please I need help with question from A to E 8. (10 points) In light of recent c

Question

please I need help with question from A to E

8. (10 points) In light of recent corporate bankruptcies and possibility of fraud, auditors are being the asked to be extra careful when considering a bankruptcy case in terms of the possibility of fraud being committed by the bankrupt firms. To evaluate this auditors have been asked to examine the nature and intensity of the cash flows of these (bankrupt) To the possibility of fraud, auditors from KPMG very large accounting firm were presented with cash flow information from an actual fraud case, and they were asked to indicate on a scale from 0 to 100 the chance of the case being fraudulent. The closer to 100 is their the the likelihood they it is a of fraud. A random sample of 36 auditors using the cash flow information yielded a sample mean fraud assessment of 36.21 with a sample standard deviation of 22.93. In another independent random sample of 36 auditors not using the cash flow information (y) yielded a sample mean fraud assessment of 47.56 with sample standard deviation of 27.56. Assume that the two population distributions are normal with equal variances. a) (2 points) State an appropriate null and alternative hypothesis for difference between the means to whether using the cash flow information has any impact on determining if a particular bankruptcy case is fraudulent. b) (2 points) Calculate the appropriate statistic to test the hypotheses in part (a) and state the critical value for a 0.05. c (2 points) Does looking at the cash flows of the potential fraud cases help in determining fraud? d) would your answers to (b and (c) change if the sample values given were the population values e) points) verify whether or not the assumption of the equality of variances is warranted at a level of significance of 2%

Explanation / Answer

Given that,
mean(x)=36.21
standard deviation , s.d1=22.93
number(n1)=36
y(mean)=47.56
standard deviation, s.d2 =27.56
number(n2)=36
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =1.994
since our test is two-tailed
reject Ho, if to < -1.994 OR if to > 1.994
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (35*525.7849 + 35*759.5536) / (72- 2 )
s^2 = 642.6693
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=36.21-47.56/sqrt((642.6693( 1 /36+ 1/36 ))
to=-11.35/5.9753
to=-1.8995
| to | =1.8995
critical value
the value of |t | with (n1+n2-2) i.e 70 d.f is 1.994
we got |to| = 1.8995 & | t | = 1.994
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -1.8995 ) = 0.0616
hence value of p0.05 < 0.0616,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.8995
critical value: -1.994 , 1.994
decision: do not reject Ho
p-value: 0.0616

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.