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 51 from the available frame. The sample mean and sample standard deviations were 48.5 and 7.5 hours, respectively. Construct a 99% confidence interval for the standard deviation of the number of hours this firm’s employees spend on work-related activities in a typical week

Answer 1

Answer:  (45.79, 51.21)

Step-by-step explanation:

Given : Significance level : $\alpha: 1-0.99=0.01$

Sample size : n= 51 , which is a large sample (n>30), so we use z-test.

Critical value: $z_{\alpha/2}=2.576$

Sample mean : $\overline{x}= 48.5\text{ hours}$

Standard deviation : $\sigma=7.5\text{ hours}$

The confidence interval for population means is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e. $48.5\pm(2.576)\dfrac{7.5}{\sqrt{51}}$

i.e.$48.5\pm2.70534112234\\\\\approx48.5\pm2.71\\\\=(48.5-2.71, 48.5+2.71)=(45.79, 51.21)$

Hence, 99% confidence interval for the standard deviation of the number of hours this firm’s employees spend on work-related activities in a typical week =  (45.79, 51.21)