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Penalty Shots in World Cup Soccer A study \(^{13}\) 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
Following these steps will lead to a conclusion about whether there is evidence that the proportion of penalty shots guessed correctly by a goalkeeper is less than \(50\%\). This conclusion will be based on comparing the P-value to the given significance level (0.05).

Step by step solution

01

Statement of Hypotheses

First we establish the null and alternative hypotheses. The null hypothesis \(H_0\) is that the goalkeeper guesses the direction of the kick correctly \(50\%\) of the time, ie. \(p=0.5\). The alternative hypothesis \(H_a\) is that the goalkeeper guesses correctly less than \(50\%\) of the time, ie. \(p<0.5\) where p is the population proportion.
02

Calculate Z-score

We can calculate the Z-score, which measures how many standard deviations an element is from the mean, using the following formula: \(Z = \frac{\(p̂ - p\)}{SE}\), where \(p̂\) is the sample proportion, \(p\) is the population proportion under the null hypothesis, and SE is the standard error. Substituting in the values from the problem: \(Z = \frac{0.41-0.5}{0.043}\).
03

Calculate P-value

The P-value, which is the probability of observing a test statistic as extreme as the one calculated, given that the null hypothesis is true, can be calculated using a Z-table or statistical software. The Z-table is left-tailed because the alternative hypothesis tests for \(p<0.5\), and we find the probability corresponding to the calculated Z-score in our normal distribution. For a calculated Z-score, infer the corresponding P-value from a statistical chart or software.
04

Decision

If the P-value is less than the significance level of \(5\%\), we reject the null hypothesis in favor of the alternative hypothesis. If the P-value is greater than or equal to the significance level, we do not reject the null hypothesis.

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Key Concepts

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

Significance Level
Understanding the significance level is crucial in hypothesis testing as it sets the threshold for making decisions. In simpler terms, the significance level, often denoted as \( \alpha \), is the probability of rejecting the null hypothesis when it is true. A common value is \( 5\% \) or 0.05. This means we are willing to accept a \( 5\% \) risk of concluding that a difference exists when there is actually no difference.
The choice of significance level affects the outcome of your hypothesis test:
  • If your P-value is less than \( \alpha \), you reject the null hypothesis.
  • If your P-value is greater or equal to \( \alpha \), you do not reject the null hypothesis.
In the context of the World Cup penalty shots, a \( 5\% \) significance level indicates we're okay with a \( 5\% \) chance of assuming goalkeepers are worse than random guessing, even if they are not. This keeps us cautious about changing our belief based on random variations.
Normal Distribution
Normal distribution is a critical concept in statistics and is often used in hypothesis testing because of its appealing properties. When data is normally distributed, it forms a bell-shaped curve. This curve reflects that most observations cluster around the central peak and fewer occur as we move away toward the tails.
The use of normal distribution in hypothesis testing is powerful, allowing us to make probabilistic inferences about population parameters. In our exercise, we expect the sampling distribution of the sample proportion to be approximately normal due to the large sample size of 138 penalty shots. This use of normal distribution provides the groundwork to calculate the Z-score, a measure of how far our sample statistic is from the assumed parameter under the null hypothesis.
P-value
The P-value is a measure that helps us understand the strength of evidence against the null hypothesis. It represents the probability of observing a test statistic at least as extreme as the one measured, assuming the null hypothesis is true.
Interpreting P-values can be simplified:
  • A small P-value (typically ≤ 0.05) suggests strong evidence against the null hypothesis, leading to its rejection.
  • A large P-value (> 0.05) suggests weak evidence against the null hypothesis, so we fail to reject it.
In the context of our soccer penalty shots, the P-value tells us whether the observed proportion of \( 41\% \) is significantly lower than the \( 50\% \) assumed under the null hypothesis. A P-value less than \( 0.05\) would indicate the goalkeepers might indeed be guessing less accurately than random chance.

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Most popular questions from this chapter

Where Is the Best Seat on the Plane? A survey of 1000 air travelers \({ }^{10}\) found that \(60 \%\) prefer a window seat. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is \(S E=0.015\). Use a normal distribution to find and interpret a \(99 \%\) confidence interval for the proportion of air travelers who prefer a window seat.

Commuting Times in St. Louis A bootstrap distribution of mean commute times (in minutes) based on a sample of \(500 \mathrm{St}\). Louis workers stored in CommuteStLouis is shown in Figure 5.13 . The pattern in this dotplot is reasonably bell-shaped so we use a normal curve to model this distribution of bootstrap means. The mean for this distribution is 21.97 minutes and the standard deviation is 0.65 minutes. Based on this normal distribution, what proportion of bootstrap means should be in each of the following regions? (a) More than 23 minutes (b) Less than 20 minutes (c) Between 21.5 and 22.5 minutes

How Often Do You Use Cash? In a survey \(^{14}\) of 1000 American adults conducted in April 2012 , \(43 \%\) reported having gone through an entire week without paying for anything in cash. Test to see if this sample provides evidence that the proportion of all American adults going a week without paying cash is greater than \(40 \%\). Use the fact that a randomization distribution is approximately normally distributed with a standard error of \(S E=0.016\). Show all details of the test and use a \(5 \%\) significance level.

Exercise and Gender The dataset ExerciseHours contains information on the amount of exercise (hours per week) for a sample of statistics students. The mean amount of exercise was 9.4 hours for the 30 female students in the sample and 12.4 hours for the 20 male students. A randomization distribution of differences in means based on these data, under a null hypothesis of no difference in mean exercise time between females and males, is centered near zero and reasonably normally distributed. The standard error for the difference in means, as estimated from the standard deviation of the randomization differences, is \(S E=2.38 .\) Use this information to test, at a \(5 \%\) level, whether the data show that the mean exercise time for female statistics students is less than the mean exercise time of male statistics students.

Exercises 5.50 to 5.55 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) when the samples have \(n_{1}=n_{2}=50, \bar{x}_{1}=35.4, \bar{x}_{2}=33.1,\) \(s_{1}=1.28,\) and \(s_{2}=1.17 .\) The standard error of \(\bar{x}_{1}-\bar{x}_{2}\) from the randomization distribution is \(0.25 .\)

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