In Exercises 5–20, conduct the hypothesis test and provide the test statistic and the P-value and, or critical value, and state the conclusion.

Police Calls Repeat Exercise 11 using these observed frequencies for police calls received during the month of March: Monday (208); Tuesday (224); Wednesday (246); Thursday (173); Friday (210); Saturday (236); Sunday (154). What is a fundamental error with this analysis?

The observed frequencies of the police calls on the 7 days of the week of in March are recorded.

It is expected that the calls occur equally frequently on the 7 days of the week.

Assume the recordings are taken from randomly selected experimental units.

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:

\(\begin{aligned}{l}{O_1} = 208\\{O_2} = 224\\{O_3} = 246\\{O_4} = 173\\{O_5} = 210\\{O_6} = 236\\{O_7} = 154\end{aligned}\)

The sum of all observed frequencies is computed below:

\(\begin{aligned}{c}n = 208 + 224 + .... + 154\\ = 1451\end{aligned}\)

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 are equal to:

\(\begin{aligned}{c}E = \frac{{1451}}{7}\\ = 207.2857\end{aligned}\)

As the expected frequencies are all greater than 5, the requirements of the test are satisfied.

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 table below shows the necessary calculations:

The value of the test statistic is equal to:

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \;\\ = 0.002461 + 1.34774 + ... + 13.69784\\ = 31.96278\end{aligned}\]

Thus,\({\chi ^2} = 31.963\).

Let k be the number of days which are 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 16.812.

The p-value is equal to,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 31.963} \right)\\ = 0.000\end{aligned}\)

Since the test statistic value is greater than the critical value and the p-value is less than 0.01, the null hypothesis is rejected.

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

A fundamental error in the given observations is that not all the days occur an equal number of times because March has 31 days. Some days will occur 4 times while others will occur 5 times in the month.

Thus, the number of calls on the days that occur 5 times will be more as compared to the other days.

So, the given analysis does not seem appropriate.