Question

In Exercises 5–20, conduct the hypothesis test and provide the test statistic and the P-value and , or critical value, and state the conclusion.

World Series Games The table below lists the numbers of games played in 105 Major League Baseball (MLB) World Series. This table also includes the expected proportions for the numbers of games in a World Series, assuming that in each series, both teams have about the same chance of winning. Use a 0.05 significance level to test the claim that the actual numbers of games fit the distribution indicated by the expected proportions.

Games Played	4	5	6	7
World Series Contests	21	23	23	38
Expected Proportion	2/16	4/16	5/16	5/16

Short answer

There is enough evidence to conclude that the number of games does not fit the expected distribution of proportions.

Step-by-step solution

Given information

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.

Check the requirements

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:

Number of Games (i)	Expected Proportions \(\left( {{p_i}} \right)\)
4	\(\frac{2}{{16}}\)
5	\(\frac{4}{{16}}\)
6	\(\frac{5}{{16}}\)
7	\(\frac{5}{{16}}\)

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.

State the hypotheses

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.

Conduct the hypothesis test

The table below shows the necessary calculations:

Games Played	O	E	\(\left( {O - E} \right)\)	\({\left( {O - E} \right)^2}\)	\(\frac{{{{\left( {O - E} \right)}^2}}}{E}\)
4	21	13.125	7.875	62.01563	4.725
5	23	26.25	-3.25	10.5625	0.402381
6	23	32.8125	-9.8125	96.28516	2.934405
7	38	32.8125	5.1875	26.91016	0.820119

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:

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

State the decision

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.

State the conclusion

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.