The manager of a state-wide discount chain would like to estimate the mean amount customers spend in the large appliance department. A random sample of n=36 sales records had a sample mean amount x =1,520 with standard deviation s=$825. Let µ be the population mean expenditure in the large appliance department. Find a 90% confidence interval for µ .
Sol:
xbar=1520
n=36
s=825
df=n-1=36-1=35
let mu be true mean expenditure in the large appliance department.
t alpha/2==T.INV(0.05;35)=1.68957
90% confidence interval for µ is
xbar-t*s/sqrt(n),xbar+t*s/sqrt(n)
1520-1.68957*825/sqrt(36),1520+1.68957*825/sqrt(36)
1287.684,1752.316
lower limit=1287.684
upper limit=1752.316
we are 90% confident that the true population mean expenditure in the large appliance department lies in between
$1287.684 to $1752.316
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