Question

6.4 The duration of customer service calls to an insurance company is normally distributed, with mean 20 minutes, and standard deviation 5 minutes. For the following sample sizes, construct a 95% confidence interval for the population mean duration of customer service calls.

(a) n = 25
(b) n = 100
(c) n = 400
(d) For each of the confidence intervals above, calculate and interpret the margin of error.

(e) Refer to part (d) above. Describe the relationship between the margin of error and sample size.

Answer 1

(a)

= Sample Mean = 20

s = Sample SD = 5

n = Sample Size = 25

SE = s/

= 5/ = 1

= 0.05

ndf = n - 1 = 25 - 1 = 24

From Table, critical values of t = 2.0639

So,

Confidence Interval:

20 (2.0639 X 1)

= (17.9361, 22.0639)

So,

Confidence Interval is:

17.9361 < < 22.0639

(b)

= Sample Mean = 20

s = Sample SD = 5

n = Sample Size = 100

SE = s/

= 5/ = 0.5

= 0.05

ndf = n - 1 = 100 - 1 = 99

From Table, critical values of t = 1.9842

So,

Confidence Interval:

20 (1.9842 X 0.5)

= 20 0.9921

= (19.0079, 20.9921)

So,

Confidence Interval is:

19.0079 < < 20.9921

(c)

= Sample Mean = 20

s = Sample SD = 5

n = Sample Size = 400

SE = s/

= 5/ = 0.25

= 0.05

ndf = n - 1 = 400 - 1 = 399

From Table, critical values of t = 1.9659

So,

Confidence Interval:

20 (1.9659 X 0.25)

= 20 0.4915

= (19.5085, 20.4915)

So,

Confidence Interval is:

19.5085 < < 20.4915

(d)

Margin of Error:

(i)

For n = 25:

MOE = 2.0639 X 1 = 2.0636

(ii)

For n = 100:

MOE = 1.9842 X 0.5 = 0.9921

(iii)

For n = 400::

MOE = 1.9659 X 0.25 = 0.4915

(e)

As the Sample Size (n) increases, the Margin of Error (MOE) decreases.