Question 977793

Poisson Probability Distribution Table


<table border=1><tr><th>k</th><th>P(X = k)</th></tr><tr><td>0</td><td>0.0183156389</td></tr><tr><td>1</td><td>0.0732625556</td></tr><tr><td>2</td><td>0.1465251111</td></tr><tr><td>3</td><td>0.1953668148</td></tr><tr><td>4</td><td>0.1953668148</td></tr><tr><td>5 or more</td><td>0.3711630648</td></tr></table>


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a)

Decision Rule: if the *[Tex \LARGE \chi^2] (chi-squared) test statistic is larger than the chi-squared critical value, then reject H0.


alpha = 0.02


critical value = 13.3882


I used <a href="https://www.swogstat.org/stat/public/chisq_calculator.htm">this calculator</a> to get the critical value
Note: df = k-1 = 6-1 = 5. The area to the left of the critical value is 0.98
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b)


with 0 applications, we expect approximately...306*0.018315639 = 5.604585534 = 5.6046
with 1 applications, we expect approximately...306*0.073262556 = 22.418342136 = 22.4183
with 2 applications, we expect approximately...306*0.146525111 = 44.836683966 = 44.8367
with 3 applications, we expect approximately...306*0.195366815 = 59.78224539 = 59.7822
with 4 applications, we expect approximately...306*0.195366815 = 59.78224539 = 59.7822
with 5 applications, we expect approximately...306*0.3711630648 = 113.5758978288 = 113.5759


Those results above form the *[Tex \LARGE f_e] column (expected frequencies column)




Now let's compute the *[Tex \LARGE \frac{(f_o-f_e)^2}{f_e}] column




0 applications:
[(observed - expected)^2]/(expected) = ((51-5.6046)^2)/(5.6046) = 367.687674617279 = 367.6877


1 applications:
[(observed - expected)^2]/(expected) = ((87-22.4183)^2)/(22.4183) = 186.044257365188 = 186.0443


2 applications:
[(observed - expected)^2]/(expected) = ((74-44.8367)^2)/(44.8367) = 18.968792683003 = 18.9688


3 applications:
[(observed - expected)^2]/(expected) = ((49-59.7822)^2)/(59.7822) = 1.94465638333819 = 1.9447


4 applications:
[(observed - expected)^2]/(expected) = ((31-59.7822)^2)/(59.7822) = 13.8572189855843 = 13.8572


5 applications:
[(observed - expected)^2]/(expected) = ((14-113.5759)^2)/(113.5759) = 87.3016182201506 = 87.3016



Final Table

<img src = "http://i150.photobucket.com/albums/s91/jim_thompson5910/7-1-2015%206-06-21%20PM_zpsucosl204.png">

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c)


Refer to part b). Add up all of the numbers in the last column.


*[Tex \Large \chi^2 = \sum \frac{(f_o-f_e)^2}{f_e}]


*[Tex \Large \chi^2 = 367.6877+186.0443+18.9688+1.9447+13.8572+87.3016]


*[Tex \Large \chi^2 = 675.8043]


*[Tex \Large \chi^2 = 675.804]


The *[Tex \LARGE \chi^2] test statistic is 675.804


Since the chi-squared test statistic (675.804) is larger than the critical value (13.3882), this means we reject the null hypothesis.


There is sufficient evidence to reject the null.


So we conclude that this observed distribution does NOT fit with the expected distribution. Therefore, the values are NOT distributed from a Poisson distribution.