In: Statistics and Probability
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.
Answer:
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 )
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