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Out of 460 overtime games, 252 of them are won by the team that has won the coin toss. It is claimed that the coin toss is fair, and an equal number of teams who win/do not win the coin toss go on to win the game.

The null hypothesis is written as follows.

The proportion of overtime wins that are won by the team that has won the coin toss is equal to 0.5.

$H_0:p=0.5$

The alternative hypothesis is written as follows.

The proportion of overtime wins that are won by the team that has won the coin toss is not equal to 0.5.

$H_1:p鈮�0.5$

The test is two-tailed.

The sample size is equal to n=460.

The sample proportion of overtime wins that are won by the team that has won the coin toss is computed below.

$$\begin{gathered} \hat{p}=\frac{252}{460} \\ =0.548 \end{gathered}$$

The population proportion of games that are won by the team that has won the coin toss is equal to 0.5.

The value of the test statistic is computed below.

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.548-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{460}}} \\ =2.05 \end{gathered}$$

Thus, z=2.05.

Referring to the standard normal table, the critical value of z at for a two-tailed test is equal to 1.96.

Referring to the standard normal table, the p-value for the test statistic value of 2.05 is equal to 0.0404.

As the p-value is less than 0.05, the null hypothesis is rejected.

There is enough evidence to reject the claim that the proportion of games that arewon by the team that has won the coin toss is equal to 0.5.

This concludes that the coin toss is not fair, and there is a greater chance of winning the game after winning the coin toss.