Using the Runs Test for Randomness. In Exercises 5–10, use the runs test with a significance level of\(\alpha \)= 0.05. (All data are listed in order by row.) Law Enforcement Fatalities Listed below are numbers of law enforcement fatalities for 20 recent and consecutive years. First find the mean, identify each value as being above the mean (A) or below the mean (B), then test for randomness above and below the mean. Is there a trend?

183	140	172	171	144	162	241	159	150	165
163	156	192	148	125	161	171	126	107	117

Data are given on the number of law enforcement fatalities for 20 years.

The mean value of the data is computed as follows:

\(\begin{array}{c}Mean = \frac{{183 + 140 + ..... + 117}}{{20}}\\ = 157.65\end{array}\)

Therefore, the mean value is equal to 157.65.

Represent values that are greater than the mean as A.

Represent values that are less than the mean as B.

The table below shows the transformed data

The null hypothesis is as follows:

The given data is random.

The alternative hypothesis is as follows:

The given data is not random.

If the value of the number of runs is less than or equal to the smaller critical value or greater than or equal to the larger critical value, the null hypothesis is rejected.

The sequence is as follows:

Now, the number of times A occurs is denoted by\({n_1}\), and the number of times B occurs is denoted by\({n_2}\).

Thus,

\(\begin{array}{l}{n_1} = 11\\{n_2} = 9\end{array}\)

The runs of the sequence are formed as follows:

\(\;\underbrace A_{{1^{st}}run}\underbrace B_{{2^{nd}}run}\underbrace {AA}_{{3^{rd}}run}\underbrace B_{{4^{th}}run}\underbrace {AAA}_{{5^{th}}run}\underbrace B_{{6^{th}}run}\underbrace {AA}_{{7^{th}}run}\underbrace B_{{8^{th}}run}\underbrace A_{{9^{th}}run}\underbrace {BB}_{{{10}^{th}}run}\underbrace {AA}_{{{11}^{th}}run}\underbrace {BBB}_{{{12}^{th}}run}\)

The number of runs denoted by G is equal to 12.

Here,\({n_1} \le 20\)and\({n_2} \le 20\).

Thus, the test statistic is G, and the level of significance\(\left( \alpha \right)\)is equal to 0.05.

The critical values of G for\({n_1} = 10\)and\({n_2} = 10\)are 6 and 16, respectively.

The value of Gequal to 12 is neither less than or equal to 6 nor greater than or equal to 16. Thus, the decision if fail to reject the null hypothesis.

There is not enough evidence to conclude that the given sample is not random.

Since the sample israndom, it can be said that it does not follow any particular trend.