Question

A Telecommunications company recently installed a new speech recognition system for its repair calls. In their old system, operators had to press keys on the numeric keypad to answer questions, which led to longer time on service calls.

The company is conducting a study to determine the effectiveness of the new system and to do so they want to estimate the average duration of service calls with the new system.

They analyzed a random sample of 13 service calls and obtained an average duration of service calls of 33.06 minutes.

Based on historical data, the standard deviation of duration of service calls is known to be σ =6.15 minutes.

What is the margin of error for a 99% confidence interval for the mean duration of service calls ? (USE 3 Decimals)

Answer 1

Solution :

Given that,

= 33.06

s = 6.15

n = 12

Degrees of freedom = df = n - 1 = 13 - 1 = 12

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,12  =3.054

Margin of error = E = t_/2,df * (s /n)

= 3.054 * (6.15 / 13)

= 5.210

Margin of error = 5.210