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

Expert verified

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

01

Given information

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

02

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.

03

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

04

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.

05

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.

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