91Ó°ÊÓ

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section. Before the overtime rule in the National Football League was changed in 2011 , among 460 overtime games, 252 were won by the team that won the coin toss at the beginning of overtime. Using a 0.05 significance level, test the claim that the coin toss is fair in the sense that neither team has an advantage by winning it. Does the coin toss appear to be fair?

Step by step solution

01

State the Hypotheses

Identify the null and alternative hypotheses. The null hypothesis (H_0) is that the coin toss is fair, so the probability (p) of winning after winning the coin toss is 0.5.\[ H_0: p = 0.5 \]The alternative hypothesis (H_a) is that the probability is not 0.5,\[ H_a: p eq 0.5 \]

02

Calculate the Test Statistic

Using the normal distribution approximation to the binomial distribution, calculate the test statistic. First, find the sample proportion (\hat{p}).\[ \hat{p} = \frac{252}{460} \approx 0.5478 \]Next, calculate the standard error (SE) using the hypothesized proportion.\[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5 \cdot 0.5}{460}} \approx 0.0233 \]Now, calculate the test statistic (z-score) using the following formula:\[ z = \frac{\hat{p} - p}{SE} = \frac{0.5478 - 0.5}{0.0233} \approx 2.06 \]

03

Determine the P-value

Using the test statistic, find the corresponding P-value from the standard normal (Z) table. For a z-score of 2.06, the P-value is approximately twice the one-tailed area:\[ P = 2 \cdot P(Z > 2.06) \approx 2 \cdot 0.0197 = 0.0394 \]

04

State Conclusion About Null Hypothesis

Compare the P-value with the significance level (\alpha = 0.05). Since the P-value (0.0394) is less than \alpha (0.05), reject the null hypothesis.\[ P-value < \alpha \Rightarrow \, Reject \, H_0 \]

05

Final Conclusion

Based on the rejection of the null hypothesis, conclude that there is sufficient evidence to suggest that the coin toss is not fair. The coin toss appears to give an advantage to the team that wins it.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

null hypothesis

In hypothesis testing, the **null hypothesis** is a statement that suggests there is no effect or no difference in a certain context. It's essentially the default assumption that there is nothing unusual happening. In the given problem, the null hypothesis \(H_0\) is that the coin toss is fair. Mathematically, this means the probability (p) of winning after winning the coin toss is 0.5: \[ H_0: p = 0.5 \] This serves as the basis for comparison against an alternative hypothesis, which we will discuss next.

alternative hypothesis

The **alternative hypothesis** is the statement you want to test against the null hypothesis. It represents the idea that there is an effect or a difference. For the coin toss problem, the alternative hypothesis \(H_a\) suggests that the probability of winning after winning the coin toss is not 0.5: \[ H_a: p eq 0.5 \] This means we suspect that the coin toss is not fair, and one team might have an advantage.

P-value

The **P-value** is a crucial concept in statistical hypothesis testing. It tells us the probability of obtaining test results at least as extreme as the one observed, assuming the null hypothesis is true. In our case, after calculating the test statistic (z-score), we find the P-value. For a z-score of 2.06, the P-value is approximately twice the one-tailed area: \[ P = 2 \cdot P(Z > 2.06) \] This results in \[ P \approx 2 \cdot 0.0197 = 0.0394 \] A low P-value (less than the significance level) indicates that the observed result is unlikely under the null hypothesis, leading us to reject the null hypothesis.

normal distribution approximation

In many statistical tests, we use the **normal distribution approximation**. This allows us to approximate the binomial distribution as a normal distribution when the sample size is large enough. In this exercise, considering the large number of games (460), we use the normal approximation to simplify our calculations. First, we find the sample proportion: \[ \hat{p} = \frac{252}{460} \approx 0.5478 \] Next, we calculate the standard error: \[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5 \cdot 0.5}{460}} \approx 0.0233 \] We then calculate the z-score: \[ z = \frac{\hat{p} - p}{SE} = \frac{0.5478 - 0.5}{0.0233} \approx 2.06 \] This z-score can be compared against a standard normal distribution to find our P-value.

significance level

The **significance level** (often denoted as \( \alpha \) ) is a threshold set for determining whether to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true (Type I error). For this problem, the significance level is set at \(0.05\). We compare the P-value to \( \alpha \). If the P-value is smaller than \( \alpha \), we reject the null hypothesis. In this case, since the P-value \( 0.0394 \) is less than \( \alpha = 0.05 \), we reject the null hypothesis. This means we have enough evidence to suggest that the coin toss is not fair.