91Ó°ÊÓ

The frequency of wins is provided for the different post positions in the 141 Kentucky Derby horse races.

Assume the experimental units are obtained by random sampling.

If the expected frequencies are larger than 5, the requirements of the test are fulfilled.

Let O denote the observed frequencies of the wins.

The following values are obtained:

\(\begin{aligned}{l}{O_1} = 19\\{O_2} = 14\\{O_3} = 11\\{O_4} = 15\end{aligned}\)

\(\begin{aligned}{l}{O_5} = 15\\{O_6} = 7\\{O_7} = 8\\{O_8} = 12\end{aligned}\)

\(\begin{aligned}{l}{O_9} = 5\\{O_{10}} = 11\end{aligned}\)

The sum of all observed frequencies is computed below:

\(\begin{aligned}{c}n = 19 + 14 + .... + 11\\ = 117\end{aligned}\)

Let E denote the expected frequencies.It is given that the number of wins is expected to be the same for all the positions.

The expected frequencies for each of the 10 positions are equal to:

\(\begin{aligned}{c}E = \frac{{117}}{{10}}\\ = 11.7\end{aligned}\)

Thus, the requirements of the test are satisfied.

The null hypothesis for conducting the given test is as follows:

\({H_0}:\)The probability of winning is the same for different post positions.

The alternative hypothesis is as follows:

\({H_a}:\)The probability of winning is not the same for different post positions.

The table below shows the necessary calculations:

The value of the test statistic is equal to:

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \;\;\\ = 4.5547 + 0.4521 + ... + 0.0419\\ = 13.855\end{aligned}\)

Thus,\({\chi ^2} = 13.855\).

Let k be the number of posts that are equal to 10.

The degrees of freedom for\({\chi ^2}\)is computed below:

\(\begin{aligned}{c}df = k - 1\\ = 10 - 1\\ = 9\end{aligned}\)

The chi-square table is used to determine the critical value of\({\chi ^2}\)at\(\alpha = 0.05\)with 9 degrees of freedom as 16.919.

The p-value is equal to,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 13.855} \right)\\ = 0.128\end{aligned}\)

Since the test statistic value is less than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to be rejected.

There is enough evidence to conclude that the probability of winning is same for different post positions.

Since there is no significant difference in the frequency of wins for the different post positions, there isn’t the need for the bettors to consider the post position of a horse in the Kentucky Derby.