Question

The table below lists the number of crimes reported at a police station on each day of the week for the past three months. 

The null hypothesis for the goodness-of-fit test is that the number of crimes reported at this police station is the same for each day of the week. What is the value of the test statistic? 

Answer 1

Assuming H0, the Expected Frequency in each Day of the Week = 140/7 = 20

Test Statistic $\chi ^{2}$ is got as follows:

Observed (O)	Expected (E)	(O - E)2/E
21	20	0.05
10	20	5.00
13	20	2.45
16	20	0.80
25	20	1.25
29	20	4.05
26	20	1.80
Total = $\chi ^{2}$ =		15.40

nfd = 7 - 1 = 6

By Technology, p - value = 0.0174

Since p - value = 0.0174 is less than $\alpha$ = 0.05, the difference is significant. Reject null hypothesis.

Conclusion:
The data do not support the claim that the number of crimes reported at this police station is the same for each day of the week.

So,

Answer to question asked:

The value of the test statistic = 15.400