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Tutorial 5 February 25, 2013

Tutorial 5 February 25, 2013. Problem 1. Use the Chi-square test with =0.05 to test whether the data in the table is uniformly distributed. Use n=10 intervals of equal length. {See XL sheet for table}. Solution 1. From the computations for chi-square test shown,  0 2 = 3.4

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Tutorial 5 February 25, 2013

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  1. Tutorial 5 February 25, 2013

  2. Problem 1 Use the Chi-square test with =0.05 to test whether the data in the table is uniformly distributed. Use n=10 intervals of equal length. {See XL sheet for table}

  3. Solution 1 • From the computations for chi-square test shown, • 02 = 3.4 • ,k-1 = 0.05,9 = 16.9 • Since 02 < 0.05,9 , the null hypothesis of no difference between the sample distribution and uniform distribution is not rejected.

  4. Problem 2:Chi-square Goodness of fit test for Poisson Distribution The number of vehicles arriving at the North-West corner of an intersection in a 5-minute period between 7:00a.m and 7:05a.m. was monitored for 5 work days over a 20 week period. The table below shows the resulting data. The 1st entry indicates that there were 12 5-minute periods during which zero vehicles arrived & so on. Assume a poisson distribution, =0.05.

  5. Solution 2 • Find mean () • Find probability density function for the Poisson distribution in which mean is known p(x) = e- x/x!, for x = 0,1,2…., 0, otherwise • 02 should not be calculated if the expected value in any category is less than 5. Hence group data accordingly.

  6. Solution 2 • 02 = 27.68 (XL sheet for table) • Degrees of freedom for the tabulated value of 2 df= k-s-1; k=12-1-4=7 s= 1, only one parameter, mean was estimated from data (If the distribution was normal, mean & variance would have to be estimated from data. Hence value of s would be:2) df = 7-1-1 = 5 • ,df = 0.05,5 = 11.1 • Since 02 > 0.05,5 , the null hypothesis is rejected.

  7. Multiple Servers Single Queue (MSSQ) • One line leads to several servers • Each server can serve any customer equally well Arrivals  Queue  • Inter-arrival and service length times have exponential distribution • Arrivals and customer service completion have Poisson Distribution Service Done Service Done Service Done

  8. Multiple Servers Single Queue • M = number of servers •  = average arrivals per unit time • µ = average number of customers a server can handle per unit time = 1/average service time •  = service utilization factor, the proportion of time on average that each server is busy •  = /Mµ • As with simple queue,  < 1 for steady state

  9. MSSQ probabilities • Probability that at any given point of time, there are no customers waiting or being served: Prob (0 in system) = 1/+ • Probability that at any given time there are n customers in the system: • If n • If n

  10. MSSQ statistics • Mean number of customers waiting in queue to be served = • Mean time customers spend in queue:  • Mean time customers spend in the system: W= • Mean number of customers in system: L=

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