Question

The accompanying table lists results of overtime football

games before and after the overtime rule was changed in the National Football League in 2011. Use a 0.05 significance level to test the claim of independence between winning an overtime game and whether playing under the old rule or the new rule. What do the results suggest about

the effectiveness of the rule change?

	Before Rule Change	After Rule Change
Overtime Coin Toss Winner Won the Game	252	24
Overtime Coin Toss Winner Lost the Game	208	23

Short answer

Winning an overtime game and whether playing under the old rule or the new rule are independent. And hence the rule change is not effective.

Step-by-step solution

Given information

The data for overtime football games before and after the overtime rule was changed in the National Football League in 2011 is provided.

Compute the expected frequencies

Theexpected frequencyis computed as,

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

The observed frequencies(O), along with row and column totals is tabulated below,

	Before Rule Change	After Rule Change	Row total
Overtime Coin Toss Winner Won the Game	252	24	276
Overtime Coin Toss Winner Lost the Game	208	23	231
Column total	460	47	507

Theexpected ( E) frequency table is represented as,

	Before Rule Change	After Rule Change
Overtime Coin Toss Winner Won the Game	250.4142	25.5858
Overtime Coin Toss Winner Lost the Game	209.5858	21.4142

Assume the subjects are randomly selected for the study.

Since all the expected values are larger than 5, the requirement for chi-square test are fulfilled.

State the null and alternate hypothesis

The claim to test the independence of the two variables; the hypotheses are formulated as follows,

\({H_0}:\)Winning an overtime game and whether playing under the old rule or the new rule are independent.

\({H_a}:\)Winning an overtime game and whether playing under the old rule or the new rule are dependent.

Compute the test statistic

The value of the test statisticis computed as,

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {252 - 250.4142} \right)}^2}}}{{250.4142}} + \frac{{{{\left( {24 - 25.5858} \right)}^2}}}{{25.5858}} + ... + \frac{{{{\left( {23 - 21.4142} \right)}^2}}}{{21.4142}}\\ = 0.2378\end{aligned}\]

Therefore, the value of the test statistic is 0.2378.

Compute the degrees of freedom

The degrees of freedomare computed as,

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

Therefore, the degrees of freedom are 1.

Compute the critical value

From chi-square table, the critical value for the row corresponding to 1 degrees of freedom and at 0.05 level of significance is 3.841.

Therefore, the critical value is 3.841.

Also, the p-value is obtained from the table as 0.626.

State the decision

Since the critical (3.841) is greater than the value of the test statistic (0.2378), in case the null hypothesis fails to be rejected.

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

State the conclusion

There issufficient evidence to support the claim that winning an overtime game and whether playing under the old rule or the new rule are independent.

The results suggest that the change in rule is not effective on the winning in over time game.