Question

The following table lists the numbers of violent crimes reported to police on randomly selected days for this year. The data was taken from 3 different cities of comparable size.

City A - 5 9 12 3 9 7 13

City B - 2 4 1 10 7 6

city C - 8 12 10 13 9 14

A partially completed ANOVA table, for the violent crime data, is below:

   

Source Degree Of freedom SS MS F

Between(Treatments)

Within(Error) 161.428

Total 18 269.789

At α ? 0.05, can the researcher conclude that there is a difference between the mean number of violent crimes reported per day for the different cities?

Answer 1

SS(between) = SS(Total) - SS(Within) = 269.789 - 161.428 = 108.361

Degrees of freedom( between) = k - 1

k = Number of groups to compare = 3

Degrees of freedom( between) = k - 1 = 3 - 1 = 2

Degrees of freedom( Within) = Degrees of freedom( Total) - Degrees of freedom( between) = 18 - 2 = 16

MS(Between) = SS(between) / Degrees of freedom( between) = 108.361 / 2 = 54.1805

MS(Within) = SS(Within) / Degrees of freedom( within) = 161.428 / 16 = 10.0893

Test statistic = F = MS(Between) / MS(Within) = 54.1805 / 10.0893 = 5.37

now,

Hence, there is enough evidence to conclude that  there is a difference between the mean number of violent crimes reported per day for the different cities at 5% level of significance.