A company is said to be out of compliance if more than 8% of all invoices contai

Question

A company is said to be out of compliance if more than 8% of all invoices contain errors, and it is considered to be seriously out of compliance if more than 12% of all invoices contain errors. Suppose an auditor randomly selects a sample of 800 invoices and found that 110 contained errors. Construct a 95% confidence interval for this company's error rate. How should the company be rated, in compliance, out of compliance, or seriously out of copliance, if statements about being out of compliance require 5% level of significance? What is the probability a company would be rated as seriously out of compliance by this test if 15% of all invoices at that company contain errors? What sample size should the auditor use to estimate the error rate to within 2% with 99% confidence if it is assumed that the error rate will be no more than 12%? Suppose the 200 erroneous invoices can be treated as a random sample from the population of all erroneous invoices. The error amounts are contained in the file http://www.UTDallas.edu/~ammann/stat3355scripts/Invoice.txt Construct a 95% confidence interval for the mean error amount. Also obtain and interpret a quantile-quantile plot of these invoice errors compared to the normal distribution

Explanation / Answer

1. company error rate = 110/800 = 0.1375

so error rate p = 0.1375

so for 95% confidence interval z - value = + - 1.96

so 95% CI = Mean error rate +- 1.96 * sqrt [ p(1-p)/n]

= 0.1375 +- 1.96 * sqrt [ 0.1375 * 0.8625/800]

= 0.1375 +- 0.01218

= (0.1253, 0.1497)

(b) So, COmpany must be rated seriously out of compliances as error rate > 0.12

(c) so here we have to calculate Norm P( p> 0.12; 0.15; sqrt(0.15* 0.85/800) =

Z - value = (0.12 - 0.15)/ sqrt(0.15* 0.85/800) = - 0.03/0.01262 = 2.37

so P( p> 0.12; 0.15; 0.01262) = 1 - 0.0089 = 0.9911

(d) Let the sample size = n

for 99% confidence Z - value = +- 2.575

here error rate will not be more than 12%, so we can say that upper limit of error rate confidence level is 12% , so p can be assumed 0.10 in this case

so 0.02 = 2 * 2.575 * sqrt [0.10 * 0.90/n]

n = 5967.56 that means n =5968 data records must be evaluated.

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