Question

Records pertaining to the monthly number of job-related injuries at an underground coal mine were being studied by a federal agency. The values for the past 100 months were as follows:

injuries per Month、 Frequency of Occurrence


0 35


1 40


2 13


3 6


4 4


5 1


6 1


(a) Apply the chi-square test to these data to test the hypothesis that the underlying distribution is Poisson. Use the level of significance α = 0. 5.


(b) Apply the chi-square test to these data to test the hypothesis that the distribution is Poisson with mean 1. 0. Again let α= 0. 5.


(c) What are the differences between parts (a) and (b), and when might each case arise?

Answer 1

Explanation:

(a) To test the hypothesis that the underlying distribution is Poisson, we need to calculate the expected frequencies for each category. The expected frequency for a Poisson distribution with parameter λ is given by:

E(x) = n * P(x) = n * (λ^x * e^(-λ)) / x!

where n is the total number of observations and P(x) is the probability of x occurrences.

Using the given data, we can calculate the expected frequencies as follows:

injuries per Month, Frequency of Occurrence, P(x), E(x)

0, 35, 0.3679, 36.79

1, 40, 0.3679, 36.79

2, 13, 0.1839, 18.39

3, 6, 0.0613, 6.13

4, 4, 0.0153, 1.53

5, 1, 0.0038, 0.38

6, 1, 0.0009, 0.09

Total, 100, -, 100.00

We can now use the chi-square test statistic to test the goodness-of-fit of the observed data to the Poisson distribution with parameter λ. The chi-square test statistic is given by:

χ² = Σ (O - E)² / E

where O is the observed frequency and E is the expected frequency.

Using the above formula, we can calculate the chi-square test statistic as follows:

χ² = [(35 - 36.79)² / 36.79] + [(40 - 36.79)² / 36.79] + [(13 - 18.39)² / 18.39] + [(6 - 6.13)² / 6.13] + [(4 - 1.53)² / 1.53] + [(1 - 0.38)² / 0.38] + [(1 - 0.09)² / 0.09]

χ² = 8.43

We can now compare this value with the critical value from the chi-square distribution with 6 degrees of freedom (7 categories - 1 estimated parameter). Using a chi-square distribution table or a calculator, we find that the critical value at α = 0.5 and 6 degrees of freedom is 12.59. Since our calculated value of χ² is less than the critical value, we fail to reject the null hypothesis that the underlying distribution is Poisson.

(b) To test the hypothesis that the distribution is Poisson with mean 1.0, we need to use a modified version of the formula for the expected frequency:

E(x) = n * P(x) = n * (1.0^x * e^(-1.0)) / x!

Using this formula, we can calculate the expected frequencies as follows:

injuries per Month, Frequency of Occurrence, P(x), E(x)

0, 35, 0.3679, 36.79

1, 40, 0.3679, 36.79

2, 13, 0.1839, 18.39

3, 6, 0.0613, 6.13

4, 4, 0.0153, 1.53

5, 1, 0.0038, 0.38

6, 1, 0.0009, 0.09

Total, 100, -, 100.00

Using the chi-square test statistic as before, we can calculate:

χ² = [(35