Payment
case week 4
David
Vaughn
Qnt/561
11/18/2019
Dr.
Heidi Carty
Payment
case week 4
Market
research and the data gathered can be used and discussed in different ways. The
discussions themselves may change based upon the presenter, but at many points
how the information is gathered can be varying. In this scenario the case uses
payment information. My purpose in this scenario is to use the gathered
information and discuss the time table or average mean of the payment times for
the companies. Based upon my research I will use a confidence interval formula
to determine rather the newly advised schedule will have the new mean µ
< 19.5 days.
The purpose is to create and interpret the confidence interval formula for a new payment schedule.
Payment schedule and interval points
Looking over the payment schedule
scenario, it says that the average of 39 days for the Stockton, Ca trucking
company is above the industry average. I believe that this may effect the
inventory turnover formula = Net sales/average total assets (Paul D. Kimmel PhD). Which may not be the problem to
solve but will help if the accountants were using any type of financial
schedule. As it would someone who is involved in investment banking, based on
the banking schedule years could be 360 days or with similar
differences (CHERYL CLEAVES). So, the idea here is to explain the
new formula for calculating payment schedules.
For the Stockton, Ca trucking company
the billing schedule was exactly equal to 39 days. Looking over the excel
spreadsheet and the information provided. We can see that there are 65 other
companies listed and each with alternate times of pay schedules. With the
information summed up there is a deviation of 4.2 days for them all, (σ = 4.2)
the current mean (average) being 39 days which is too high for the
industry. From this I will be able to schedule a confidence interval that may
satisfy the demands of the company with the information given.
Confidence
interval and 4.2 days
Beginning with the formula for statistic intervals; to
compute this accurately it is then changed to an algebra formula (Black). The formula will then be expressed as
algebra
µ
= z( ) I
used parentheses for multiplication. The formula from here is simple, just plug
in the numbers to find the new average. The new formula will look this way; µ =
39 ± 1.96 (). Once this is calculated we can then
see the upper and lower bounds of the scheduling and be able to determine if µ
< 19.5 days because the new schedule is said to change by 50%.
Accurate and detailed I looked over the formula and
put in place not just the given mean of the Stockton, Ca trucking (39 days) company but the
averages the research was said to use(30 days). I find that using hypothesis
testing will reveal that the alternative hypotheses has obtained a significant
result. Rejected and is accepted. This is based upon estimating the
population mean Using the z statistic (Black)
p.232; and rearranging them to form a algebra problem
µ = z( )
yields this result µ = 39 + 1.96() gives the result of 40.02 the upper
bound limit. Using the same formula and subtracting in place of addition states
µ = 37.97. Both of which are still substantially higher than the industry
average of 30 days.
Conclusion
37.97< µ < 40.02 for the distribution
of sample means making accepted. With the 99% confidence interval it is very unlikely that
the answer will be any different based on previous test. I however complete the
formula and can show it as = 38.993 < µ <
39 + 2.575 (.01/2) 4.2/ = 39.0067. This
makes the new statement µ ≤ 19.5 days inaccurate.
If the population mean was 19.5 what
is the probability of observing a sample invoice of 65 less than or equal to
18.1007 days. The sample mean formula was used this will relate
as = 64.42%