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The frequencies of baseball players born in different months are provided.

Let O denote the observed frequencies of the players born in the 12 months.

The sum of all observed frequencies is computed below:

\(\begin{aligned}{c}n = 387 + 329 + ...... + 371\\ = 4515\end{aligned}\)

Let E denote the expected frequencies.

It is given that the number of births is expected to occur with equal frequency in all of the 12 months.

The expected frequency for each of the 12 months is the same and is equal to:

\(\begin{aligned}{c}E = \frac{{4515}}{{12}}\\ = 376.25\end{aligned}\)

As the expected value is greater than 5, the requirements for the test are satisfied.

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

\({H_0}:\)The frequency of baseball players鈥� births is equal in different months of the year.

The alternative hypothesis is as follows:

\({H_a}:\)The frequency of baseball players鈥� births is not equal in different months of the year.

The test is right-tailed.

The table below shows the necessary calculations:

The value of the test statistic is equal to:

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = 0.307143 + 5.933721 + ... + 0.073256\\ = 93.07176\end{aligned}\]

Thus,\({\chi ^2} = 93.072\).

Let k be the number of months, which is 12.

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

\(\begin{aligned}{c}df = k - 1\\ = 12 - 1\\ = 11\end{aligned}\)

The critical value of\({\chi ^2}\)at\(\alpha = 0.05\)with 11 degrees of freedom is equal to 19.675.

The p-value is equal to 0.000.

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 thatbaseball players are not born with the same frequency in different months of the year.

The sample data for verifying the author鈥檚 claim,

The sum is higher in case of births before july 31.

Thus, it does not support the claim of the author.