Answer to Question #232911 in Statistics and Probability for Veyy

Question

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.

Answer 1

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);