Question

Senior management of a consulting services firm is concerned about a growing decline in the firm’s weekly number of billable hours. The firm expects each professional employee to spend at least 40 hours per week on work. In an effort to understand this problem better, management would like to estimate the standard deviation of the number of hours their employees spend on work-related activities in a typical week. Rather than reviewing the records of all the firm’s full-time employees, the management randomly selected a sample of size 50 from the available frame. The sample mean and sample standard deviations were 46.4 and 7.2 hours, respectively.

Construct a 97% confidence interval for the average of the number of hours this firm’s employees spend on work-related activities in a typical week.

PLACE LOWER LIMIT,in hours, rounded to 1 decimal place.

PLACE UPPER LIMIT, in hours, rounded to 1 decimal place

Answer 1

Level of Significance ,    α =    0.03          
degree of freedom=   DF=n-1=   49          
't value='   tα/2=   2.235   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   7.2000   / √   50   =   1.0182
margin of error , E=t*SE =   2.2351   *   1.0182   =   2.2759
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    46.40   -   2.275878   =   44.1
Interval Upper Limit = x̅ + E =    46.40   -   2.275878   =   48.7
97%   confidence interval is (   44.1 < µ <   48.7 )