Question

There is an average of 22.455 employees at 22 office furniture dealers in a major metropolitan area with a standard deviation of 18.52. Construct a 99% confidence interval for the true mean number of full-time employees at office furniture dealers.

Answer 1

We have given that,          
          
Sample mean =22.455  
Sample standard deviation =18.52
Sample size =22  
Level of significance=0.01  
Degree of freedom =21  
          
t critical value is (by using t table)=2.8314
          
Confidence interval formula is

=(11.28,33.63)          
          
          
          
Lower confidence limit=   11.28  
          
Upper confidence limit=   33.63