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Home Field Advantage Is there a home field advantage in soccer? We are specifically interested in the Football Association (FA) premier league, a football (soccer) league in Great Britain known for having especially passionate fans. We took a sample of 120 matches (excluding all ties) and found that the home team was victorious in 70 cases \(^{56}\) (a) What is the population of interest? What is the specific population parameter of interest? (b) Estimate the population parameter using the sample. (c) Using StatKey or other technology, construct and interpret a \(90 \%\) confidence interval. (d) Using StatKey or other technology, construct and interpret a \(99 \%\) confidence interval. (e) Based on this sample and the results in parts (c) and (d), are we \(90 \%\) confident a home field advantage exists? Are we \(99 \%\) confident?

Short Answer

Expert verified
The population of interest is all matches (excluding ties) in the Football Association Premier league with the specific population parameter of interest being the true proportion of matches that the home team wins. The estimated parameter based on the sample is 0.583. The 90% confidence interval for the parameter is approximately (0.51, 0.66), and the 99% confidence interval is approximately (0.47, 0.69). There is sufficient evidence to suggest a home field advantage at both 90% and 99% confidence levels as neither intervals include the proportion of 0.5, representing an equal chance of winning. Therefore, we can be fairly confident that a home field advantage exists.

Step by step solution

01

Identify the Population and Parameter

Population of interest for this study is all matches (excluding ties) in the Football Association Premier league. The parameter of interest is the true proportion of matches where the home team is victorious (p).
02

Estimate the Population Parameter

An estimate of the population parameter can be obtained using the sample proportion (\(\hat{p}\)). It is calculated as the number of successes (in this case victories by the home team) divided by the total number of observations. Here, \(\hat{p} = 70 / 120 = 0.583\). This is an estimate of the true proportion of matches won by the home team.
03

Construct a 90% Confidence Interval

A 90% confidence interval can be computed using the formula: \( \hat{p} ± Z ∗ \sqrt{(\hat{p}(1−\hat{p})/n)} \), where Z is the z-value from a standard normal distribution that corresponds to the desired confidence level (1.645 for 90%). So the interval will be \(0.583 ± 1.645 ∗ \sqrt{(0.583 ∗ (1 - 0.583) / 120)}\). When these calculations are carried out, the resulting 90% confidence interval is approximately (0.51, 0.66).
04

Construct a 99% Confidence Interval

Applying the same concept for a 99% confidence interval, with the z-value of 2.576. Thus, the interval will be approximately \(0.583 ± 2.576 ∗ \sqrt{(0.583 ∗ (1 - 0.583) / 120)}\), which yields the interval approximately (0.47, 0.69).
05

Interpretation and Conclusion

Based on the sample, the 90% confidence interval indicates that we are 90% confident that the true proportion of soccer games won by the home team in the Football Association Premier League lies between 0.51 and 0.66. Similarly, the 99% confidence interval indicates that we are 99% confident that the true proportion falls between 0.47 and 0.69. Because neither interval contains 0.5 (indicating no specific home advantage), we can infer that there is statistical evidence to suggest a home field advantage in soccer.

