You can use a chi-square test for goodness of fit. The null hypothesis is

H0:

where is the probability of an accident in the ith hourly interval for i = 1, 2, ...... 8.

The chi-square test statistic will have 8 - 1 = 7 degrees of freedom.

Now calculate the value of and test it for significance at the 0.01 level:

where n is the total number of accidents and is the number of accidents in the ith hourly interval (you should be familiar with this formula and where it comes from).