Question

The auditor for a large corporation routinely monitors cash disbursements. As part of this process, the auditor examines check request forms to determine whether they have been properly approved. Improper approval can occur in several ways. For instance, the check may have no approval, the check request might be missing, the approval might be written by an unauthorized person, or the dollar limit of the authorizing person might be exceeded.

(a) Last year the corporation experienced a 5 percent improper check request approval rate. Since this was considered unacceptable, efforts were made to reduce the rate of improper approvals. Letting p be the proportion of all checks that are now improperly approved, set up the null and alternative hypotheses needed to attempt to demonstrate that the current rate of improper approvals is lower than last year's rate of 5 percent. (Round your answers to 2 decimal places.)

H₀: p > ______ versus H_a: p < ______.

(b) Suppose that the auditor selects a random sample of 617 checks that have been approved in the last month. The auditor finds that 19 of these 617 checks have been improperly approved. Calculate the test statistic. (Round your answers to 2 decimal places. Negative value should be indicated by a minus sign.)

z

(c) Find the p-value for the test of part b. Use the p-value to carry out the test by setting a equal to .10, .05, .01, and .001. Interpret your results. (Round your answer to 4 decimal places.)

p-value                 

Reject H₀ at α = (Click to select).1, and .05/ .1, .05 and .01/ .10, .05, .01, and .001/ none.

Answer 1

1. Consider a sample of size, n, employed to investigate whether the proportion of a particular feature possessed by the population members is different from the presumed proportion p. Let be the sample proportion.

   So, the null hypothesis in this case is

   against one of the following alternative hypotheses:

   

   In such a situation, the test statistic to test the hypothesis is

   .

   At a level of significance, , we have the decision criteria to reject the null hypothesis if p-value is less than. Here, the p-value is defined as the probability, computed assuming that the null hypothesis is true, or observing a value of the test statistic that is at least as contradictory to and supportive of as the value actually computed from the sample data.

   Another way to define the decision criteria is to compare the absolute value of the test statistic with the absolute value of. The value of is such that. We can reject the null hypothesis if.

2. If the is rejected at level of significance of

3. • 0.10, we have some evidence that is false.

   • 0.05, we have strong evidence that is false.

   • 0.01, we have very strong evidence that is false.

   • 0.001, we have extremely strong evidence that is false.

4. (a) Let the current proportion of the improper check request approval be denoted by p.

   We are interested in testing whether the current proportion has significantly reduced from the observed proportion of 5 percent during the last audit. Let us construct null and alternative hypotheses that serve the purpose.

   

   In other words, the proportion of the improper check request approvals has remained constant since the last audit.

   

   In other words, the proportion of the improper check request approvals has reduced since the last audit.

5. H0 : P0 >= 0.05

   Ha : P < 0.05

6. (b) The auditor has found that 18 of 625 checks have been approved improperly. We are interested in testing the hypotheses stated in the solution of Part (a).

7. )p = 19/617

   = 0.030794

   Let us compute the value of the test statistic using formula.

8. Z = (0.030794 - 0.05)/(sqrt(0.05*0.95/617)) = -1.685

   We can reject null hypothesis if , . Let us consider various levels of significance and conclude whether the proportion has reduced significantly.

9. Table1:

   		Decision When Z = -1.685
   0.1	-1.2816	Reject
   0.05	-1.6449	Reject
   0.01	-2.3263	Reject
   0.001	-3.0902	Do not Reject

   So, from Table 1, we can observe that the minimum level of significance at which we have rejected the null hypothesis is 0.01. Hence, we have very strong evidence that is false.

   Thus, we can conclude that the proportion of improper check request approvals has significantly reduced since the last audit.

10. c) Let us now compute the p-value for the observed test statistic. The decision criterion is to reject the null hypothesis if the p-value is less than. As per the definition, the p-value for the given alternative test statistic would be

11. Let us consider various levels of significance and conclude whether the proportion has reduced significantly.

    So, , we can observe that the minimum level of significance at which we have rejected the null hypothesis is 0.01. Hence, we have very strong evidence that is false.

    Thus, we can conclude that the proportion of improper check request approvals has significantly reduced the since last audit.

    critical values :

    for 0.01 is -1.32

    for 0.001 is -2.3263

    p value is : 0.0307

    hence we reject H0 at alpha = 0.1,0.05 , 0.01 , 0.001