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The number of games played in 105 Major League Baseball (MLB) World Series is provided. The expected proportions for the number of games in a World Series are also given.

Let O denote the observed frequencies of the games.

The following values are obtained:

\(\begin{aligned}{l}{O_4} = 21\\{O_5} = 23\\{O_6} = 23\\{O_7} = 38\end{aligned}\)

The sum of all observed frequencies is computed below:

\(\begin{aligned}{c}n = 21 + 23 + 23 + 38\\ = 105\end{aligned}\)

Let E denote the expected frequencies.

Let pi be the expected proportions for the ith game.

The values of pi are tabulated below:

The expected frequencies for each number of games are equal to:

\(\begin{aligned}{c}{E_1} = n{p_1}\\ = 105\left( {\frac{2}{{16}}} \right)\\ = 13.125\end{aligned}\)

\(\begin{aligned}{c}{E_2} = n{p_2}\\ = 105\left( {\frac{5}{{16}}} \right)\\ = 26.25\end{aligned}\)

\(\begin{aligned}{c}{E_3} = n{p_3}\\ = 105\left( {\frac{5}{{16}}} \right)\\ = 32.8125\end{aligned}\)

\(\begin{aligned}{c}{E_4} = n{p_4}\\ = 105\left( {\frac{5}{{16}}} \right)\\ = 32.8125\end{aligned}\)

As all the expected values are larger than 5, the requirements for the test are fulfilled if it is assumed that sampling is conducted in a random manner.

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

\({H_0}:\)The number of games fits the expected distribution of proportions.

The alternative hypothesis is as follows:

\({H_a}:\)The number of games does not fit the expected distribution of proportions.

The test is right-tailed.

The table below shows the necessary calculations:

The value of the test statistic is equal to:

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \;\\ = 4.725 + 0.402381 + 2.934405 + 0.820119\\ = 8.881905\end{aligned}\]

Thus,\({\chi ^2} = 8.882\).

Let k be the games played which are 4.

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

\(\begin{aligned}{c}df = k - 1\\ = 4 - 1\\ = 3\end{aligned}\)

The critical value of\({\chi ^2}\)at\(\alpha = 0.05\)with 9 degrees of freedom is equal to 7.815.

The p-value is,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 8.882} \right)\\ = 0.031\end{aligned}\).

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

There is enough evidence to conclude that the actual number of games does not fit the distribution of expected proportions at the given level of significance.