Question

Winning team data were collected for teams in different sports, with the results given in the table on the top of the next page (based on data from “Predicting Professional Sports Game Outcomes fromIntermediateGame Scores,” by Copper, DeNeve, and Mosteller, Chance,Vol. 5, No. 3–4). Use a 0.10significance level to test the claim that home/visitor wins are independent of the sport. Given that among the four sports included here, baseball is the only sport in which the home team canmodify field dimensions to favor its own players, does it appear that baseball teams are effective in using this advantage?

	Basketball	Baseball	Hockey	Football
Home Team Wins	127	53	50	57
Visiting Team Wins	71	47	43	42

Short answer

Home/visitor wins are independent of the sport.

Thus, the advantage to the baseball teams is not effective.

Step-by-step solution

Given information

The data for different sports and home/Visiting team wins is provided.

The level of significance 0.10.

Compute the expected frequencies

The formula forexpected frequencyis,

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The table with row and column total for observed counts is represented as,

	Basketball	Baseball	Hockey	Football	Row total
Home Team Wins	127	53	50	57	287
Visiting Team Wins	71	47	43	42	203
Column total	198	100	93	99	490

Theexpected frequency tableis represented as,

	Basketball	Baseball	Hockey	Football
Home Team Wins	115.9714	58.5714	54.4714	57.9857
Visiting Team Wins	82.0286	41.4286	38.5286	41.0143

State the null and alternate hypothesis

The hypotheses are formulated as,

\({H_0}:\)Home/visitor wins are independent of the sport.

\({H_1}:\)Home/visitor wins are dependent on the sport.

Compute the test statistic

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {127 - 115.9714} \right)}^2}}}{{115.9714}} + \frac{{{{\left( {53 - 58.5714} \right)}^2}}}{{58.5714}} + ... + \frac{{{{\left( {42 - 41.0143} \right)}^2}}}{{41.0143}}\\ = 4.7372\end{aligned}\]

Therefore, the value of the test statistic is 4.7372.

Compute the degrees of freedom

The degrees of freedomare computed as,

\(\begin{aligned}{c}\left( {r - 1} \right)\left( {c - 1} \right) = \left( {2 - 1} \right)\left( {4 - 1} \right)\\ = 3\end{aligned}\)

Therefore, the degrees of freedom are 3.

Compute the critical value

From the chi-square table, the critical value to 3 degrees of freedom and at 0.10 level of significance 6.2514.

Therefore, the critical value is 6.2514

State the decision

Since the critical value (6.2514) is greater than the value of the test statistic (4.7372). In this case, the null hypothesis fails to be rejected.

Therefore, the decision is to fail to reject the null hypothesis.

State the conclusion

There issufficient evidence to support the claim that Home/visitor wins are independent of the sport.

Thus, the wins are not dependent on the sport.

Thus, the advantage is not effective for the wins of baseball teams.