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 45.5 and 7.5 hours, respectively. Construct a 92% confidence interval for the average number of hours this firm’s employees spend on work-related activities in a typical week.

Place your LOWER limit, in hours, rounded to 1 decimal place, in the first blank. For example, 6.7 would be a legitimate entry

Place your UPPER limit, in hours, rounded to 1 decimal place, in the first blank

Answer 1

Given Sample Mean = 45.5

Sample Standard Deviation S = 7.5

Sample Size n = 50

Here we need to calculate 92% confidence interval

Confidence level = 92% = 0.92

= 1 - confidence level

= 1 - 0.92

= 0.08

/2 = 0.04

So Z/2 will be a z-score that has an area of 0.04 to its right or 0.96 to its left

The z-score that has an area of 0.96 to its left is 1.750686

92% confidence Interval for avergae number of hours this firm’s employees spend on work-related activities in a typical week =   Z/2 * ( S / )

= 45.5 1.750686 * (7.5 / )

= 45.5 1.750686 * (7.5 / 7.071068)

= 45.5 1.750686 * 1.06066

= 45.5 1.856883

= (43.64312, 47.35688)

= (43.6, 47.4) rounded to 1 decimal place

So Lower limit, in hours = 43.6

Upper limit, in hours = 47.4