To evaluate if auditors might be helped in determining the chances of fraud if t
ID: 3314286 • Letter: T
Question
To evaluate if auditors might be helped in determining the chances of fraud if they carefully measure cash flow, samples of midlevel auditors from CPA firms were asked to indicate the chance of material fraud on a scale from 0 to 100 for a case. A random sample of 41 auditors used the cash-flow information. Their mean assessment was 32.88, and the sample standard deviation was 22.26. For an independent random sample of 41 auditors not using the cash-flow information, the sample mean and standard deviation were, respectively, 49.83 and 26.99. Assuming that the two population distributions are normal with equal variances, test against a two-sided alternative the null hypothesis that the population means are equal. (Use =0.10.) Click the icon to view a table of critical values for the Students t-distribution. Let be the mean assessment for auditors who measured cash flow and y be the mean assessment for auditors not using the cash flow information. Determine the null and alternative hypotheses. Choose the correct answer below. The test statistic is t.(Round to three decimal places as needed.) The critical value(s) is(are) (Round to three decimal places as needed. Use a comma to separate answers as needed.) There is evidence of a difference in the two population means Since the test statistic is less than-tnx + ny-2, /2 greater than-tn,' ny-2, greater than tn,' ny-2, , less than tn,' ny-2, , less than-tn,' ny_2. greater than tnx + ny-2, /2 Click to select your answ between-tn-n,-2, /2 a nd tnx + ny-2,a/2Explanation / Answer
Given that,
mean(x)=32.88
standard deviation , s.d1=22.26
number(n1)=41
y(mean)=49.83
standard deviation, s.d2 =26.99
number(n2)=41
null, Ho: ux - uy = 0
alternate, ux - uy != 0
level of significance, = 0.1
from standard normal table, two tailed t /2 =1.664
since our test is two-tailed
reject Ho, if to < -1.664 OR if to > 1.664
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (40*495.5076 + 40*728.4601) / (82- 2 )
s^2 = 611.9839
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=32.88-49.83/sqrt((611.9839( 1 /41+ 1/41 ))
to=-16.95/5.4638
to=-3.1022
| to | =3.1022
critical value
the value of |t | with (n1+n2-2) i.e 80 d.f is 1.664
we got |to| = 3.1022 & | t | = 1.664
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -3.1022 ) = 0.0026
hence value of p0.1 > 0.0026,here we reject Ho
ANSWERS
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a. Option E, null, Ho: ux - uy = 0, alternate, H1: ux - uy != 0
test statistic: -3.102
critical value: -1.664 , 1.664
greater than, less than -t(nx+ny-2),alpha, suffcient
decision: reject Ho
p-value: 0.0026