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1. To compare the stock-picking abilities of two brokerage firms, we compared th

ID: 3175502 • Letter: 1

Question

1.         To compare the stock-picking abilities of two brokerage firms, we compared the annual gain (excluding brokerage fees) for a $1000 investment in each of 30 stocks listed on each of the two firm's "most recommended" lists of stocks. The means and standard deviations (in dollars) for each of the two samples are shown in the accompanying table. Do these data provide sufficient evidence to indicate a difference in the mean return per recommended stock for the two brokerage firms?

                             

                             

FIRM

SAMPLE INFORMATION

ABC

XYZ

Sample Size

30

30

Sample Mean

269

199

Sample Standard deviation

107

111

a.         State the null and alternative hypotheses that will best answer this question.

                        Ho :

                        H1 :

            b.         Calculate the test statistic for the above test.

            c.         Give the decision rule for the test using = .01 on the graph. (Indicate the critical value(s) and the rejection and acceptance regions.)

            d.         Conclusion.

                             

                             

FIRM

SAMPLE INFORMATION

ABC

XYZ

Sample Size

30

30

Sample Mean

269

199

Sample Standard deviation

107

111

Explanation / Answer

Given that,
mean(x)=269
standard deviation , s.d1=107
number(n1)=30
y(mean)=199
standard deviation, s.d2 =111
number(n2)=30
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.756
since our test is two-tailed
reject Ho, if to < -2.756 OR if to > 2.756
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =269-199/sqrt((11449/30)+(12321/30))
to =2.487
| to | =2.487
critical value
the value of |t | with min (n1-1, n2-1) i.e 29 d.f is 2.756
we got |to| = 2.48682 & | t | = 2.756
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.4868 ) = 0.019
hence value of p0.01 < 0.019,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.487
critical value: -2.756 , 2.756
decision: do not reject Ho
p-value: 0.019

no sufficient evidence to indicate a difference in the mean return per recommended stock
for the two brokerage firms