Question

At a large grocery store, based on a sample of size 15 , the cashier error has a sample mean of   27.6 dollars per day and a standard deviation of   5.0 dollars. Find a 87% confidence interval for the population cashier error. You may assume that the cashier errors are normally distributed.

Answer: We are 87% confident that the true population mean of cashier error lies somewhere between        and       

Answer 1

We have given that,              
              
Sample mean =27.6      
Sample standard deviation =5  
Sample size =15      
Level of significance=1-0.87=0.13      
Degree of freedom =14      
              
t critical value is (by using t table)=   1.609      
$\bar{x}\pm t*\frac{S}{\sqrt{n}}= 27.6 \pm 1.609 *(\frac { 5 } {\sqrt{ 15 }})=( 25.5228 , 29.6772 )$

We are 87% confident that the true population mean of cashier error lies somewhere between 25.5 and 29.7