1 answer

2. (25 P) A simulation team collected and sorted the data below for the number of...

Question:

2. (25 P) A simulation team collected and sorted the data below for the number of arrivals of customers for a restaurant for

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. 1 2 3 1 3 4 6 7 4 2 3 4 6 A 5 2 3 4 6 8 6 3 4 5 6 8 7 4 4 5 6 8 8 4 4 5 6 8 9 2 4 5 7 9 10 3 4 5 7 9 11 3 4 5 7 3 4 5 7 9 12 13 Perform x goodness of fit test for Poisson Distribution at a= 0.01 (state the hypothesis first). 14 15 16

Answers

Answer:-

GIven that:-

We have to perform chi-square test for goodness of fit to test that poisson distribution is good fit for the data or not at 0.01 level of significance.

Hypothesis -

он - Poisson distribution is good fit for given data.

H - Poisson distribution is not good fit for given data.

Test statistic -

\chi^{2}=\sum (\frac{(O_{i}-E_{i})^{2}}{E_{i}})

Test criterion -We have to reject H0 if \

\chi ^{2} \geq \chi ^{2}\alpha,n-p-k-1

Where, n - number of classes, p - numbe

r of parameters estimated, k - number of pullings

Now, to find the probability for each observation, we have to use PMF of poisson distribution -

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!}:x=0,1,2,......,n;\lambda>0

= ; otherwise

Now, we first estimate value of \lambda . We know that sample mean is an unbiased estimator of \lambda

So, \lambda=\overline{x}

Observation table -

X Frequency(Oi) xf
1 1 1
2 3 6
3 7 21
4 12 48
5 7 35
6 6 36
7 6 42
8 4 32
9 4 36
Total 50 257


Calculations -
\overline{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{257}{50}=5.14

So, \lambda = \overline{x} = 5.14

Now, we can calculate probabilities by using below formula -

P(X=x)=\frac{e^{-5.14}(5.14)^{x}}{x!}

Observation table -

X Frequency(Oi) probability(Pi) Ei Ei Oi (Oi-Ei)^2/Ei
1 1 0.0301 1.505 - - -
2 3 0.0774 3.87 5.375 4 0.3517
3 7 0.1326 6.63 6.63 7 0.0206
4 12 0.1704 8.52 8.52 12 1.4214
5 7 0.1751 8.755 8.755 7 0.3518
6 6 0.15 7.5 7.5 6 0.3
7 6 0.1102 5.51 5.51 6 0.0436
8 4 0.0708 3.54 5.56 8 1.0708
9 4 0.0404 2.02 - - -
Total 50 0.957 47.85 - - 3.5599


If expected frequency is less than 5, then we have to do pullings by adding above or below expected frequency to that less than 5 expected frequency.

Calculations -

\chi^{2}=\sum (\frac{(O_{i}-E_{i})^{2}}{E_{i}})=3.5599

Critical value -

\chi ^{2}_{\alpha,n-p-k-1}= \chi ^{2}_{0.01,10-1-2-1} = \chi ^{2}_{0.01,6 }= 16.812

Where, n - number of classes = 10, p - number of parameters estimated = 1 , k - number of pullings = 2

conclusion -

Since, \chi ^{2} (3.5599) < \chi ^{2} _{\alpha,n-p-k-1} (16.812) So, we have to accept H0 at 0.01 level of significance.

Result -

There is sufficient evidence that poisson distribution is good fit for given data at 0.01 level of significance.

Degrees of Freedom 0.99 0.01 10 11 12 13 14 15 0.975 0.001 0.051 0.216 0484 0.831 1.237 1.690 2.180 2.700 3.247 3.816 4.404 5

.

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