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.)

Testing for Randomness of Super Bowl Victories Listed below are the conference designations of conference designations that won the Super Bowl, where N denotes a team from the NFC and A denotes a team from the AFC. Do the results suggest that either conference is superior?

N

N

A

A

A

N

A

A

A

A

A

N

A

A

A

N

N

A

N

N

N

N

N

N

N

N

N

N

N

N

N

A

A

N

A

A

N

A

A

A

A

N

A

N

N

N

A

N

A

Data are given on the conference designations of teams (NFC and AFC) that won the Super Bowl.

Here, N represents the team from NFC, and A represents the team from AFC.

The null hypothesis is as follows:

The given sequence of conference designations is random.

The alternative hypothesis is as follows:

The given sequence of conference designations is not random.

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,

\(\begin{array}{l}{n_1} = 26\\{n_2} = 23\end{array}\)

The runs of the sequence are formed as follows:

\(\begin{array}{l}\underbrace {NN}_{{1^{st}}run}\underbrace {AAA}_{{2^{nd}}run}\underbrace N_{{3^{rd}}run}\underbrace {AAAAA}_{{4^{th}}run}\underbrace N_{{5^{th}}run}\underbrace {AAA}_{{6^{th}}run}\underbrace {NN}_{{7^{th}}run}\underbrace A_{{8^{th}}run}\underbrace {NNNNNNNNNNNNN}_{{9^{th}}run}\underbrace {AA}_{{{10}^{th}}run}\\\underbrace N_{{{11}^{th}}run}\underbrace {AA}_{{{12}^{th}}run}\underbrace N_{{{13}^{rtrh}}run}\underbrace {AAAA}_{{{14}^{th}}run}\underbrace N_{{{15}^{th}}run}\underbrace A_{{{16}^{th}}run}\underbrace {NNN}_{{{17}^{th}}run}\underbrace A_{{{18}^{th}}run}\underbrace N_{{{19}^{th}}run}\underbrace A_{{{20}^{th}}run}\end{array}\)

The number of runs denoted by G is equal to 20.

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:

\(\begin{array}{c}{\mu _G} = \frac{{2{n_1}{n_2}}}{{{n_1} + {n_2}}} + 1\\ = \frac{{2\left( {26} \right)\left( {23} \right)}}{{26 + 23}} + 1\\ = 25.41\end{array}\)

The standard deviation of G is computed as follows:

\(\begin{array}{c}{\sigma _G} = \sqrt {\frac{{2{n_1}{n_2}\left( {2{n_1}{n_2} - {n_1} - {n_2}} \right)}}{{{{\left( {{n_1} + {n_2}} \right)}^2}\left( {{n_1} + {n_2} - 1} \right)}}} \\ = \sqrt {\frac{{2\left( {26} \right)\left( {23} \right)\left( {2\left( {26} \right)\left( {23} \right) - 26 - 23} \right)}}{{{{\left( {26 + 23} \right)}^2}\left( {26 + 23 - 1} \right)}}} \\ = 3.45\end{array}\)

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

\(\begin{array}{c}z = \frac{{G - {\mu _G}}}{{{\sigma _G}}}\\ = \frac{{20 - 25.41}}{{3.45}}\\ = - 1.568\end{array}\)

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

If the value of the test statistic 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 value of z equal to -1.568 is neither less than or equal to -1.96 nor greater than or equal to 1.96. Thus, the decision is fail to reject the null hypothesis.

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

Thus, the sequence of conference designations is random.

Since the sequence is randomly arranged, it can be said that none of the conferences is superior.