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?

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.

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.

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.

The table below shows the necessary calculations:

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.

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.