Payment Time Confidence Interval Case Study

Payment Time Case Study

Electronic billing systems were created to make life easier for companies. Technically speaking, the new billing system is supposed to significantly reduce the payment time process. For this case study, the datasets provided in the assignment will be used and the concepts of sampling distributions and confidence intervals of 95% and 99% will be applied to the analysis. Analysis will be conducted to determine whether the new electronic billing system that was developed does in fact reduce the payment time. In this case, the new system was developed for a trucking company based in Stockton, CA. By completing this assignment, I will demonstrate abilities in using datasets to apply the concepts of sampling distributions and confidence intervals to make sound management decisions.

Constructing Confidence Interval Estimate

Data

Population Standard Deviation 4.2

Sample Mean 18.1077

Confidence Level 95%

Intermediate Calculations

Standard error of the Mean 0.5209

Z Value 1.96

Interval half width 1.021

Confidence Interval

Interval lower limit 17.0867

Interval upper limit 19.1287

Interval estimate

17.0867__________ 19.1287 This has an implication that the new billing system was effective.

Interpretation Of 95% Confidence Interval

The provided standard deviation of payment times for all the companies is 4.2 days. Using this and a 95% confidence interval, the author will determine the new billing system’s effectiveness at improving payment times. The author utilized the sample of 65 various payment times extracted from 7,823 invoices. The mean that was calculated from the samples is 18.11 days.

Determination of the Population Mean

The following formula will help to determine the population mean:

CI=X ± Z×α/ √ N Confidence Interval Estimate for the Mean:

Data

Population Standard Deviation σ = 4.2

Sample Mean x = 18.1077

Sample Size n = 65

Confidence Level 95%

Intermediate Calculations

Standard Error of the Mean 0.5209

z (α/2) Value 1.9600

Interval Half Width 1.0210

Confidence Interval

Interval Lower Limit 17.0867

Interval Upper Limit 19.1287

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

95% CI = (17.0867, 19.1287) less than 19.5.

We conclude that 95% confident that µ ≤ 19.5 days.

99% Confidence Interval

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

Confidence Interval Estimate for the Mean is arrived at below:

Data

Population Standard Deviation σ = 4.2

Sample Mean x = 18.1077

Sample Size n = 65

Confidence Level 99%

Intermediate Calculations

Standard Error of the Mean 0.5209

z (α/2) Value 2.5760

Interval Half Width 1.3419

Confidence Interval

Interval Lower Limit 16.7658

Interval Upper Limit 19.4496

99% CI = (16.7658, 19.4496) less than 19.5.

These results clearly show the importance of the sample size, as a bigger sample size will give more accurate results than a smaller one, “as we will get wider narrower limit and large one in calculating the confidence interval” (Black, 2017). Thus, it can be stated that it is very useful to undertake utilization of data sets in the application of the concepts of sample distributions as well as confidence intervals for the purpose of making management decisions. The report will give a description of the interpretation of 95% confidence interval and mention whether the billing system was effective. Besides, with the use of the 95% confidence interval, we can conclude that 95% confident that µ ≤ 19.5 days.  In addition, it can be concluded 99% confident that µ ≤ 19.5 day. If the population mean payment time is 19.5 days, the probability of observing a sample mean payment time of 65 invoices less than or equal to 18.1077 days

Z value for 18.1077 is z = (18.1077-19.5)/0.5209 = -2.67

P (mean x <18.1077) = P (z < -2.67) =0.0038

The Test

The null hypothesis in which (Ho) symbolizes is that the mean is greater than or equal to 19.5 days, while the alternate hypothesis (Ha) is that the mean is less than 19.5 days. The ‘alpha’ is set at 0.05 since a 95 percent confidence level is required. The test is a “z” test since the sample size is greater than 30. The ‘fdr’ is to discard the null if “z” is less than the critical value (CRV) of 1.645, this is the CRV used for a one-tail test. The statistic will be as follows;

z=´x−µσ√n=18.11−19.54.2 /√ 65=−1.390.520946 =− 2.67265

The result is since; “z” is -2.67 and it is less than the CRV of 1.645, the null is excluded. The assumption is that the new billing system made a statistically substantial difference in reducing the payment time. Consequently, the approval is to continue with marketing the system to other trucking firms.

Conclusion

Consulting Firm, LLC established a system intended to reduce the payment time of A Trucking Company’s invoices. However, statistical examination was mandatory to estimate the success of the new system in order to further market the system to other trucking firms. While the sample of 65 invoices seems to be fairly small linked to the 7,823 total invoices, they were verified with a 95 percent confidence level that the new system did make a statistically substantial decrease in the payment time. Similarly, it is likely to see the new system operated by several trucking firms all over.

References

Black, K. (2017). Business statistics for contemporary decision making (9th ed.). Hoboken, NJ: John Wiley & Sons, Inc