Question

Police Calls The police department in Madison, Connecticut, released the following numbers of calls for the different days of the week during February that had 28 days: Monday (114); Tuesday (152); Wednesday (160); Thursday (164); Friday (179); Saturday (196); Sunday (130). Use a 0.01 significance level to test the claim that the different days of the week have the same frequencies of police calls. Is there anything notable about the observed frequencies?

Short answer

There is enough evidence to conclude that the police calls do not occur equally frequently on the different days of the week.

The observed frequencies increase from Monday to Saturday and then decrease on Sunday.

Step-by-step solution

Given information

The observed frequencies of the police calls on the seven days of the week are recorded.

It is expected that the calls occur equally frequently on the seven days of the week.

Check the requirements

Let the serial numbers from 1 to 7 denote the seven days of the week starting from Monday.

Let O denote the observed frequencies of the police calls.

The following values are obtained:

\(\begin{aligned}{l}{O_1} = 114\\{O_2} = 152\\{O_3} = 160\\{O_4} = 164\\{O_5} = 179\\{O_6} = 196\\{O_7} = 130\end{aligned}\)

The sum of all observed frequencies is computed below:

\(\begin{aligned}{c}n = 114 + 152 + .... + 130\\ = 1095\end{aligned}\)

Let E denote the expected frequencies. It is given that the days are expected to occur with the same frequency on each day.

The expected frequencies for each of the 7 days is equal to:

\(\begin{aligned}{c}E = \frac{{1095}}{7}\\ = 156.4286\end{aligned}\)

Assuming the experimental units are selected randomly and since the expected values are larger than 5, the requirements for the test are met.

State the hypotheses

The null hypothesis for conducting the given test is as follows:

The police calls occur equally frequently on the different days of the week.

The alternative hypothesis is as follows:

The police calls do not occur equally frequently on the different days of the week.

The test is right-tailed.

If the absolute value of the test statistic is greater than the critical value, the null hypothesis is rejected.

Compute the test statistic

The table below shows the necessary calculations:

Day	O	E	\(\left( {O - E} \right)\)	\({\left( {O - E} \right)^2}\)	\(\frac{{{{\left( {O - E} \right)}^2}}}{E}\)
Monday	114	156.4286	-42.4286	1800.184	11.5080
Tuesday	152	156.4286	-4.4286	19.61224	0.12537
Wednesday	160	156.4286	3.5714	12.7551	0.0815
Thursday	164	156.4286	7.5714	57.3265	0.3665
Friday	179	156.4286	22.5714	509.4694	3.2569
Saturday	196	156.4286	39.5714	1565.898	10.0103
Sunday	130	156.4286	-26.4286	698.4694	4.4651

The value of the test statistic is equal to:

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = 11.5080 + 0.12538 + ... + 4.4651\\ = 29.8137\end{aligned}\)

Thus,\({\chi ^2} = 29.8137\).

Let k be the number of days, which are equal to 7.

The degrees of freedom for\({\chi ^2}\)is computed below:

\(\begin{aligned}{c}df = k - 1\\ = 7 - 1\\ = 6\end{aligned}\)

The critical value of\({\chi ^2}\)at\(\alpha = 0.01\)with 6 degrees of freedom is equal to 16.8119.

The p-value is equal to,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 16.812} \right)\\ = 0.000\end{aligned}\)

Since the test statistic value is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

State the conclusion

There is enough evidence to conclude that the police calls do not occur equally frequently on the different days of the week.

The trend observed in the observed frequencies is that the calls increase from Monday through Saturday and decrease drastically on Sunday.