/*! 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 294 How big is the home field advant... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

How big is the home field advantage in the National Football League (NFL)? In Exercise 6.240 on page 419 , we examine a difference in means between home and away teams using two separate samples of 80 games from each group. However, many factors impact individual games, such as weather conditions and the scoring of the opponent. It makes more sense to investigate this question using a matched pairs design, using scores for home and away teams matched for the same game. The data in NFLScores2011 include the points scored by the home and away team in 256 regular season games in \(2011 .\) We will treat these games as a sample of all NFL games. Estimate average home field scoring advantage and find a \(90 \%\) confidence interval for the mean difference.

Short Answer

Expert verified
The average home field scoring advantage and its 90% confidence interval can be calculated using the steps provided. The values will depend on the actual scores from the games in the 2011 NFL regular season.

Step by step solution

01

Identify the matched pairs

A matched pair in this context refers to the scores for home and away teams in the same game. These pairs should be listed with each containing the score of the home team and away team.
02

Calculate the mean difference

The mean difference is found by subtracting the away scores from the home scores for each game and then averaging these differences. This will give the average home field scoring advantage.
03

Find the standard deviation of the differences

The standard deviation can be calculated using the differences in scores for each matched pair. This provides a measure of the spread or variability in the scores.
04

Calculate the standard error of the mean difference

The standard error of the mean difference is calculated by dividing the standard deviation of the differences by the square root of the number of matched pairs.
05

Find the 90% confidence interval for the mean difference

The 90% confidence interval of the mean difference can be calculated using the formula: Mean difference ± (Critical value * Standard error of mean difference). The critical value for a 90% confidence interval can be found from a standard table or statistical calculator. This range will give the interval within which we are 90% confident the true mean difference lies.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Home Field Advantage
In the context of sports, the home field advantage refers to the phenomenon where sports teams tend to perform better when playing on their home ground compared to when they are away. This advantage can stem from numerous factors, like familiar playing conditions, the support of local fans, and reduced travel stress for the home team.

When evaluating the home field advantage, especially in the NFL as described in the exercise, it's crucial to consider the structure of the games being analyzed. By using a matched pairs design, we are effectively comparing each team's performance at home with their performance away for the same games, canceling out many confounding variables like team strategy or opponent strength, making for a fairer comparison.

Analyzing this advantage through statistical methods provides more than just anecdotal evidence, giving us concrete data to back up what many sports fans believe—that playing at home truly offers an edge.
Mean Difference
The mean difference in sports statistics, especially in a matched pairs context, is a key metric to quantify any observed effect. In the case of the NFL data, the mean difference is calculated by looking at all the pairings of home and away games, subtracting the away score from the home score for each game.

This mathematical operation gives us differences for all games, illustrating the score gap for each particular instance. Finding the average of these differences across all games tells us the typical scoring advantage that teams have when playing at home. A greater mean would indicate a stronger advantage, while a smaller value suggests a milder home advantage. Calculating the mean difference thus simplifies complex game scores into a single, expressive number, which makes it easier to understand and compare.
Confidence Interval
Confidence intervals are a statistical tool used to estimate the range within which we expect the true mean of a population to fall. In the context of the NFL home field advantage study, a confidence interval provides insights into how precise our estimate of the mean difference might be.

For example, a 90% confidence interval means that if we were to repeat the analysis multiple times, we'd expect the true mean difference to fall within this calculated range 90% of the time. To compute this, we take the mean difference, find its standard error, and apply a critical value that corresponds to 90% confidence. This interval gives a snapshot not only of the estimate but also its reliability. A narrow confidence interval would indicate a high level of precision, while a wider interval suggests more variability in the data. Having these insights helps in understanding not just the nature of the home field advantage but also the certainty with which we can make conclusions based on the data.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

On page 1 1 in Section \(1.1,\) we describe studies to investigate whether there is evidence of pheromones (subconscious chemical signals) in female tears that affect sexual arousal in men. In one of the studies, \(^{71} 50\) men had a pad attached to the upper lip that contained either female tears or a salt solution dripped down the same female's face. Each subject participated twice, on consecutive days, once with tears and once with saline, randomized for order, and doubleblind. Testosterone levels were measured before sniffing and after sniffing on both days. While normal testosterone levels vary significantly between different men, average levels for the group were the same before sniffing on both days and after sniffing the salt solution (about \(155 \mathrm{pg} / \mathrm{mL}\) ) but were reduced after sniffing the tears (about \(133 \mathrm{pg} / \mathrm{mL}\) ). The mean difference in testosterone levels after sniffing the tears was 21.7 with standard deviation \(46.5 .\) (a) Why did the investigators choose a matchedpairs design for this experiment? (b) Test to see if testosterone levels are significantly reduced after sniffing tears? (c) Can we conclude that sniffing female tears reduces testosterone levels (which is a significant indicator of sexual arousal in men)?

