Many people argue that the home team has an advantage in sporting events. Referring specifically to the NFL, playing on your home field is a game changer. The benefits of playing on your home court include the added energy of the crowd, the ability to follow the team's normal routine, and the benefit of not having to travel. In fact, these advantages are so significant to some people that a three-point advantage has been associated with home court. Essentially, the home team in a football game has a one-goal advantage. NFL fans are more than aware that a field goal can be the deciding point of a game. This article will examine statistics from the 2016 NFL season to determine whether the data supports the hypothesis that the home team has an advantage. Say no to plagiarism. Get a tailor-made essay on "Why Violent Video Games Shouldn't Be Banned"? Get an Original Essay The 2016 NFL Excel Data Sheet will be used to examine this question. The specific variables of interest are “Winner To,” “Points Won,” and “Points Lost.” This analysis first requires determining the home team's point margin. In case of victory, the team margin will be a positive value, while in case of defeat, the team margin will be a negative value. Generating descriptive statistics for the point margin yields a mean of 3.1176, a median of 3, and a mode of 3. All three statistics greatly reflect this assumed three-point advantage. Furthermore, the standard deviation calculated from this dataset is 14.1942. To determine how this data might reflect all NFL games, rather than strictly during the 2016 season, it is necessary to use a confidence interval for the population proportion. This data will be treated as a sample of 255 NFL games. The data shows that the home team has won 153 of the 255 total matches. The following formula will be used: In this case p=153/225, which equals 0.6 or 60%. This means that, in this sample, the home team won 60% of the time. This can be used to estimate the home team's true winning percentage across the entire NFL. For a 95% confidence interval, the critical value used is 1.96. Using the data provided, the formula produces a range of (.5399, .6601) or (53.99%, 66.01%). This means that we are 95% sure that this range includes the true proportion of the population. This data can also be used to test the hypothesis of whether the average home team advantage is actually worth a field goal for all NFL games. The null hypothesis for this test will be μ=3. Since the standard deviation of all NFL games is not known, the t-test must be performed. The formula for the t-statistic within the t-test is as follows: The value of μ was determined by the null hypothesis. The remaining values ​​to complete the formula are provided by the descriptive statistical analysis completed in Excel. Sample Mean x¯=3.117641614 Sample Standard Deviation S=14.1941614 Sample Size=255 The solution to this formula using the data is t=0.1323. This test will also be conducted using a 95% confidence interval. In the t distribution table, it is recognized that as the number of elements in a sample increases, t approaches z. Since the NFL sample contains 255 games, the critical t value of this test is equivalent to the z value (1.96). To determine whether the null is rejected, the t-statistic must be compared to the critical value. The rule for deciding whether to reject the null hypothesis is that the null hypothesis is rejected if the magnitude of the t-statistic is greater than.