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 618 checks that have been approved in the last month. The auditor finds that 17 of these 618 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 3 decimal places.)

p-value                  

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

Answer 1

Answer)

A)

Ho : P >= 0.05.

Ha : P < 0.05.

B)

N = 618

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 30.9

N*(1-p) = 587.1

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed P = 17/618.

N = 618

Claimed P = 0.05

After substitution.

Test statistics z = -2.57

C)

From z table, P(z<-2.57) = 0.0051.

We reject the null hypothesis when p-value is less than alpha

For 0.01, 0.05 and 0.1

We will reject Ho.