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In the years between 1974 and 2011, out of 460 overtime games, 252 games were won by the team that won the coin toss.

In the first 3 years beginning from 2012, there were 47 overtime games, and 24 of those games were won by the team that won the coin toss.

It is given that the rule that was applied earlier (between the years 1974 and 2011) was unfair and led to a greater proportion of teams who won the game as well as the coin toss. The rules were changed to correct this practice.

An appropriate claim would be that the overtime rule change was effective in decreasing the bias in winning the game.

Corresponding to the given claim, the following hypotheses are set up:

Null Hypothesis: The proportion of overtime wins after the rules were changed is equal to the proportion of overtime wins before the rules were changed.

\({H_0}:{p_1} = {p_2}\)

Alternative Hypothesis: The proportion of overtime wins before the rules were changed is greater than the proportion of overtime wins after the rules were changed.

\({H_1}:{p_1} > {p_2}\)

The test is right-tailed.

If the test statistic value is greater than the critical value, the null hypothesis is to be rejected.

Let\({\hat p_1}\)denote the sampleproportion of overtime wins before the rules were changed.

\(\begin{aligned} {{\hat p}_1} &= \frac{{252}}{{460}}\\ &= 0.548\end{aligned}\)

Let\({\hat p_2}\)denote the sampleproportion of overtime wins after the rules were changed.

\(\begin{aligned} {{\hat p}_2} &= \frac{{24}}{{47}}\\ &= 0.511\end{aligned}\)

\({n_1}\)is equal to 460 and\({n_2}\)is equal to 47.

The value of the pooled sample proportion is computed as follows:

\(\begin{aligned} \bar p &= \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}}\\ &= \frac{{252 + 24}}{{460 + 47}}\\ &= 0.544\end{aligned}\)

\(\begin{aligned} \bar q &= 1 - \bar p\\ &= 1 - 0.544\\ &= 0.456\end{aligned}\)

The value of the test statistic is computed below:

\(\begin{aligned} z &= \frac{{\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - \left( {{p_1} - {p_2}} \right)}}{{\sqrt {\frac{{\bar p\bar q}}{{{n_1}}} + \frac{{\bar p\bar q}}{{{n_2}}}} }}\;\;\;\;{\rm{where}}\left( {{p_1} - {p_2}} \right) = 0\\ &= \frac{{\left( {0.548 - 0.511} \right) - 0}}{{\sqrt {\frac{{\left( {0.544} \right)\left( {0.456} \right)}}{{460}} + \frac{{\left( {0.544} \right)\left( {0.456} \right)}}{{47}}} }}\\ &= 0.488\end{aligned}\)

Thus, z=0.488.

Refer to the standard normal table:

The critical value of z corresponding to\(\alpha = 0.05\)for a right-tailed test is equal to 1.645.

The corresponding p-value obtained using the test statistic is equal to 0.3128.

Since the z-score value is less than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to reject.

The following formula of the confidence interval is used to estimate the difference in the two proportions:

\(CI = \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\)where

• \({\hat p_1}\)denotes the sampleproportion of people who received a $1 bill and spent it.
• \({\hat p_2}\)denotes the sampleproportion of people who received 4 quarters and spent the money.
• E is the margin of error

The level of significance is equal to 0.05. Thus, the corresponding confidence level is equal to 95%.

Refder to standard normal table, the value of\({z_{\frac{\alpha }{2}}}\)when\(\alpha = 0.05\)is equal to 1.96.

Thus, the margin of error is equal to:

\(\begin{aligned} E &= {z_{\frac{\alpha }{2}}}\sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{{{n_2}}}} \\ &= \left( {1.96} \right)\sqrt {\frac{{\left( {0.548} \right)\left( {1 - 0.548} \right)}}{{460}} + \frac{{\left( {0.511} \right)\left( {1 - 0.511} \right)}}{{47}}} \\ &= 0.14998\end{aligned}\)

Thus, the 95% confidence interval is equal to:

\(\begin{aligned} CI &= \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\\ &= \left( {\left( {0.548 - 0.511} \right) - 0.14998,\left( {0.548 - 0.511} \right) + 0.14998} \right)\\ &= \left( { - 0.113,0.187} \right)\end{aligned}\)

Therefore, the confidence interval to estimate the difference in the two proportions of overtime wins is equal to (-0.113, 0.187).

The null hypothesis fails to be rejected as the confidence interval constructed contains the null value 0.

Thus, there is not enough evidence to support the claim thatthe overtime rule change was effective in decreasing the bias in winning the game.