An inspector for the IRS is auditing a mid-sized firm. She only has half a day t

Question

An inspector for the IRS is auditing a mid-sized firm. She only has half a day to go through all the books and decides to inspect a sample of the invoices to see if she should start a fraud investigation (which would free up additional auditing resources). As a first step, she pulls 300 invoices at random for further inspection. 12 of these show inaccuracies.

1. Can she assume that the sample distribution is (approximately) normal?

2. Let’s assume the IRS normally only launches a full scale investigation when they suspect more than 3% of all invoices contain errors. Given the 12 invoices she found in her sample, what is the probability that more than 3% of all invoices have inaccuracies?

3. The IRS agent wants to report the 99.5% confidence interval for the proportion of incorrect invoices. Should she use a z table or t table and what z or t value should she use?

4. Calculate the 99.5% confidence interval for the proportion of incorrect invoices?

Explanation / Answer

1. Can she assume that the sample distribution is (approximately) normal?

Yes , she assume because the sample is greater than 30 and proportion is 12/300 = 0.04 is significant

2)

Probability that more than 3% of all invoices have inaccuracies is 0.04

3)

use Z value because is a proportion sample so it is normally distributed

4)

alpha / 2 = 0.0025    Z=2.80

0.04 +/- 2.80 SRQT (0.04*0.96 / 300 )

0.04 +/- 0.0317

0.0083 < p < 0.0717

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