Answer to Question #277908 in Statistics and Probability for Amon

Question

A hospital employs 250 nurses. In a random sample of 81 of these, the mean number of

hours overtime billed in a particular week was 12.3, and the sample standard deviation was

8.4 hours. Find a 98% confidence interval for the mean number of hours overtime billed

per nurse in this hospital that week?

Answer 1

Standard error of sample is $s_s = \frac{\sigma}{\sqrt{n}} = \frac{8.4}{\sqrt{81}} \approx 0.93$

From Student's distribution with 11 degrees of freedom and 98% probability get $t_{cr} = 2.718t$

​therefore;$\epsilon = t_{cr} s_s = 2.718 * 0.93= 2.528$

and the confidence interval is (12.3-2.528, 12.3+2.528)= (9.772, 14.828).