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6.44 Proportion of home team wins in soccer, using \(\hat{p}=0.583\) with \(n=120\)

Short Answer

Expert verified
The home team won approximately 70 games out of total 120 games played.

Step by step solution

01

Understanding the given information

The given details are that the home team won 0.583 of the time over a total of 120 games. This means, the symbol \(\hat{p}\) represents the proportion of wins by the home team, which is 0.583. The variable \(n\) represents the total number of games, which is 120.
02

Calculate the precise number of wins

This can be calculated by multiplying the total number of games, \(n\), by the proportion of wins, \(\hat{p}\). So the equation is : \(\hat{p} * n = 0.583 * 120\).
03

Evaluate the result

By multiplying 0.583 by 120, the result is 69.96. This indicates that the home team won approximately 70 games (since we typically round to whole games in this context).

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Key Concepts

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

Binomial Proportions
When we talk about the proportion of home team wins in soccer, we're often discussing binomial proportions. These are outcomes from repeated trials (such as soccer matches) that have two possible results: a win or a loss. For home team wins, we can use binomial proportions to estimate the likelihood of the home team winning a game.

In our case, the proportion of home team wins, denoted by \(\hat{p}\), is the number of successes (wins) divided by the total number of trials (games). With \(\hat{p}=0.583\) for 120 games, we're witnessing an empirical probability. But how did we arrive at this value? If we declare a 'win' as the 'success' in a binomial distribution, then out of 120 games played (our number of trials), the number of games won by the home team can be considered our number of successes.

This basic understanding is crucial because the calculation of binomial proportions involves estimating the likelihood of a given number of successes over a series of trials, assuming each trial is independent and has the same probability of winning. To break down further:
Statistical Significance
Let's delve into the notion of statistical significance. It's about determining whether a result—like the proportion of wins for the home team—is due to some real effect or simply random chance. Here's where your math classes come into play with a concept called the p-value, which helps in making such determinations.

A low p-value indicates that the observed data is unlikely to occur if there were no real effect. For example, if we observed a high proportion of home team wins, we might want to know: is this because the home team really does have an advantage, or could this just be a coincidence?

To check for statistical significance, we calculate the p-value and compare it to a predefined significance level, typically set at 0.05. If the p-value is less than the significance level, we can confidently say that the proportion of home team wins is statistically significant, meaning there's a real effect happening, not just luck or random chance at play.
Hypothesis Testing
Now, we arrive at hypothesis testing, a structured method to decide whether to accept or reject a hypothesis based on sample data. In the realm of soccer, for instance, we could establish a hypothesis that the home team has an advantage (and not merely due to chance).

In hypothesis testing, we start with two hypotheses: the null hypothesis (\(H_0\)), which typically posits that there is no effect or no difference, and the alternative hypothesis (\(H_1\) or \(H_a\)), which suggests that there is a significant effect or a difference.

If we were to apply hypothesis testing to our example with the proportion of home team wins, we might propose a null hypothesis stating, 'The home team wins 50% of the games,' and an alternative hypothesis saying, 'The home team wins more than 50% of the games.' We would then use our sample data (\(\hat{p}=0.583\) with \(n=120\)) to calculate the test statistic and the corresponding p-value.

Finally, based on our results and the p-value, we would make a decision: If our p-value is low (typically less than 0.05), we reject the null hypothesis in favor of the alternative hypothesis. This indicates that the home team's higher win rate is unlikely to be due to random variation alone and that there might be a real home advantage at play.

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Most popular questions from this chapter

In the dataset ICUAdmissions, the variable Infection indicates whether the ICU (Intensive Care Unit) patient had an infection (1) or not (0) and the variable Sex gives the gender of the patient \((0\) for males and 1 for females.) Use technology to test at a \(5 \%\) level whether there is a difference between males and females in the proportion of ICU patients with an infection.

A sample with \(n=10, \bar{x}=508.5,\) and \(s=21.5\)

According to the 2006 Australia Census, \(^{43} 25.5 \%\) of Australian women over the age of 25 had a college degree, while the percentage for Australian men was \(21.4 \% .\) Suppose we select random samples of 200 women and 200 men from this population and look at the differences in proportions with college degrees, \(\hat{p}_{f}-\hat{p}_{m}\), in those samples. (a) Describe the distribution (center, spread,shape) for the difference in sample proportions. Include a rough sketch of the distribution with values labeled on the horizontal axis. (b) What is the chance that the proportion with college degrees in the men's sample is actually more than the proportion in the women's sample? (Hint: Think about what must be true about \(\hat{p}_{f}-\hat{p}_{m}\) when this happens.)

Use a t-distribution. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=8\) and \(n_{2}=10\)

Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=-2.6, s_{d}=4.1\) \(n_{d}=18\)

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