Question

The customer expectation when phoning a customer service line is that the average amount of time from completion of dialing until they hear the message indicating the time in queue is equal to 55.0 seconds (less than a minute was the response from customers surveyed, so the standard was established at 10% less than a minute).  

You decide to randomly sample at 20 times from 11:30am until 9:30pm on 2 days to determine what the actual average is.  The actual data was as follows:

54.1, 53.3, 56.1, 55.7, 54.0, 54.1, 54.5, 57.1, 55.2, 53.8,54.1, 54.1, 56.1, 55.0, 55.9, 56.0 ,54.9, 54.3, 53.9, 55.0

What is a 95% confidence interval for the true mean call completion time?

What is the 95% confidence interval on the standard deviation

The customer expectation when phoning a customer service line is that the average amount of time from completion of dialing until they hear the message indicating the time in queue is equal to 55.0 seconds (less than a minute was the response from customers surveyed, so the standard was established at 10% less than a minute).  

You decide to randomly sample at 20 times from 11:30am until 9:30pm on 2 days to determine what the actual average is.  The actual data was as follows:

54.1, 53.3, 56.1, 55.7, 54.0, 54.1, 54.5, 57.1, 55.2, 53.8,54.1, 54.1, 56.1, 55.0, 55.9, 56.0 ,54.9, 54.3, 53.9, 55.0

What is a 95% confidence interval for the true mean call completion time?

What is the 95% confidence interval on the standard deviation

Answer 1

95% Confidence interval for the true mean call completion times:

From the data (calculations are given in the end): = 54.86, s = 1.0081, n = 20

Since population standard deviation is unknown, the tcritical (2 tail) for = 0.05, for df = n -1 = 19, is 2.093

The Confidence Interval is given by ME, where

The Lower Limit = 54.86 - 0.472 = 54.388

The Upper Limit = 54.86 + 0.472 = 55.332

The 95% Confidence Interval is (54.388 , 55.332)

If Required to 2 decimals the CI will be (54.39, 55.33)

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The 95% CI for Standard Deviation:

s = 1.0081, s² = (1.0081)², n = 20, and degrees of freedom = n - 1 = 19. = 0.05

We find critical values for /2 = 0.025 and 1- /2 = 1 – 0.025 = 0.975

The lower critical value = 32.852 and the upper critical critical value = 8.907

The confidence interval is given by:

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Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)².

#	X	Mean	(x - mean)2
1	54.1	54.86	0.5776
2	53.3	54.86	2.4336
3	56.1	54.86	1.5376
4	55.7	54.86	0.7056
5	54	54.86	0.7396
6	54.1	54.86	0.5776
7	54.5	54.86	0.1296
8	57.1	54.86	5.0176
9	55.2	54.86	0.1156
10	53.8	54.86	1.1236
11	54.1	54.86	0.5776
12	54.1	54.86	0.5776
13	56.1	54.86	1.5376
14	55	54.86	0.0196
15	55.9	54.86	1.0816
16	56	54.86	1.2996
17	54.9	54.86	0.0016
18	54.3	54.86	0.3136
19	53.9	54.86	0.9216
20	55	54.86	0.0196

n	20
Sum	1097.2
Average	54.86
SS(Sum of squares)	19.308
Variance = SS/n-1	1.016
Std Dev=Sqrt(Variance)	1.0081