The text discusses a statistical analysis of fatalities in cavalry regiments over a 20-year period. It hypothesizes that if the fatalities occur independently and with equal probability for each regiment, the deaths per regiment per year should follow a Poisson distribution. The data is presented to check this hypothesis. Below is the table illustrating the number of fatalities per regiment-year:| Number of deaths (X) | Number of regiment-years ||----------------------|--------------------------|| 0 | 109 || 1 | 65 || 2 | 22 || 3 | 3 || 4 | 1 || >4 | 0 || **Total** | **200** |This table is sourced from Bortkiewicz, 1898.**Tasks:**a. Determine the mean number of deaths from horse kicks per regiment-year.b. Evaluate whether a Poisson distribution appropriately fits the data presented. **Text:**15. In the 19th century, cavalries were still an important part of the European military complex. While horses have many wonderful qualities, they can be dangerous beasts, especially if poorly treated. The Prussian army kept track of the number of fatalities caused by horse kicks to members of 10 of**Explanation:**The text discusses the role of cavalries in the 19th-century European military. It highlights the dual nature of horses as valuable yet potentially dangerous animals, particularly if not treated well. The Prussian army's efforts to monitor fatalities from horse kicks are mentioned, indicating a systematic approach to documenting such incidents for safety or analytical purposes.

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Description: The text discusses a statistical analysis of fatalities in cavalry regiments over a 20-year period. It hypothesizes that if the fatalities occur independently and with equal probability for each regiment, the deaths per regiment per year should foll

Given that : The number of fatalities caused by horse kicks to members of 10 of their cavalary regiments over a 20 - year time span. Fatalities occured independently and with equal probability for each regiment. The number of deaths by horse kicks per regiment per year should follow a poisson distribution. The mean number of deaths from the horse kicks per regiment - yeara. Therefore, we get the mean number of deaths from the horse kicks per regiment - year b. To test whether a poisson distribution fits this data the following are the steps : The mean of the poisson distribution is denoted by . The test of hyposthesis is given by : Since we are testing for poisson distribution it is possible that the ( mean of poisson) can be less or greater than the actual mean. Therefore, it is a two tailed test. which gives the following probabilities and expected frequencies (200* probability) for each value of X 4. The test statistic is Note: The observed frequeny is equal to the actual frequency and E is the expected value which we calculate above.. Here the ( O- E ) is the observed frequency - the expected value which can be done in excel using command : = cell value of observed freq - cell value of E like this we can calculate ( O- E) for all value of X. On the same line, we can find the ( O- E) * ( O-E) and ( O-E) *( O-E) /2 Therefore, the test statistic we obtain is 0.78 which nothing but the sum of all the observation of column ( O-E)* ( O-E) /E 4. Decision rule If test statistic > critcal value Reject ho If test statistic < critcal value do not reject Ho chi - square critical can be found using excel command : where, α = 0.05, n -2 is the degree of freedom and cumulative = TRUE = 0.00030734 the test statistic is greater than the critical value therefore Reject Ho Hence it is suggest that the number of deaths by horse kick per regiment per year does not have a poisson distribution

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their cavalry regiments over a 20-year time span. If these fatalities occurred independently and with equal probability for each regiment, then the number of deaths by horse kick per regiment per year should follow a Poisson distribution. On the other hand, if some regiments during some years consisted of particularly bad horsemen," then the events would not occur with equal probability. in which case we would expect a frequency dis- tribution different from the Poisson distribution. The following table shows the data, expressed as the number of fatalities per regiment-year (Bort- kiewicz 1898). Number of deaths (X) Number of regiment-years 109 65 22 3 3. 4. >4 Total 200 a. What is the mean number of deaths from horse kicks per regiment-year? b. Test whether a Poisson distribution fits these data.

their cavalry regiments over a 20-year time span. If these fatalities occurred independently and with equal probability for each regiment, then the number of deaths by horse kick per regiment per year should follow a Poisson distribution. On the other hand, if some regiments during some years consisted of particularly bad horsemen," then the events would not occur with equal probability. in which case we would expect a frequency dis- tribution different from the Poisson distribution. The following table shows the data, expressed as the number of fatalities per regiment-year (Bort- kiewicz 1898). Number of deaths (X) Number of regiment-years 109 65 22 3 3. 4. >4 Total 200 a. What is the mean number of deaths from horse kicks per regiment-year? b. Test whether a Poisson distribution fits these data.

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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