Letter: S
Question
Stocks are classified as conservative (C), moderately aggressive (M), or aggressive (A). Suppose a random sample of stocks from each of these categories was selected and the returns tracked over a six-month period.
(a) Examine the data between conservative (C) and moderately aggressive (M) to determine which test would be more appropriate.
(b) Regardless of your answer in (a), use a parametric test to test whether the data show evidence of real differences in average returns between conservative and moderately aggressive stocks. Use a level of significance of 5%. Show your intermediate calculations.
(c) Perform the appropriate non-parametric test using Minitab to test for differences in the median returns between conservative and moderately aggressive stocks. Use a level of significance of 5%. Remember to do a complete test and use Minitab only as a calculator.
returns type return A Return C Return M -32 c 27 -32 -2 57 c -27 57 -29 -4 c -11 -4 18 17 c -17 17 -25 33 c -19 33 -30 22 c -5 22 -58 -2 M 6 -6 -29 M 22 -56 18 M -33 -25 M -30 M -58 M -6 M -56 M 27 A -27 A -11 A -17 A -19 A -5 A 6 A 22 A -33 AExplanation / Answer
1. Here we wish to compare the means of several populations i.e. for different stock returns M,A & C. Thus we can use the Kruskal Wallis test for the given problem.(non-parametric approach)
2. For parametric set up we may use the ANOVA technique to deal with the problem.
Then we assume each population to be normal and we wish to test the hypothesis
H: m1=m2=m3
k: atleast one inequality holds.
then the test statistic is given by
F= (ESS/SSRES)*(23-3)/(3-1)
FROM THE GIVEN DATA,
the explaines sum of squares is obtained as 5082.15
the error or residual sum of squares is obtained as 13209.500
The value of the test statistic is given by 3.847 with p value 0.039
hence the null hypothesis is rejected at 5% level of significance.
i.e the mean returns for the three categories are significantly different from each other
3. we can use the extended median test for the given problem.
the procedure is as follows:
1. arrange the entire data in ascending order
2. determine the median from the combined data
3. count the number of observations above and below the median respectively and compute a 2*2 contingency table.
4. calculate the expected frequencies from the table
5. apply the chi square test statistic, compare it with the chi sq value at al[pha level with k-1 degrees of freedom take decisions accordingly.where k is the no. of classes