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The manager of a state-wide discount chain would like to estimate the mean amount customers spend...

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 .

Math   Statistics-And-Probability   
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Answer #1

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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