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Chapter 18: Problem 34

Football 2016 During the first 15 weeks of the 2016 season, the home team won 137 of the 238 regular-season National Football League games. Is this strong evidence of a home field advantage in professional football? Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.

Short Answer

Expert verified
Based on the hypothesis test, if the computed z-score is greater than the critical value or the p-value is less than alpha level, it leads to the conclusion that there is strong evidence that home field advantage exists in professional football. The exact numbers for z-score and p-value will depend on the calculations performed according to Step 3.

Step by step solution

01

Set up null and alternate hypothesis

First, set up the null and alternate hypotheses. The null hypothesis \(H_0\) is that there is no home field advantage and the probability of the home team winning is 0.5. The alternative hypothesis \(H_1\) is that there is a home field advantage and the probability of the home team winning is greater than 0.5. Mathematically this can be shown as: \(H_0: p = 0.5 \) and \(H_1: p > 0.5\). 'p' stands for proportion in this context.
02

Check conditions

Check if the conditions are met for performing the test. The conditions are: 1. Randomness: Assuming that each game is independent of one another, randomness can be considered satisfied. 2. Normal: In order for the sampling distribution to be approximately normal, we need to fulfill: \(np > 5\) and \(n(1-p) > 5 \) where n is the sample size and p is the population proportion. Here, n=238 and p=0.5 so both conditions are satisfied.
03

Compute test statistic and compare to a critical value

Calculate the test statistic for one-sample z test for proportions, which is: \( z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}\), where \(z\) is the observed statistic (z-score), \(\hat{p}\) is the observed proportion and n is the sample size. From the problem, we have \(\hat{p} = \frac{137}{238} =.575 \), so we can substitute all the values into the formula: \( z = \frac{.575 - .5}{\sqrt{\frac{.5(1-.5)}{238}}} \).Now, compare this value to a critical value or determine a p-value. If the z-score lies beyond the critical value or the p-value is less than alpha level (typically 0.05), we reject the null hypothesis.
04

Conclude

Based on your tests, conclude whether there is a significant statistical evidence of home field advantage in professional football or not. If you reject the null hypothesis, it suggests that there is a home field advantage.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis serves as a starting assumption that there is no effect or no difference. It is denoted by \( H_0 \). For our football game statistical analysis, the null hypothesis implies no home field advantage. This means the probability that a home team wins a game is equal to 0.5. Essentially, this suggests games are fair and the location of the game does not affect the outcome. By setting up the null hypothesis, we establish a baseline to compare against our data.
Alternative Hypothesis
The alternative hypothesis, denoted by \( H_1 \), contradicts the null hypothesis. It proposes that there is a significant effect or difference. In the context of the football season example, the alternative hypothesis suggests that there is a home field advantage. Specifically, it states that the probability of the home team winning is greater than 0.5. By testing against this hypothesis, we investigate if the home team's winning proportion significantly exceeds what would be expected by chance.
Z-Test for Proportions
The Z-Test for Proportions is a statistical test used to determine if there is a significant difference between an observed sample proportion and a known or hypothesized population proportion. This test is appropriate when checking if the home team in football games has a higher chance to win than expected. We calculate the test statistic using the formula: \[z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}\]Here, \( \hat{p} \) is the sample proportion of home team wins, \( p \) is the known population proportion of 0.5, and \( n \) is the sample size, 238 games in this case. The resulting \( z \)-score helps us assess if the observed proportion of wins deviates significantly from the hypothesized 0.5.
Statistical Significance
Statistical significance is a term used to indicate whether the result of a study is likely due to chance or to some factor of interest. In hypothesis testing, significance is typically assessed with a p-value, which tells us the probability of observing the test results given that the null hypothesis is true.
  • If the p-value is less than a predefined threshold \( \alpha \) (commonly 0.05), we reject the null hypothesis, suggesting that the results are significant.
  • If the p-value is higher, we fail to reject the null hypothesis, indicating insufficient evidence against it.
For the football analysis, assessing statistical significance helps determine if the perceived home field advantage is not merely due to random factors.
Sampling Distribution
Sampling distribution refers to the probability distribution of a statistic obtained through random sampling from a population. In the context of our hypothesis test on football games, the sampling distribution of the sample proportion of home wins is crucial to evaluating our null hypothesis.
  • When conditions of randomness and normality are met, the sampling distribution of the proportion is approximately normal.
  • This allows us to use the Z-test for proportions to determine if the observed home win proportion significantly deviates from 0.5.
Understanding the sampling distribution gives us the foundation to apply our statistical tests confidently and effectively.

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Most popular questions from this chapter

Parameters and hypotheses For each of the following situations, define the parameter (proportion or mean) and write the null and alternative hypotheses in terms of parameter values. Example: We want to know if the proportion of up days in the stock market is \(50 \% .\) Answer: Let \(p=\) the proportion of up days. \(\mathrm{H}_{0}: p=0.5 \mathrm{vs} . \mathrm{H}_{\mathrm{A}}: p \neq 0.5\) a. A casino wants to know if their slot machine really delivers the 1 in 100 win rate that it claims. b. Last year, customers spent an average of \(\$ 35.32\) per visit to the company's website. Based on a random sample of purchases this year, the company wants to know if the mean this year has changed. c. A pharmaceutical company wonders if their new drug has a cure rate different from the \(30 \%\) reported by the placebo. d. A bank wants to know if the percentage of customers using their website has changed from the \(40 \%\) that used it before their system crashed last week.

Relief A company's old antacid formula provided relief for \(70 \%\) of the people who used it. The company tests a new formula to see if it is better and gets a P-value of \(0.27 .\) Is it reasonable to conclude that the new formula and the old one are equally effective? Explain.

Seeds A garden center wants to store leftover packets of vegetable seeds for sale the following spring, but the center is concerned that the seeds may not germinate at the same rate a year later. The manager finds a packet of last year's green bean seeds and plants them as a test. Although the packet claims a germination rate of \(92 \%\), only 171 of 200 test seeds sprout. Is this evidence that the seeds have lost viability during a year in storage? Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.

Obesity 2016 In \(2016,\) the Centers for Disease Control and Prevention reported that \(36.5 \%\) of adults in the United States are obese. A county health service planning a new awareness campaign polls a random sample of 750 adults living there. In this sample, 228 people were found to be obese based on their answers to a health questionnaire. Do these responses provide strong evidence that the \(36.5 \%\) figure is not accurate for this region? Correct the mistakes you find in a student's attempt to test an appropriate hypothesis. $$ \begin{array}{l} \mathrm{H}_{0}: \hat{p}=0.365 \\ \mathrm{H}_{\mathrm{A}}: \hat{p}<0.365 \\ \mathrm{SRS}, 750 \geq 10 \\ \frac{228}{750}=0.304 ; S D(\hat{p})=\sqrt{\frac{(0.304)(0.696)}{750}}=0.017 \end{array} $$ \(z=\frac{0.304-0.365}{0.017}=-3.588\) $$ \mathrm{P}=P(z>-3.588)=0.9998 $$ There is more than a \(99.98 \%\) chance that the stated percentage is correct for this region.

Women executives A company is criticized because only 13 of 43 people in executive-level positions are women. The company explains that although this proportion is lower than it might wish, it's not a surprising value given that only \(40 \%\) of all its employees are women. What do you think? Test an appropriate hypothesis and state your conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.
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