Question

• assuming the standard deviation of the payment times for all payments is 4.2 days, construct a 95% confidence interval estimate to determine whether the new billing system was effective. state the interpretation of 95% confidence interval and state whether or not the billing system was effective.

Answer 1

Assuming the standard deviation of the payment times for all payments is 4.2 days, construct a 95% confidence interval estimate to determine whether the new billing system was effective. State the interpretation of 95% confidence interval and state whether or not the billing system was effective.

Using the 95% confidence interval, can we be 95% confident that µ ≤ 19.5 days?

Using the 99% confidence interval, can we be 99% confident that µ ≤ 19.5 days?

If the population mean payment time is 19.5 days, what is the probability of observing a sample mean payment time of 65 invoices less than or equal to 18.1077 days?

Case Study – Payment Time Case Study

Therefore, if µ denotes the new mean payment time, the consulting firm believes that µ will be less than 19.5 days. Therefore, to assess the system’s effectiveness (whether µ < 19.5 days), the consulting firm selects a random sample of 65 invoices from the 7,823 invoices processed during the first three months of the new system’s operation. Whereas this is the first time the consulting company has installed an electronic billing system in a trucking company, the firm has installed electronic billing systems in other types of companies.

Analysis of results from these other companies show, although the population mean payment time varies from company to company, the population standard deviation of payment times is the same for different companies and equals 4.2 days. The payment times for the 65 sample invoices are manually determined and are given in the Excel® spreadsheet named “The Payment Time Case”. If this sample can be used to establish that new billing system substantially reduces payment times, the consulting firm plans to market the system to other trucking firms.

Data Set as provided

PayTime

22

19

16

18

13

16

29

17

15

23

18

21

16

10

16

22

17

25

15

21

20

16

15

19

18

15

22

16

24

20

17

14

14

19

15

27

12

17

25

13

17

16

13

18

19

18

14

17

17

12

23

24

18

16

16

20

15

24

17

21

15

14

19

26

21

$X bar (Sample Mean)= \frac{\sum x}{n}$

A. Confidence Interval at 95%

The confidence level at 95% is:

$CI = 17.08663837 < \mu < 19.12874624$

The confidence level is used to make an inference about the population mean, μ, with the help of the sample mean (also known as a point estimate).

With the help of the confidence interval constructed above, we can say that the mean payment time was somewhere between 17.08663837 and 19.12874624 days.

Since the upper bound of the confidence interval, 19.12874624 days is less than the previous population mean at 19.5, we can conclude that the mean payment time was less than 19.5 days and therefore the new billing system was effective.

We begin by calculating the mean of all the 65 observations by using the formula:

$X bar (Sample Mean)= \frac{\sum x}{n}$

$X bar (Sample Mean)= \frac{1177}{65}$

X bar (Sample Mean) = 18.10769231

Since the population standard deviation is given, we can use the following formula to construct the Confidence Interval (CI).

$CI = X bar \pm Z\frac{\sigma }{\sqrt{n}}$

where

CI = confidence interval

Z = two tailed Z score at 95% confidence level

σ = population standard deviation

n = sample size.

Substituting the values in the equation above, we get,

$CI = 18.10769231 \pm 1.96\frac{4.2}{\sqrt{65}}$

$CI = 18.10769231 \pm 1.96* 0.520945885}$

$CI = 18.10769231 \pm 1.021053935$

$CI = 17.08663837 < \mu < 19.12874624$

B. If the population mean payment time is 19.50, the probability of observing a sample mean payment time of 18.1077 days in a sample of 65 invoices is 0.3763%.

We use the following formula to calculate the Z score:

$Z = \frac{X bar - \mu }{\frac{\sigma }{\sqrt{n}}}$

Substituting the values in the equation above, we get,

$Z = \frac{18.1077 - 19.5 }{\frac{4.2 }{\sqrt{65}}}$

$Z = \frac{-1.3923}{0.520945885}}$

$Z = -2.672638444$

By looking up the probabilities for standardized z score, we get

$P(Z\leq -2.672638444) = 0.0037629$ or

P(Z ≤ -2.672638444) = 0.3763%.