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.)

Baseball World Series Victories Test the claim that the sequence of World Series wins by American League and National League teams is random. Given on the next page are recent results, with A = American League and N = National League.

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The teams that won the World Series and belong to two different Leagues are provided.

The researcher wants to check the claim that sequence of wins by American league and national league teams is random.

The null hypothesis is as follows:

The sequence of wins by American league and national league teams is random.

The alternative hypothesis is as follows:

The sequence of winsby American league and national league teamis not random.

If the value of the test statistic is less than or equal to a smaller critical value or greater than or equal to a larger critical value, the null hypothesis is rejected.

The sequence is as follows:

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

Thus,

\(\begin{array}{l}{n_1} = 47\\{n_2} = 63\end{array}\)

The runs of the sequence are formed as follows:

\(\underbrace A_{{1^{st}}run}\underbrace N_{{2^{nd}}run}\underbrace A_{{3^{rd}}run}\underbrace {NNN}_{{4^{th}}run}\underbrace {AAAA}_{{5^{th}}run}\underbrace N_{{6^{th}}run}\underbrace {AAAA}_{{7^{th}}run}\underbrace N_{{8^{th}}run}\underbrace A_{{9^{th}}run}\underbrace N_{{{10}^{th}}run}\underbrace {AA}_{{{11}^{th}}run}\underbrace {NN}_{{{12}^{th}}run}\underbrace A_{{{13}^{th}}run}\)

\(\underbrace {AA}_{{{14}^{th}}run}\underbrace N_{{{15}^{th}}run}\underbrace A_{{{16}^{th}}run}\underbrace {NN}_{{{17}^{th}}run}\underbrace {AAAAA}_{{{18}^{th}}run}\underbrace N_{{{19}^{th}}run}\underbrace A_{{{20}^{th}}run}\underbrace N_{{{21}^{st}}run}\underbrace A_{{{22}^{nd}}run}\underbrace N_{{{23}^{rd}}run}\underbrace A_{{{24}^{th}}run}\underbrace N_{{{25}^{th}}run}\underbrace {AAAAA}_{{{26}^{th}}run}\)

\(\underbrace A_{{{27}^{th}}run}\underbrace {NN}_{{{28}^{th}}run}\underbrace A_{{{29}^{th}}run}\underbrace N_{{{30}^{th}}run}\underbrace A_{{{31}^{st}}run}\underbrace {NN}_{{{32}^{nd}}run}\underbrace {AA}_{{{33}^{rd}}run}\underbrace {NNN}_{{{34}^{th}}run}\underbrace A_{{{35}^{th}}run}\underbrace N_{{{36}^{th}}run}\underbrace A_{{{37}^{th}}run}\underbrace N_{{{38}^{th}}run}\underbrace A_{{{39}^{th}}run}\underbrace N_{{{40}^{th}}run}\underbrace {AA}_{{{41}^{st}}run}\underbrace N_{{{42}^{nd}}run}\)

\(\underbrace {AA}_{{{43}^{rd}}run}\underbrace {NNNN}_{{{44}^{th}}run}\underbrace {AAA}_{{{45}^{th}}run}\underbrace N_{{{46}^{th}}run}\underbrace A_{{{47}^{th}}run}\underbrace N_{{{48}^{th}}run}\underbrace A_{{{49}^{th}}run}\underbrace N_{{{50}^{th}}run}\underbrace {AAA}_{{{51}^{th}}run}\underbrace N_{{{52}^{nd}}run}\underbrace A_{{{53}^{rd}}run}\underbrace N_{{{54}^{th}}run}\underbrace {AAA}_{{{55}^{th}}run}\)

\(\underbrace A_{{{56}^{th}}run}\underbrace N_{{{57}^{th}}run}\underbrace {AA}_{{{58}^{th}}run}\underbrace N_{{{59}^{th}}run}\underbrace A_{{{60}^{th}}run}\underbrace N_{{{61}^{st}}run}\underbrace A_{{{62}^{nd}}run}\underbrace {NNN}_{{{63}^{rd}}run}\underbrace A_{{{64}^{th}}run}\underbrace N_{{{65}^{th}}run}\)

The total number of runs is denoted by G is equal to 6.

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:

\(\begin{array}{c}{\mu _G} = \frac{{2{n_1}{n_2}}}{{{n_1} + {n_2}}} + 1\\ = \frac{{2\left( {47} \right)\left( {63} \right)}}{{47 + 63}} + 1\\ = 54.84\end{array}\)

The standard deviation of G is computed as follows:

\(\begin{array}{c}{\sigma _G} = \sqrt {\frac{{2{n_1}{n_2}\left( {2{n_1}{n_2} - {n_1} - {n_2}} \right)}}{{{{\left( {{n_1} + {n_2}} \right)}^2}\left( {{n_1} + {n_2} - 1} \right)}}} \\ = \sqrt {\frac{{2\left( {47} \right)\left( {63} \right)\left( {2\left( {47} \right)\left( {63} \right) - 47 - 63} \right)}}{{{{\left( {47 + 63} \right)}^2}\left( {47 + 63 - 1} \right)}}} \\ = 5.108\end{array}\)

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

\(\begin{array}{c}z = \frac{{G - {\mu _G}}}{{{\sigma _G}}}\\ = \frac{{65 - 54.84}}{{5.108}}\\ = 1.989\end{array}\)

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

The value of z equal to 1.9089 does not lie between the two critical values. Thus, the decision is to reject the null hypothesis.

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