Question

2. A simple random sample of 50 calls is monitored at an in-bound call center and the average length of the calls is 6 minutes. The population standard deviation σ is unknown. Instead the sample standard deviation s is also calculated from the sample and is found to be 4 minutes.

a. Construct a 99% confidence interval (using the t-distribution) for the average length of inbound calls.

b. What is the margin of error at the 95% confidence level?

Answer 1

Solution:

Given that,

   = 6 ....... Sample mean

s = 4 ........Sample standard deviation

n = 50 ....... Sample size

Note that, Population standard deviation() is unknown..So we use t distribution.

a)

Our aim is to construct 99% confidence interval.   

c = 0.99

= 1 - c = 1 - 0.99 = 0.01

  /2 = 0.01 2 = 0.005

Also, d.f = n - 1 = 50 - 1 = 49

    =    =  _0.005,49 = 2.680

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.680 * (4 / 50)

= 1.5160

Now , confidence interval for mean() is given by:

( - E ) <   <  ( + E)

(6 - 1.5160)   <   <  (6 + 1.5160)

4.4840 <   < 7.5160

Required 99% confidence interval is (4.4840 , 7.5160)

b)

For 95% confidence level ,

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, d.f = n - 1 = 50 - 1 = 49

    =    =  _0.025,49 = 2.010  

( use t table or t calculator to find this value..)

The margin of error is given by

E =  /2,d.f. * ( / n )

= 2.010* (4 / 50 )

= 1.1370
At 95% level , the margin of error is 1.1370