Question

In: Statistics and Probability

In examining the credit accounts of a department store, an auditor would like to estimate the...

In examining the credit accounts of a department store, an auditor would like to estimate the true mean account error (book value – audited value). To do so, the auditor selected a random sample of 50 accounts and found the mean account error to be $60 with a standard deviation of $30.
a. Construct a 95% confidence interval for the population mean account error. What conclusion can be made from this confidence

interval?

b. How large a sample is actually needed to ensure with 95% confidence in estimating the population mean account error to within ±$5?

c. Among the 50 selected accounts, the auditor found that there were 6 of them consisted of an account error of more than $100. Based on this information, construct a 99% confidence interval for the population proportion of accounts with an error of more than $100. What conclusion can be made from this confidence interval?

Solutions

Expert Solution

SolutionA:

n=50

xbar=60

sample stddev=s=30

df=degrees of freedom=n-1=50-1=49

alpha=1-0.95=0.05

alpha/2=0.05/2=0.025

t crit=T.INV(0.025;59)=2.00958

95% confidence interval for the population mean account error.

xbar-t*s/sqrt(n),xbar+t*s/sqrt(n)

60-2.00958*30/sqrt(50),60-2.00958*30/sqrt(50)

51.474,68.526

95% lower limit=51.474

95% upper limit=68.526

b. How large a sample is actually needed to ensure with 95% confidence in estimating the population mean account error to within ±$5?

z crit for 95%=1.96

E=margin of error=5

sigma=s=30

n=required sample size

=(z*sigma/E)^2

=(1.96*30/5)^2

=138.2976

=138

required sample size=n=138

Solutionc:

sample proportion=p^=x/n=6/50=0.12

z crit for 99%=2.576

99% confidence interval for the population proportion of accounts with an error of more than $100

=p^-z*sqrt(p^(1-p^)/n,p^+z*sqrt(p^(1-p^)/n

=0.12-2.576*sqrt(0.12*(1-.12)/50,0.12+2.576*sqrt(0.12*(1-.12)/50)

=0.00162,0.23838

0.00162<p<0.23838

99% lower limit=0.00162

99% upper limit=0.23838

we are 99% confident that the true population proportion of accounts with an error of more than $100.

lies in between 0.00162 and 0.23838

orchestra answered 1 year ago

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