# Qnt/561 Development of money: kewlbeer’d

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%