The Poisson distribution is a discrete distribution that expresses the probability of a fixed number of events occurring in a fixed interval.For example,suppose we want to model the number of arrivals per minute at the campus dining hall during lunch.We observe the actual arrivals in 200 one-minute periods in 1 week.The sample mean is 3.8 and the results are shown below. The probabilities based on a Poisson distribution with a mean of 3.8 are shown below. Perform a formal test to determine if the observed counts are compatible with the Poisson distribution with a mean of 3.8 and a significance level of .05.

A) The P-value is very small;therefore the observed counts are compatible with the Poisson distribution.
B) The P-value is large;therefore the observed counts are compatible with the Poisson distribution.
C) The P-value is very small;therefore the observed counts are not compatible with the Poisson distribution.
D) The P-value is large;therefore the observed counts are not compatible with the Poisson distribution.

Question

The Poisson distribution is a discrete distribution that expresses the probability of a fixed number of events occurring in a fixed interval.For example,suppose we want to model the number of arrivals per minute at the campus dining hall during lunch.We observe the actual arrivals in 200 one-minute periods in 1 week.The sample mean is 3.8 and the results are shown below.

The probabilities based on a Poisson distribution with a mean of 3.8 are shown below.

Perform a formal test to determine if the observed counts are compatible with the Poisson distribution with a mean of 3.8 and a significance level of .05.

A) The P-value is very small;therefore the observed counts are compatible with the Poisson distribution.
B) The P-value is large;therefore the observed counts are compatible with the Poisson distribution.
C) The P-value is very small;therefore the observed counts are not compatible with the Poisson distribution.
D) The P-value is large;therefore the observed counts are not compatible with the Poisson distribution.

Michelle Livingston · Accepted Answer

Final Answer : C, Explanation :  We can perform a goodness-of-fit test using the chi-square distribution. The null hypothesis is that the observed counts are compatible with the Poisson distribution with a mean of 3.8. The test statistic is calculated as follows:X^2 = Σ [(Observed Count - Expected Count)^2 / Expected Count]Using the probabilities based on a Poisson distribution with a mean of 3.8, we can calculate the expected counts for each number of arrivals. The expected counts and the chi-square contributions are shown in the table below.Number of Arrivals | Observed Count | Expected Count | (Observed - Expected)^2 / Expected0                 | 13            | 14.59          | 0.271                 | 38            | 55.52          | 6.362                 | 47            | 70.41          | 4.823                 | 44            | 53.23          | 1.554                 | 32            | 31.85          | 0.015                 | 19            | 16.12          | 0.346                 | 6             | 6.12           | 0.027                 | 1             | 1.60           | 0.488 or more         | 0             | 0.18           | 0.18The total chi-square value is X^2 = 13.75 with 6 degrees of freedom. The P-value for this test is less than .05, indicating that the observed counts are not compatible with the Poisson distribution with a mean of 3.8 at the .05 level of significance. Therefore, we reject the null hypothesis and conclude that there is evidence of a departure from the Poisson distribution for the number of arrivals at the campus dining hall during lunch.

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