The following data is to be used for a simulation model to represent the service

Question

The following data is to be used for a simulation model to represent the service/process times. Based on the management-team advice, is tested for Uniform Distribution. Complete the Goodness of fitness tests you see appropriate. State the hypothesis tests needed and Complete the test State your results as well as State your conclusion. (Use 95% confidence interval) Simulate the following real time scenario using ARENA. Three different parts are processed via 3 different machines. Parts are moved via a transporter and follow the given sequences, Transporter speed is 1 foot/minute. Machine 3 fails sometimes, uptime is expo(50) and down time is expo (2). Use the frequency statistics to tabulate/plot the number of parts waiting in front of the machine #2. Run simulation for 10 week, 8hrs per day. Use minutes as base time. Explain the modules you use step by step. Run the simulation with faster transporters 2 feet/minute. Tabulate the results. Management wishes to compare the effect of having transporter with speed of 1 foot/minute vs 2 feet/minute Each of the 10 sample is obtained from simulation runs of 40,000 and total statistics are measured. The resulting data are shown below: A manager claims that the true mean of the total time in system with the slower transporter exceeds that for the speedy transporter. Are the differences in results statistically significant? It is reasonable to assume that the total time in system over replications follow a Normal distribution. Formulate your hypotheses and explain why you chose them. Does the above data support the management's claim at alpha = 0.05? Compute the Confidence Interval and compare Compute the t statistics and the P values to compare

Explanation / Answer

Dear student, I am only answering the first question as per Chegg Guidelines.

Question2:

(a)

H0: The given data is uniformly distributed

Ha: The given data is not uniformly distributed

(b)

So the calculated chi-squared value is 12.37

For df=n-1=8 and 0.05 level of significance the p-value at 12.37 is 0.1354.

(c) The null hypothesis is not rejected as the p-value is not less than 0.1354

(d) The given distribution is not significantly different from uniform distribution.

x Obs Freq (O) Expected Freq (E) (O-E)^2 (O-E)^2/E 3 12 13.22222222 1.49382716 0.1129785247 4 13 13.22222222 0.04938271605 0.003734827264 2 17 13.22222222 14.27160494 1.079365079 1 18 13.22222222 22.82716049 1.726423903 9 10 13.22222222 10.38271605 0.7852474323 8 18 13.22222222 22.82716049 1.726423903 6 17 13.22222222 14.27160494 1.079365079 7 7 13.22222222 38.71604938 2.928104575 5 7 13.22222222 38.71604938 2.928104575 Total: 119 119 12.3697479

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.