/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Penalty Shots in World Cup Socce... [FREE SOLUTION] | 91Ó°ÊÓ

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Penalty Shots in World Cup Soccer A study \(^{11}\) of 138 penalty shots in World Cup Finals games between 1982 and 1994 found that the goalkeeper correctly guessed the direction of the kick only \(41 \%\) of the time. The article notes that this is "slightly worse than random chance." We use these data as a sample of all World Cup penalty shots ever. Test at a \(5 \%\) significance level to see whether there is evidence that the percent guessed correctly is less than \(50 \%\). The sample size is large enough to use the normal distribution. The standard error from a randomization distribution under the null hypothesis is \(S E=0.043 .\)

Short Answer

Expert verified
Yes, there is evidence at the 5% significance level that the percent guessed correctly by the goalkeeper is less than 50%.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis \(H_0\) is that the goalkeeper guesses the direction correctly 50% of the time, i.e., the population proportion \(p = 0.50\). The alternative hypothesis \(H_a\) is that the population proportion is less than 50%, i.e., \(p < 0.50\).
02

Calculate the test statistic

We have a sample proportion \(\hat{p} = 0.41\), a null proportion \(p_0 = 0.50\), and a standard error SE = 0.043. The test statistic (z) for the sample proportion under the null hypothesis is calculated by the formula: \(z = (\hat{p} - p_0) / SE\). Substituting the given values, we get \(z = (0.41 - 0.50) / 0.043 = -2.09\) approximately.
03

Compute the p-value

We are testing a less than alternative, thus our p-value is the probability that a standard normal random variable is less than -2.09. Using the standard normal table, we find that the area to the left of -2.09 is approximately 0.018.
04

Conclusion

We conclude by comparing the p-value to the significance level (α = 0.05). Since the p-value (0.018) is less than the significance level, there is strong evidence to reject the null hypothesis in favor of the alternative hypothesis.

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Key Concepts

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

Normal Distribution
In statistics, the normal distribution is a fundamental concept. It's often called a "bell curve" due to its shape, which looks like a bell. This distribution is symmetrical, meaning it has the same shape on either side of the center. As you learn more about statistics, you'll find that many natural phenomena tend to follow this shape.
The normal distribution is characterized by two main parameters:
  • Mean (\(\mu\)): The average or center of the distribution.
  • Standard deviation (\(\sigma\)): This indicates how spread out the data is around the mean. A small standard deviation means the data points are close to the mean, while a larger standard deviation indicates the data is more spread out.
The concept of normality is crucial when conducting hypothesis testing, especially in large samples, as it allows statisticians to apply various statistical methods that rely on this distribution. In testing our goalkeeper problem, assuming the normal distribution helps us understand how the data (penalty shot guesses) should generally behave.
P-value Calculation
The p-value is a core aspect of hypothesis testing. It helps you determine the strength of your results. In simple terms, the p-value tells you how likely it is to obtain your test results under the assumption that the null hypothesis is true.
If you have a small p-value, it means that the observed data is unlikely to have happened purely by random chance under the null hypothesis. This indicates potential evidence against the null hypothesis. Here's how to interpret p-values:
  • P-value < 0.05: Strong evidence against the null hypothesis.
  • P-value > 0.05: Weak evidence against the null hypothesis.
In the context of the exercise, the p-value we calculated was approximately 0.018. This is less than 0.05, suggesting strong evidence against the null hypothesis, which means it's unlikely that the guessing accuracy is 50% by random chance. Calculating the p-value typically involves using a statistical function or table, as we did when looking at the standard normal distribution to find the area to the left of our test statistic.
Statistical Significance
Statistical significance is another key concept in hypothesis testing. When you say a result is statistically significant, you're claiming that it's unlikely the observed effect is due to chance alone. This judgment is always against a predetermined threshold known as the significance level. In hypothesis testing, the most common significance level used is 0.05. This translates to a 5% risk of concluding that a result exists when there is none. In our exercise about the penalty shot guesses, this level was explicitly set at 5%.
To determine statistical significance:
  • Compare the p-value with the significance level (α).
  • If the p-value is smaller than α, the result is statistically significant, meaning you have enough evidence to reject the null hypothesis.
In our penalty shots example, the p-value of 0.018 is lower than the significance level of 0.05, implying the results are statistically significant. Thus, we conclude there's strong evidence to suggest the goalkeeper's guessing rate is indeed less than 50%.

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