NBA Free Throws In Exercise 6.10, we learn that the percent of free throws made in the \(\mathrm{NBA}\) (National Basketball Association) has been about \(75 \%\) for the last 50 years. If we take random samples of free throws in the NBA and compute the proportion of free throws made, what percent of samples of size \(n=200\) will have a sample proportion greater than \(80 \%\) ? Use the fact that the sample proportions are normally distributed and compute the mean and standard deviation of the distribution.

The AllCountries dataset shows that the percent of the population living in rural areas is \(57.3 \%\) in Egypt and \(21.6 \%\) in Jordan. Suppose we take random samples of size 400 people from each country, and compute the difference in sample proportions \(\hat{p}_{E}-\hat{p}_{J}\) where \(\hat{p}_{E}\) represents the sample proportion living in rural areas in Egypt and \(\hat{p}_{J}\) represents the sample proportion living in rural areas in Jordan. (a) Find the mean and standard deviation of the distribution of differences in sample proportions, \(\hat{p}_{E}-\hat{p}_{J}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. (c) Using the graph drawn in part (b), are we likely to see a difference in sample proportions as large in magnitude as \(0.4 ?\) As large as \(0.5 ?\) Explain.

Drinking tea appears to offer a strong boost to the immune system. In a study introduced in Exercise 3.82 on page \(203,\) we see that production of interferon gamma, a molecule that fights bacteria, viruses, and tumors, appears to be enhanced in tea drinkers. In the study, eleven healthy non-tea- drinking individuals were asked to drink five or six cups of tea a day, while ten healthy nontea- and non-coffee-drinkers were asked to drink the same amount of coffee, which has caffeine but not the \(L\) -theanine that is in tea. The groups were randomly assigned. After two weeks, blood samples were exposed to an antigen and production of interferon gamma was measured. The results are shown in Table 6.23 and are available in ImmuneTea. The question of interest is whether the data provide evidence that production is enhanced in tea drinkers. (a) Is this an experiment or an observational study? (b) What are the null and alternative hypotheses? (c) Find a standardized test statistic and use the t-distribution to find the p-value and make a conclusion. (d) Always plot your data! Look at a graph of the data. Does it appear to satisfy a normality condition? (e) A randomization test might be a more appropriate test to use in this case. Construct a randomization distribution for this test and use it to find a p-value and make a conclusion. (f) What conclusion can we draw? $$ \begin{array}{lrrrrrr} \hline \text { Tea } & 5 & 11 & 13 & 18 & 20 & 47 \\ & 48 & 52 & 55 & 56 & 58 & \\ \hline \text { Coffee } & 0 & 0 & 3 & 11 & 15 & 16 \\ & 21 & 21 & 38 & 52 & & \\ \hline \end{array} $$

Exercise B.5 on page 305 introduces a study examining the effect of diet cola consumption on calcium levels in women. A sample of 16 healthy women aged 18 to 40 were randomly assigned to drink 24 ounces of either diet cola or water. Their urine was collected for three hours after ingestion of the beverage and calcium excretion (in mg) was measured. The summary statistics for diet cola are \(\bar{x}_{C}=56.0\) with \(s_{C}=4.93\) and \(n_{C}=8\) and the summary statistics for water are \(\bar{x}_{W}=49.1\) with \(s_{W}=3.64\) and \(n_{W}=8\). Figure 6.26 shows dotplots of the data values. Test whether there is evidence that diet cola leaches calcium out of the system, which would increase the amount of calcium in the urine for diet cola drinkers. In Exercise B.5, we used a randomization distribution to conduct this test. Use a t-distribution here, after first checking that the conditions are met and explaining your reasoning. The data are stored in ColaCalcium.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.