The operations manager of a sugar mill in Chesterville wants to estimate the average size of
an order received. An order is measured in the number of pallets shipped from overseas. A
random sample of 84 orders from customers had a sample mean value of 121.6 pallets.
Assume that the population standard deviation is 20 pallets and that order size is normally
distributed. Estimate, with 92% confidence, the mean size of orders received from all the mill’s
customers.
When the population standard deviation is known, the formula for a confidence interval (CI) for a population mean is x̄ ± z* σ/√n, where x̄ is the sample mean, σ is the population standard deviation, n is the sample size, and z* represents the appropriate z*-value from the standard normal distribution for your desired confidence level.
We have: "\\bar{x}=121.6" - the sample mean;
"\\sigma =20" - population standard deviation ;
n=84- size of sample;
z*=norminv(0.96,0,1)=1.75- z value for 0.92 confidence interval, this value we calculate with Excel;
Lower bound of confidence interval is
x1=121.6-1.75"\\cdot 20\\cdot \\frac{1}{\\sqrt{84}}=117.78"
Upper bound is
x2=121.6+1.75"\\cdot 20\\cdot \\frac{1}{\\sqrt{84}}=125.42"
Thus 92% confidence interval for population mean of order size is
(117.78; 125.62);
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