Problem 37 - An Introduction to Mathematical Statistics and Its Applications

Question

In athletic contests, a wide-spread conviction exists that the home team has an advantage. However, one explanation for this is that a team schedules some much weaker opponents to play at home. To avoid this bias, a study of college football games considered only games between teams ranked in the top twenty- five. For three hundred seventeen such games, the margin of victory \((y=\) home team score-visitors score) was recorded. For these data, \(\bar{y}=4.57\) and \(s=18.29\). Does this study confirm the existence of a home field advantage? Test \(H_{0}: \mu=0\) versus \(H_{1}: \mu>0\) at the \(0.05\) level of significance.

Step-by-Step Solution

Answer

Yes, this study does confirm the existence of a home field advantage at the 0.05 level of significance.

Step 1: Calculate the Test Statistic

First, we'll calculate the test statistic for this problem. The test statistic in this case is a z-score, which is calculated by the formula: \(z = (\bar{y} - μ)/(s/√n)\). Here, \(μ = 0\) (from null hypothesis), \(\bar{y} = 4.57\), \(s = 18.29\), and \(n = 317\). Thus, plugging values into the formula, we get: \(z = (4.57 - 0)/(18.29/√317) = 1.758\).

Step 2: Compare with Critical Value

Now, we'll compare this test statistic with the critical z-value. For a one-sided test at a 0.05 significance level, the critical z-value is 1.645 (you can find this from a z-table or using a calculator). Since our calculated z-score (1.758) is greater than the critical z-score (1.645), we would reject the null hypothesis.

Step 3: Interpret the Result

In rejecting the null hypothesis, we're saying that there is sufficient evidence to conclude that the mean difference in scores between the home team and the visiting team is greater than 0, which suggests a home field advantage.

Key Concepts

Home Field Advantage

In sports, we often hear about the 'home field advantage.' This is a common belief that teams playing on their home ground have a better chance of winning. But why does this happen? Various factors come into play:

• Familiarity: Teams know their home field's conditions better.
• Fan Support: Home crowds can boost team morale.
• Reduced Travel Fatigue: Playing at home eliminates travel stress.

However, it’s crucial to analyze data properly to confirm this advantage, taking biases into account. The study done on college football games tries to eliminate biases by focusing on games between equally ranked teams. By examining the mean margin of victory — the home team score minus the visitors' score — researchers analyze whether playing at home significantly increases a team's score on average.

Z-Test

A z-test is a statistical method used to determine whether there's a significant difference between an observed sample mean and a theoretical population mean. In this study, the researchers use a one-sample z-test to see if playing at home provides a statistically significant advantage.

• The formula used is: \[ z = \frac{\bar{y} - \mu}{s/\sqrt{n}} \]
• Where \(\bar{y}\) is the sample mean, \(\mu\) is the hypothesized population mean, \(s\) is the standard deviation, and \(n\) is the sample size.

The z-test helps quantify how far, in standard deviations, the sample mean is from the hypothesized mean. In this football example, using the sample data (\(\bar{y} = 4.57\), \(s = 18.29\), \(n = 317\)), we find a z-score of 1.758, indicating how much the observed home team's score diverges from the expectation under the null hypothesis.

Level of Significance

The level of significance, often denoted by \(\alpha\), is a threshold set by researchers that guides decision-making in statistical hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true.

• Common levels are 0.05 and 0.01.
• A lower \(\alpha\) indicates stricter criteria for rejecting the null hypothesis.

In this study, the researchers choose \(\alpha = 0.05\). This level implies a 5% risk of concluding that there is a home field advantage when, in fact, there might not be any. Deciding on \(\alpha\) involves balancing the risks of Type I error (false positives) against the desire to detect a true effect.

Critical Value

The critical value is a point on the test distribution that is compared to the test statistic to decide whether to reject the null hypothesis. It serves as a cutoff.

• For a one-tailed test at \(\alpha = 0.05\), you find a critical value of 1.645 in a z-table.
• If the calculated z-score exceeds this critical value, the null hypothesis is rejected.

In this case, the z-score is 1.758, which is greater than the critical value of 1.645. This indicates that the observed data provides enough evidence to suggest that the home team does indeed have an advantage over visiting teams at a 5% significance level.