In: Statistics and Probability
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?
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