Question

Runs Test with Large Samples. In Exercises 9–12, use the runs test with a significance level of\(\alpha \)= 0.05. (All data are listed in order by row.)

Stock Market: Testing for Randomness Above and Below the Median Listed below are the annual high values of the Dow Jones Industrial Average for a recent sequence of years. Find the median, then test for randomness below and above the median. What does the result suggest about the stock market as an investment consideration?

969	995	943	985	969	842	951	1036	1052	892	882	1015
1000	908	898	1000	1024	1071	1287	1287	1553	1956	2722	2184
2791	3000	3169	3413	3794	3978	5216	6561	8259	9374	11568	11401
11350	10635	10454	10855	10941	12464	14198	13279	10580	11625	12929	13589
16577	18054

Short answer

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

There appears to be an upward trend,and investment in the stock market seems profitable.

Step-by-step solution

Given information

The annual high values of Dow Jones Industrial Average are provided.

Identify the hypothesis

The null hypothesis is as follows:

The given sequence of high values is random.

The alternative hypothesis is as follows:

The given sequence of high values is not random.

Calculate the median

The total number of observations in the dataset is equal to 50.

The following table shows the data values arranged in ascending order:

842	995	1553	5216	11401
882	1000	1956	6561	11568
892	1000	2184	8259	11625
898	1015	2722	9374	12464
908	1024	2791	10454	12929
943	1036	3000	10580	13279
951	1052	3169	10635	13589
969	1071	3413	10855	14198
969	1287	3794	10941	16577
985	1287	3978	11350	18054

Since the number of observations is odd, the following formula is used to compute the median value:

\(\begin{array}{c}Median = \frac{{{{\left( {\frac{n}{2}} \right)}^{th}}obs + {{\left( {\frac{n}{2} + 1} \right)}^{th}}obs}}{2}\\ = \frac{{{{25}^{th}}obs + {{26}^{th}}obs}}{2}\\ = \frac{{2791 + 3000}}{2}\\ = 2895.5\end{array}\)

Thus, the median is equal to 2895.5.

Data transformation

Assign letter A to all values above the median value.

Assign letter B to all values below the median value.

The following table shows the values along with the symbols:

969	B	1000	B	5216	A	11625	A
995	B	1024	B	6561	A	12929	A
943	B	1071	B	8259	A	13589	A
985	B	1287	B	9374	A	16577	A
969	B	1287	B	11568	A	18054	A
842	B	1553	B	11401	A
951	B	1956	B	11350	A
1036	B	2722	B	10635	A
1052	B	2184	B	10454	A
892	B	2791	B	10855	A
882	B	3000	A	10941	A
1015	B	3169	A	12464	A
1000	B	3413	A	14198	A
908	B	3794	A	13279	A
898	B	3978	A	10580	A

Calculate the test statistic

The sequence of high values is as follows:

B

B

B

B

B

B

B

B

B

B

B

B

B

B

B

B

B

B

B

B

B

B

B

B

B

A

A

A

A

A

A

A

A

A

A

A

A

A

A

A

A

A

A

A

A

A

A

A

A

A

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

Thus,

\(\begin{array}{l}{n_1} = 25\\{n_2} = 25\end{array}\)

The runs of the sequence are formed as follows:

\(\underbrace {BBBBBBBBBBBBBBBBBBBBBBBBB}_{{1^{st}}run}\underbrace {AAAAAAAAAAAAAAAAAAAAAAAA}_{{2^{nd}}run}\)

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

Here,\({n_1} > 20\)and\({n_2} > 20\). The value of the test statistic z needs to be calculated.

The mean value of G is calculated as follows:

\(\begin{array}{c}{\mu _G} = \frac{{2{n_1}{n_2}}}{{{n_1} + {n_2}}} + 1\\ = \frac{{2\left( {25} \right)\left( {25} \right)}}{{25 + 25}} + 1\\ = 26\end{array}\)

The standard deviation of G is computed as follows:

\(\begin{array}{c}{\sigma _G} = \sqrt {\frac{{2{n_1}{n_2}\left( {2{n_1}{n_2} - {n_1} - {n_2}} \right)}}{{{{\left( {{n_1} + {n_2}} \right)}^2}\left( {{n_1} + {n_2} - 1} \right)}}} \\ = \sqrt {\frac{{2\left( {25} \right)\left( {25} \right)\left( {2\left( {25} \right)\left( {25} \right) - 25 - 25} \right)}}{{{{\left( {25 + 25} \right)}^2}\left( {25 + 25 - 1} \right)}}} \\ = 3.499\end{array}\)

Thus, the test statistic (G) is computed below:

\(\begin{array}{c}z = \frac{{G - {\mu _G}}}{{{\sigma _G}}}\\ = \frac{{2 - 26}}{{3.499}}\\ = - 6.859\end{array}\)

Determine the critical value and the conclusion of the test

The critical values of z at\(\alpha = 0.05\)are -1.96 and 1.96.

The value of z equal to -6.859 is less than the smaller critical value of -1.96. Thus, the null hypothesis is rejected.

There is enough evidence to conclude that the given sequence of high values is not random.

Determine the trends

It can be observed that the values are constantly rising.

Thus, it can be said that there is an upward trend, and investing in the stock market appears to be profitable for long term.