Question

A regional CPA firm conducted an audit for a discount chain. One part of the audit involved developing an estimate for the mean dollar error in total charges that occur during the checkout process. They wish to develop a 90% confidence interval estimate for the population mean. A simple random sample of n = 20 is selected and the data is given in file “audit.csv”.

Develop a 90% confidence interval estimate for the population mean.

Q1. What is the lower bound?
a) 0.30 b) 0.50 c) 1.23 d) 3

Q2. What is the upper bound? a) 0.31 b) 0.50 c) 1.23 d) 3

Data: Data: https://docs.google.com/spreadsheets/d/1FlddRpUPLjcl0Lx0gd8iK6HdzKR1ssawfssVZ12ObKA/edit?usp=sharing

Answer 1

Mean of the charges = Sum of all the charnges / Number of charges

= 15.41 / 20

= 0.7705

Standard Deviation S = [ (xi - )2 ] / n -1

= 27.0691 / 19

= 1.424689

= 1.193603

90% confidence interval estimate for the population mean = t-score * (S / )

n = 20

Degrees of freedom = n -1

= 19

The t-score for 90% confidence level and 19 degrees of freedom from the below attached t-table = 1.729

90% confidence interval estimate for the population mean = t-score * (S / )

= 0.7705 1.729 * (1.193603 / )

= 0.7705 1.729 * 0.266898

= 0.7705 0.461466

= (0.309034, 1,231966)

Question (1)

So the lower bound is 0.30

Answer is Option A

Question (1)

the upper bound is 1.23

Answer is Option C