Answer to Question #229016 in Statistics and Probability for nat

Question

1.1 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

The critical value for $\alpha=0.08$ is $z_c=z_{1-\alpha/2}=1.7507.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-z_c\times\dfrac{\sigma}{ \sqrt{n}}, \bar{X}+z_c\times\dfrac{\sigma}{ \sqrt{n}})$$

$$=(121.6-1.7507\times\dfrac{20}{ \sqrt{84}}, 121.6+1.7507\times\dfrac{20}{ \sqrt{84}})$$

$$=(117.78,125.42)$$

Therefore, based on the data provided, the 92 % confidence interval for the population mean is $117.78<\mu<125.42,$ which indicates that we are 92% confident that the true population mean $\mu$ is contained by the interval $(117.78,125.42).$