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 said 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 finds that 104 contained errors.

a) Construct a 90% confidence interval for this company's error rate.

b) How should the company be rated if statements about being out of compliance or seriously out of compliance require 5% level of significance?

c) 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?

d) What sample size should the auditor use to estimate the error rate to within 2% with 95% confidence if it is assumed that the error rate will be no more than 15%?

e) Suppose the 104 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/InvoiceErr.txt
Note: since this file just contains a single set of numeric values, you can use the scan() function in R to read this data. For example,

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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