CH05_SEC1_EX31_P201_HorseKicks

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Chapter 5.1, Exercise 31, Page 201

In one of the first studies of the Poisson distribution, von Bortkiewicz considered the frequency of deaths from kicks in the Prussian army corps.
From the study of 14 corps over a 20-year period, he obtained the data shown.

Source: Ladislaus von Bortkiewicz, Das Gesetz der kleinen Zahlen [The law of small numbers] (Leipzig, Germany: B.G. Teubner, 1898).
4. Beispiel: Die durch Schlag eines Pferdes im preussischen Heere Getöteten.
[4th Example: Those killed in the Prussian army by a horse’s kick.]
Retrieved from https://archive.org/details/dasgesetzderklei00bortrich/page/n63 (Page 24 in source.)

Fit a Poisson distribution to this data and see if you think that the Poisson distribution is appropriate.

Table of Annual Deaths (observed)

NumberOfDeaths = 0:4    # Values per year, from above table, range from 0 through 4

# Actual values, tallied from Bortkiewicz's table
actuals = c(144,91,32,11,2)

# Total number of individuals who died from horse kicks
TotalDeaths = as.numeric(NumberOfDeaths %*% actuals)   # 196

# Number of observed army corps (14) * years (20)
NumberOfObs = sum(actuals)                             # 280

# Poisson lambda -- expected (average) deaths per year, in a typical corps
lambda = TotalDeaths/NumberOfObs                       # 196/280 = 0.7

# Result of poissondistribution
predicted = round(dpois(NumberOfDeaths,lambda)*NumberOfObs)

table1 = cbind(NumberOfDeaths,actuals)
colnames(table1)[2] = "Number of corps with x deaths in a given year"

## display table1
table1 %>% 
  kable()   %>% 
  column_spec(.,column = 1, width = "10em")   %>% 
  column_spec(.,column = 2, width = "30em")   %>% 
  kable_styling(c("striped", "bordered"))

NumberOfDeaths	Number of corps with x deaths in a given year
0	144
1	91
2	32
3	11
4	2

The total number of deaths is 196 .

The total number of observations is 14 * 20 = 280
(i.e, 14 Prussian military corps observed over 20 years, from 1875 through 1894) .

Predicted

table2 = cbind(table1, predicted)
colnames(table2)[3]="Poisson prediction"
table2 %>% 
  kable()   %>% 
  column_spec(column = 1:3, width = "10em")   %>% 
  kable_styling(c("striped", "bordered"))

NumberOfDeaths	Number of corps with x deaths in a given year	Poisson prediction
0	144	139
1	91	97
2	32	34
3	11	8
4	2	1

Barplot

barplot(t(table2[,2:3]),
        beside=T, col=c("blue","green"), names.arg=NumberOfDeaths,
        xlab="Number of Deaths in year (in a single corp)",
        ylab="Number of corps with x deaths in year",
        main = "Deaths from horse kicks in Prussian army corps, 1875-1894",
        legend.text = c("Actual","Predicted from Poisson(lambda=0.7)"))

### Chi-Squared test

chisq=chisq.test(actuals,p=predicted,rescale.p = T)
chisq

## 
##  Chi-squared test for given probabilities
## 
## data:  actuals
## X-squared = 2.78008849, df = 4, p-value = 0.59527466

Because the p-value from the chi-squared test is high, we fail to reject the null hypothesis, which is:

\(H_{0}\): The poisson distribution with lambda = 0.7 is appropriate for Bortkiewicz’s data set on deaths from horsekicks.

Conclusion

The Poisson distribution fits the data very well.