In: Statistics and Probability
Coal mining records from the period 1851 to 1962 revealed 186 explosions that killed 10 or more workers. If the distribution of accidents by day of the week is uniform, then approximately one-sixth of the accidents could be expected to have occurred on any workday. Do the following data support the uniform hypothesis?
day | mon | tue | wed | thu | fri | sat |
no. of explosion | 19 | 34 | 33 | 36 | 35 | 29 |
Let the event be the explosions that killed 10 or more coal mining workers.
We want to test if the distribution of accidents by day of the week is uniform with probability = 1/6 or
P(mon) = P(tue) = P(wed) = P(thu) = P(fri) = P(sat) = 1/6
We want to test the following hypothesis
We use chi-square test to test the goodness of fit of uniform distribution to describe the accidents
If the probability of accidents happening on any day is the same then the number of accidents on any of the weekdays from Monday to Saturday = 186*1/6 = 31 accidents. That means we could have expected that the number of accidents happened on each of Monday to Saturday is 31. We call this expected frequency. The observed frequencies are given in the table. For example on Mondays 19 accidents were recorded but in null hypothesis were true then the expected number on Mondays is 31 accidents
The chi-square statistics is calculated as
where is the expected frequency
is the observed frequency
The following shows the calculation
The formula used is
The chi-square value of the sample is 6,516 and the degrees of freedom is (number of classes -1) = (6-1) = 5
We get the p-value of this chi-square statistics using excel function =CHISQ.DIST.RT(6.516,5) and get the p-value as 0.259. We compare this against the level of significance used to test this hypothesis, which is 0.05. Since the p-value is greater than the significance level we cannot reject the null hypothesis and say that
at 0.05 level of significance, the distribution of accidents by day of the week is uniform with probability = 1/6