Answer to Question #349640 in Statistics and Probability for PEECAY

Question

The operations manager of a sugar mill in Durban wants to estimate the average size of an order by constructing a 90% confidence interval. An order is measured in the number of pallets shipped. A random sample of 87 orders from customers had an average of 131.6 pallets. Based on information from past records, the operations manager assumes that the number of pallets shipped is normally distributed with a standard deviation of 25 pallets.

What is the confidence interval constructed by the operations manager?

(Round off to the nearest whole numbers.)

Answer 1

The critical value for $\alpha = 0.1$ is $z_c = z_{1-\alpha/2} = 1.6449.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}},\bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$ $$=(131.6-1.6449\times\dfrac{25}{\sqrt{87}},131.6+1.6449\times\dfrac{25}{\sqrt{87}})$$ $$=(127, 136)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $127 < \mu < 136 ,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(127,136).$