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Key Concepts

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

Population Parameter
A population parameter is a crucial concept in statistics. It refers to a numerical summary of a characteristic for the entire population. Imagine you want to know some true characteristic about every match in the FA Premier League where the outcome isn't a tie. That's your population, consisting of all such matches.
For instance, in the given exercise, the population parameter of interest is the true proportion of matches where the home team wins. This is represented by the symbol \( p \). It's essential to understand that while the value of \( p \) is fixed, it is often unknown because collecting data from each member of the population can be practically impossible.
Instead, statisticians use methods to estimate this parameter using samples. In our case, the sample consists of 120 matches, out of which we observed 70 victories by the home team. By studying these samples, we aim to make inferences about the larger population.
Sample Proportion
The sample we collect from a population allows us to calculate what's known as a sample proportion. This is often denoted by \( \hat{p} \), where the hat symbol indicates it's an estimate of the true proportion \( p \). In our scenario, \( \hat{p} \) represents the proportion of matches won by the home team from our sample of 120 matches.
To determine \( \hat{p} \), we divide the number of successes in our sample (victories by the home team) by the total number of observations (the matches sampled). So, \( \hat{p} = \frac{70}{120} = 0.583 \). This value of 0.583 tells us that in about 58.3% of matches, the home team was victorious in our sample.
Keep in mind, this sample proportion is a way to approximate the true underlying population parameter and provides a foundation for further statistical analysis, like constructing confidence intervals to make more precise inferences.
Statistical Evidence
The process of constructing confidence intervals allows us to determine statistical evidence regarding a hypothesis. When we talk about statistical evidence, we're talking about how likely it is that our observations are reflective of the true state of the population.
By calculating confidence intervals, such as the 90% or 99% intervals done in this exercise, we're essentially saying, "Here's a range where we expect the true population parameter to be." The intervals provide boundaries based on our sample data, helping us infer about the whole population.
If a confidence interval for a parameter, like the proportion of home wins, doesn't include certain critical values (like 0.5 in our case), it provides evidence of a real effect or trend. Since neither the 90% nor the 99% confidence interval includes 0.5, this suggests that the home teams win more often than 50% of the time, giving us statistical evidence to suggest a home field advantage exists.

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Most popular questions from this chapter

Student Misinterpretations Suppose that a student is working on a statistics project using data on pulse rates collected from a random sample of 100 students from her college. She finds a \(95 \%\) confidence interval for mean pulse rate to be (65.5,71.8) . Discuss how each of the statements below would indicate an improper interpretation of this interval. (a) I am \(95 \%\) sure that all students will have pulse rates between 65.5 and 71.8 beats per minute. (b) I am \(95 \%\) sure that the mean pulse rate for this sample of students will fall between 65.5 and 71.8 beats per minute. (c) I am 95\% sure that the confidence interval for the average pulse rate of all students at this college goes from 65.5 to 71.8 beats per minute. (d) I am sure that \(95 \%\) of all students at this college will have pulse rates between 65.5 and 71.8 beats per minute. (e) I am \(95 \%\) sure that the mean pulse rate for all US college students is between 65.5 and 71.8 beats per minute. (f) Of the mean pulse rates for students at this college, \(95 \%\) will fall between 65.5 and 71.8 beats per minute. (g) Of random samples of this size taken from students at this college, \(95 \%\) will have mean pulse rates between 65.5 and 71.8 beats per minute.

Do You Find Solitude Distressing? "For many people, being left alone with their thoughts is a most undesirable activity," says a psychologist involved in a study examining reactions to solitude. \({ }^{26}\) In the study, 146 college students were asked to hand over their cell phones and sit alone, thinking, for about 10 minutes. Afterward, 76 of the participants rated the experience as unpleasant. Use this information to estimate the proportion of all college students who would find it unpleasant to sit alone with their thoughts. (This reaction is not limited to college students: in a follow-up study involving adults ages 18 to 77 , a similar outcome was reported.) (a) Give notation for the quantity being estimated, and define any parameters used. (b) Give notation for the quantity that gives the best estimate, and give its value. (c) Give a \(95 \%\) confidence interval for the quantity being estimated, given that the margin of error for the estimate is \(8 \%\).

Do You Prefer Pain over Solitude? Exercise 3.58 describes a study in which college students found it unpleasant to sit alone and think. The same article describes a second study in which college students appear to prefer receiving an electric shock to sitting in solitude. The article states that "when asked to spend 15 minutes in solitary thought, 12 of 18 men and 6 of 24 women voluntarily gave themselves at least one electric shock." Use this information to estimate the difference between men and women in the proportion preferring pain over solitude. The standard error of the estimate is 0.154 (a) Give notation for the quantity being estimated, and define any parameters used. (b) Give notation for the quantity that gives the best estimate, and give its value. (c) Give a \(95 \%\) confidence interval for the quantity being estimated. (d) Is "no difference" between males and females a plausible value for the difference in proportions?

In Exercises 3.51 to 3.56 , information about a sample is given. Assuming that the sampling distribution is symmetric and bell-shaped, use the information to give a \(95 \%\) confidence interval, and indicate the parameter being estimated. \(\hat{p}_{1}-\hat{p}_{2}=0.08\) and the margin of error for \(95 \%\) confidence is \(\pm 3 \%\).

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