Question

In order to ensure efficient usage of a server, it is necessary to estimate the mean number of concurrent users. According to 100 randomly selected times of day, the mean number of concurrent users is 37.7 and the standard deviation is 9.2.

a. Find the 99% confidence interval for the mean number of concurrent users.

b. Write a statement explaining your confidence interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 37.7

Population standard deviation =    = 9.2
Sample size = n =100

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

Margin of error = E = Z/2* ( /n)

=2.576 * (9.2 / 100)

= 2.3699

At 99% confidence interval estimate of the population mean is,

- E < < + E

37.7-2.3699 < < 37.7+2.3699

35.3301< < 40.0699

(35.3301<,40.0699 )

At 99% confidence interval estimate of the population mean is,(35.3301<,40.0699 )