Question

2. (25 P) A simulation team collected and sorted the data below for the number of arrivals of customers for a restaurant for 50 days. 3 3 4 4 6 6 1 2 2 3 7 7 8 4 6 3 4 4 6 4 6 5 5 5 5 00 00 00 4 4 4 6 7 9 انحنا | ال 4 5 5 7 7 9 9 4 4 3 7 9 Perform x goodness of fit test for Poisson Distribution at a= 0.01 (state the hypothesis first).

Answer 1

First step is to combine the same number of arrivals of customers and make a table then proceed as follows:

No of days No. of customers (Frequency 1433 P(x= x;) E; Expected for 1 1 0.0301 1.50 2 3 0.0774 3.87 3 6.63 7 0-1325 4 8.52 12 0.1703 5 0.1751 8.76 6 6 0.1500 7.50 7 0 1101 5.51 3.54 4 0.0707 9 2.02 0.0404 50 Mean = Efix; Efi 5.14 = 5.14 -5-14 P(X=x) i é e x(5.147% 2= 1, 2, 3, 9 el n! E;= P(x= x;) X E fi

Now, X square table 2. (0.-€:) xi Ei Op = fi 1 10.-E;) (02-E;) EU 0.167 1 1.50 -0.5 0.25 2 3 3.87 -0.87 0.757 0.195 3 7 6.63 0.37 0.137 0.020 12 8.52 3.48 12.11 1.421 5 7 8.76 -1.76 3.09 0.354 6 6 7-50 2.25 0.300 7 6 0.49 0.24 0.043 5.51 3.54 8 다 0.46 0.060 6.211 3.92 co 2.02 1.98 1-940 4.502 50 Hypothesis :- Null hypothesis (Ho): Number of arrivals of customes is Poisson Alternative hypothesis (Ha): Number of arrivals of customers is not Poisson Significance level, a = 0.01 Degree of Freedom, V = 9-2= 7 1 for Porasan distribution

We got x² = 4.502 a=0.1 vo O X 4 = 18.96 For DOF=7 x=0.01 Gitical value. 18.46 17) = 4:502, p <0.01 x² < critical value Accept the hull hypothesis Ho.