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9.36 Goodness of fit to a Poisson distribution. Refer to Example 5.30 (page 329) where a Poisson distribution is described as a model for the number of dropped calls on your cellphone per day. The mean number of calls is 2.1. In this setting, the probabilities for 0, 1, 2, and 3 or more dropped calls are 0.1225, 0.2572, 0.2700, and 0.3503, respectively. Suppose that you record the number of dropped calls per day for the next 100 days. Your observed counts of dropped calls are 11, 22, 28, and 39, respectively. Use a chi-square goodness of fit test to test the hypothesis that your calls are distributed according to this Poisson distribution.

9.36 Goodness of fit to a Poisson distribution. Refer to Example 5.30 (page 329) where a Poisson distribution is described as a model for the number of dropped calls on your cellphone per day. The mean number of calls is 2.1. In this setting, the probabilities for 0, 1, 2, and 3 or more dropped calls are 0.1225, 0.2572, 0.2700, and 0.3503, respectively. Suppose that you record the number of dropped calls per day for the next 100 days. Your observed counts of dropped calls are 11, 22, 28, and 39, respectively. Use a chi-square goodness of fit test to test the hypothesis that your calls are distributed according to this Poisson distribution.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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9.36 Goodness of fit to a Poisson distribution. Refer to Example 5.30 (page 329) where a Poisson distribution is described as a model for the number of dropped calls on your cellphone per day. The mean number of calls is 2.1. In this setting, the probabilities for 0, 1, 2, and 3 or more dropped calls are 0.1225, 0.2572, 0.2700, and 0.3503, respectively. Suppose that you record the number of dropped calls per day for the next 100 days. Your observed counts of dropped calls are 11, 22, 28, and 39, respectively. Use a chi-square goodness of fit test to test the hypothesis that your calls are distributed according to this Poisson distribution.
Transcribed Image Text:9.36 Goodness of fit to a Poisson distribution. Refer to Example 5.30 (page 329) where a Poisson distribution is described as a model for the number of dropped calls on your cellphone per day. The mean number of calls is 2.1. In this setting, the probabilities for 0, 1, 2, and 3 or more dropped calls are 0.1225, 0.2572, 0.2700, and 0.3503, respectively. Suppose that you record the number of dropped calls per day for the next 100 days. Your observed counts of dropped calls are 11, 22, 28, and 39, respectively. Use a chi-square goodness of fit test to test the hypothesis that your calls are distributed according to this Poisson distribution.
EXAMPLE 5.30 Number of dropped calls. Suppose that the number of dropped calls on your cell phone varies, with an average of 2.1 calls per day. If we assume that the Poisson setting is reasonable for this situation, we can model the daily count of dropped calls X using the Poisson distribution with = 2.1. What is the probability of having no more than two dropped calls tomorrow? We can calculate P(X ≤ 2) either using software or the Poisson probabil- ity formula. Using the probability formula: P(X ≤2) = P(X= 0) + P(X = 1) + P(X = 2) e-2.1 (2.1)⁰ e 2.1 (2.1)¹ e 2.1 (2.1)² + + 1! 2! = = = O! 0.1225 +0.2572 +0.2700 0.6497 Using the R software, the probability is dpois (0,2.1) + dpois(1,2.1) + dpois(2,2.1) [1] 0.6496314
Transcribed Image Text:EXAMPLE 5.30 Number of dropped calls. Suppose that the number of dropped calls on your cell phone varies, with an average of 2.1 calls per day. If we assume that the Poisson setting is reasonable for this situation, we can model the daily count of dropped calls X using the Poisson distribution with = 2.1. What is the probability of having no more than two dropped calls tomorrow? We can calculate P(X ≤ 2) either using software or the Poisson probabil- ity formula. Using the probability formula: P(X ≤2) = P(X= 0) + P(X = 1) + P(X = 2) e-2.1 (2.1)⁰ e 2.1 (2.1)¹ e 2.1 (2.1)² + + 1! 2! = = = O! 0.1225 +0.2572 +0.2700 0.6497 Using the R software, the probability is dpois (0,2.1) + dpois(1,2.1) + dpois(2,2.1) [1] 0.6496314
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