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Chapter 3: Problem 129

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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Most popular questions from this chapter

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