/*! 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 55 Home Field Advantage in Baseball... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Home Field Advantage in Baseball 2009 There were 2430 Major League Baseball (MLB) games played in \(2009,\) and the home team won the game in \(54.9 \%\) of the games. \({ }^{15}\) If we consider the games played in 2009 as a sample of all MLB games, test to see if there is evidence, at the \(1 \%\) level, that the home team wins more than half the games. Show all details of the test.

Short Answer

Expert verified
To determine whether the home team wins more than half the games, the hypotheses were tested using appropriate statistical techniques. The test statistic was calculated and compared to the critical z-score for a one-tailed test at the 1% significance level. The decision about the hypotheses is based on whether the test statistic exceeds the critical value.

Step by step solution

01

State the Hypotheses

The null hypothesis \(H_0\) is: p = 0.5 (the home team wins half of the games) and the alternative hypothesis \(H_1\) is: p > 0.5 (the home team wins more than half the games).
02

Calculate the Test Statistic

The test statistic z for the sample proportion is calculated using the formula: \( z = ( \widehat{p} - p_{0} ) / \sqrt{ p_{0}( 1 - p_{0} ) / n }\) where \(\widehat{p} = 0.549\), \(p_{0} = 0.5\), and \(n = 2430\). By performing the calculations, the value of the test statistic can be found.
03

Determine the Critical Value

For a significance level of 1%, and since this is a one-tailed test (as we are interested in whether the home team wins 'more than' half the games), the critical z-score from the standard normal distribution table is found to be approximately 2.33.
04

Make the Decision

If the calculated test statistic from Step 2 is greater than the critical value from Step 3, the null hypothesis is rejected, indicating that there is sufficient evidence to suggest that the home team wins more than half the games at the 1% level. Otherwise, the null hypothesis is not rejected.

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.

Null and Alternative Hypotheses
The pillars of hypothesis testing are the null and alternative hypotheses, which provide the framework for decision-making in statistics. The null hypothesis (H_0) usually postulates that there is no effect or no difference - in the case of the baseball study, that the home team wins half (50%) of the games. On the contrary, the alternative hypothesis (H_1) suggests that there's an effect we're aiming to detect, such as the home team winning more than half of the games. In a hypothesis test, we gather data and analyze it to determine if we can reject the null hypothesis in favor of the alternative.

For the 2009 MLB games, we articulate H_0 as the home team's win rate being 50% (p = 0.5), and H_1 as the home team's win rate being greater than 50% (p > 0.5). These hypotheses set the foundation for further steps in our analysis. Establishing clear and precise hypotheses is crucial, as this guides the entire testing process.
Significance Level
The significance level, denoted by \(\alpha\), is the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true. In simpler terms, it's the threshold at which we decide whether to consider our findings as evidence against H_0. For the baseball game analysis at the 1% level (\(\alpha = 0.01\)), it means we're accepting a 1% chance of incorrectly claiming that the home team wins more than half the games if, in reality, they don't.

The significance level is predetermined before analyzing the data, and it's a critical factor because it dictates the rigor of our test. Choosing a low significance level, like 1%, indicates that we require strong evidence before we reject H_0. It's our benchmark for determining the critical value, which we'll compare against our test statistic.
Test Statistic
Once we establish our hypotheses and significance level, we compute the test statistic. The test statistic is a numerical value that summarizes the sample data and helps us decide whether to reject H_0. It combines the sample's results and measures the distance in standard error units from what we would expect if H_0 were true.

In the context of the MLB example, the test statistic we use is the z-value, calculated from the sample proportion (\(\widehat{p} = 0.549\)), hypothesized proportion (\(p_0 = 0.5\)), and sample size (\(n = 2430\)). The formula \( z = (\widehat{p} - p_0) / \sqrt{ p_0(1 - p_0)/n }\) produces a z-value indicating how many standard deviations the sample proportion is from the hypothesized proportion. This statistic plays a key role in comparing against the critical value to make our statistical decision.
P-value
The p-value is another critical component in hypothesis testing, representing the probability of obtaining test results at least as extreme as those observed during the test, assuming that the null hypothesis is correct. It's a way of measuring the strength of the evidence against H_0 and determining whether it's sufficient to reject it.

Suppose we calculate a p-value for our MLB data and find it to be lower than our \(\alpha\) level of 0.01; this would mean that such a result would be very unlikely if H_0 were true. Consequently, a low p-value provides strong evidence against the null hypothesis, and we would reject it, inferring that the home team indeed wins more than half of the games. Understanding the p-value helps us use a scientific approach to interpreting our hypothesis test, ensuring that conclusions are not based on chance results.
Standard Normal Distribution
The standard normal distribution is a key concept in statistics, often used in hypothesis testing. It's a specialized case of the normal distribution with a mean of 0 and a standard deviation of 1. When we convert our sample data into a z-score, we're effectively using the standard normal distribution to standardize our results.

In the baseball study, when we calculate the z-score to obtain our test statistic, we can look up this value in standard normal distribution tables—or use statistical software—to find out how unusual our result is compared to what we'd expect if the null hypothesis were true. The area to the right of our observed z-score gives us the p-value. The farther our test statistic is into the tail of the distribution, the stronger the evidence against H_0. Familiarity with the standard normal distribution is imperative as it provides the theoretical basis for many of the calculations and decisions in hypothesis testing.

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

Metal Tags on Penguins and Breeding Success Data 1.3 on page 10 discusses a study designed to test whether applying metal tags is detrimental to penguins. Exercise 6.148 investigates the survival rate of the penguins. The scientists also studied the breeding success of the metal- and electronic-tagged penguins. Metal-tagged penguins successfully produced offspring in \(32 \%\) of the 122 total breeding seasons, while the electronic-tagged penguins succeeded in \(44 \%\) of the 160 total breeding seasons. Construct a \(95 \%\) confidence interval for the difference in proportion successfully producing offspring \(\left(p_{M}-p_{E}\right)\). Interpret the result.

Split the Bill? Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The data in SplitBill includes the cost of what each person ordered (in Israeli shekels) and the payment method (Individual or Split). Some summary statistics are provided in Table 6.20 and both distributions are reasonably bell-shaped. Use this information to test (at a \(5 \%\) level ) if there is evidence that the mean cost is higher when people split the bill. You may have done this test using randomizations in Exercise 4.118 on page 302 .

Metal Tags on Penguins and Arrival Dates Data 1.3 on page 10 discusses a study designed to test whether applying a metal tag is detrimental to a penguin, as opposed to applying an electronic tag. One variable examined is the date penguins arrive at the breeding site, with later arrivals hurting breeding success. Arrival date is measured as the number of days after November \(1^{\text {st }}\). Mean arrival date for the 167 times metal-tagged penguins arrived was December \(7^{\text {th }}\left(37\right.\) days after November \(\left.1^{\text {st }}\right)\) with a standard deviation of 38.77 days, while mean arrival date for the 189 times electronic-tagged penguins arrived at the breeding site was November \(21^{\text {st }}(21\) days after November \(\left.1^{\text {st }}\right)\) with a standard deviation of \(27.50 .\) Do these data provide evidence that metal tagged penguins have a later mean arrival time? Show all details of the test.

In Exercises 6.32 and 6.33, find a \(95 \%\) confidence interval for the proportion two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the normal distribution and the formula for standard error. Compare the results. Proportion of home team wins in soccer, using \(\hat{p}=0.583\) with \(n=120\)

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To study the effect of women's tears on men, levels of testosterone are measured in 50 men after they sniff women's tears and after they sniff a salt solution. The order of the two treatments was randomized and the study was double-blind.

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.