Question

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.)

Newspapers Media experts claim that daily print newspapers are declining because of Internet access. Listed below are the numbers of daily print newspapers in the United States for a recent sequence of years. First find the median, then test for randomness of the numbers above and below the median. What do the results suggest?

1611	1586	1570	1556	1548	1533	1520	1509	1489	1483	1480
1468	1457	1456	1457	1452	1437	1422	1408	1387	1382

Short answer

It can be concluded that the given sample of the number of daily print newspapers is random.

The result suggests a downward trendas the number of daily print newspapers isdeclining with each passing year.

Step-by-step solution

Given information

Data areprovided on the number of newspapers printed daily for a series of years.

Identify the hypothesis

The null hypothesis is as follows:

The given sequence of the number of print newspapers is random.

The alternative hypothesis is as follows:

The given sequence of the number of print newspapers 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.

Calculate the median

The number of daily print newspapers arranged in ascending orderistabulated below:

S.No.	Values
1	1382
2	1387
3	1408
4	1422
5	1437
6	1452
7	1456
8	1457
9	1457
10	1468
11	1480
12	1483
13	1489
14	1509
15	1520
16	1533
17	1548
18	1556
19	1570
20	1586
21	1611

The total number of values given is equal to 21.

As the number of values is odd, the following formula is used to compute the median:

\(\begin{array}{c}Median = {\left( {\frac{{n + 1}}{2}} \right)^{th}}obs\\ = {\left( {\frac{{22}}{2}} \right)^{th}}obs\\ = {11^{th}}obs\\ = 1480\end{array}\)

Symbolize the values above the median using A and below the median using B.

If a value is equal to the median, exclude it.

The table below shows the indicated symbols:

1611	A
1586	A
1570	A
1556	A
1548	A
1533	A
1520	A
1509	A
1489	A
1483	A
1468	B
1457	B
1456	B
1457	B
1452	B
1437	B
1422	B
1408	B
1387	B
1382	B

Step 4:Calculate the test statistic

The sequence is as follows:

A

A

A

A

A

A

A

A

A

A

B

B

B

B

B

B

B

B

B

B

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,

\(\begin{array}{l}{n_1} = 10\\{n_2} = 10\end{array}\)

The runs of the sequence are formed as follows:

\(\underbrace {AAAAAAAAAA}_{{1^{st}}run}\underbrace {BBBBBBBBBB}_{{2^{nd}}run}\)

The number of runs denoted by G is equal to 2.

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

Thus, the test statistic is equal to G =2.

Determine the critical value and the conclusion of the test

The critical values of G for\({n_1} = 10\)and\({n_2} = 10\)at\(\alpha = 0.05\)are 6 and 16.

As the value of G is less than the smaller critical value of 6, the null hypothesis is rejected.

There is enough evidence to conclude that the given sample of the number of daily print newspapers is not random.

Thus, the sequenceof the number of print newspapers is not random.

It can be observed that all values below the median appear at the end, and those above the median appear at the beginning.

As there is a downward trend in the values, it suggests that the number of daily print newspapers is declining with each passing year, and the claim of the experts is true.