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Chapter 11: Problem 25

The article "Footwear, Traction, and the Risk of Athletic Injury" (January \(2016,\) www.lermagazine.com/article/footwear -traction-and-the-risk-of-athletic-injury, retrieved December \(15,\) 2016) describes a study in which high school football players were given either a conventional football cleat or a swivel disc shoe. Of 2373 players who wore the conventional cleat, 372 experienced an injury during the study period. Of the 466 players who wore the swivel disc shoe, 24 experienced an injury. The question of interest is whether there is evidence that the injury proportion is smaller for the swivel disc shoe than it is for conventional cleats. a. What are the two treatments in this experiment? b. The article didn't state how the players in the study were assigned to the two groups. Explain why it is important to know if they were assigned to the groups at random. c. For purposes of this example, assume that the players were randomly assigned to the two treatment groups. Carry out a hypothesis test to determine if there is evidence that the injury proportion is smaller for the swivel disc shoe than it is for conventional cleats. Use a significance level of 0.05 .

Short Answer

Expert verified
The experiment's two treatments are the conventional football cleat and the swivel disc shoe. Random assignment is important to minimize biases and establish a causal relationship between the shoe type and injury risk. A hypothesis test at a 0.05 significance level was conducted, resulting in the rejection of the null hypothesis. Thus, there is sufficient evidence to conclude that the injury proportion is smaller for the swivel disc shoe compared to conventional cleats.

Step by step solution

01

a. Identifying the treatments.

The two treatments in this experiment are the conventional football cleat and the swivel disc shoe. Players were given either one of these shoes.
02

b. Importance of random assignment

Knowing if the players were assigned to the groups at random is important because random assignment minimizes biases and ensures that each player has an equal chance of being in either group. This helps to establish a causal relationship between the type of shoe and injury risk, and it allows us to generalize the results of the study to the larger population of high school football players.
03

c. Hypothesis test

In this part, we will perform a hypothesis test to determine if there is evidence that the injury proportion is smaller for the swivel disc shoe than it is for conventional cleats, assuming random assignment and using a significance level of \(0.05\). Step 1: Hypotheses \(H_0\): The injury proportion is the same for both types of shoes. \(H_1\): The injury proportion is smaller for the swivel disc shoe. Step 2: Test Statistic and Distribution We will use a two-sample Z-test for the difference in proportions. Let \(p_1\) be the proportion of injuries in the conventional cleat group, and \(p_2\) the proportion of injuries in the swivel disc shoe group. Additionally, let \(\hat{p}_1\) and \(\hat{p}_2\) be the respective sample proportions. The test statistic is calculated as follows: \(Z = \frac{(\hat{p}_1 - \hat{p}_2) - 0}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}\), where \(\hat{p}\) is the pooled proportion, \(\hat{p} = \frac{X_1+X_2}{n_1+n_2}\). Step 3: Data and Test Statistic Calculation For the conventional cleat group, \(n_1 = 2373\) and \(X_1 = 372\) injuries. For the swivel disc shoe group, \(n_2 = 466\) and \(X_2 = 24\) injuries. Calculating \(\hat{p}_1\), \(\hat{p}_2\), \(\hat{p}\), and the test statistic \(Z\), \(\hat{p}_1 = \frac{372}{2373} = 0.1567\), \(\hat{p}_2 = \frac{24}{466} = 0.0515\), \(\hat{p} = \frac{372+24}{2373+466} = 0.1331\), \(Z = \frac{(0.1567 - 0.0515) - 0}{\sqrt{\frac{0.1331(1-0.1331)}{2373}+\frac{0.1331(1-0.1331)}{466}}} = 5.415\). Step 4: Decision Using a significance level of \(0.05\) and a one-tailed Z-test, we have a critical Z-value of \(1.645\). Since our calculated Z value (\(5.415\)) is greater than the critical Z value (\(1.645\)), we reject the null hypothesis \(H_0\). Conclusion: Based on the hypothesis test, there is sufficient evidence at the \(0.05\) significance level to conclude that the injury proportion is smaller for the swivel disc shoe than it is for conventional cleats.

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Key Concepts

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

The Significance of Random Assignment in Studies
Understanding the significance of random assignment in statistical studies is crucial for establishing the reliability and validity of the experiment's results. In the context of the study described in the exercise, the random assignment of football players to receive either conventional football cleats or swivel disc shoes is fundamental.

Random assignment plays a pivotal role because it ensures that each participant has an equal chance of being placed in any of the treatment groups, thus minimizing confounding variables. Confounding variables are extraneous factors that could influence the outcome of the study. By randomly assigning subjects, researchers can be more confident that the results, such as the rate of injury in this case, can be attributed to the variable of interest鈥攖he type of shoe鈥攔ather than other hidden factors.

Moreover, random assignment helps in balancing out participant characteristics across different groups, leading to each group being comparable in terms of potential risk factors. This is particularly important when researchers aim to generalize the findings to a broader population. When study participants represent the larger population fairly, it strengthens the study's external validity. Therefore, knowing whether random assignment was used in the study about football cleats is critical to evaluating the study's conclusions about injury risk.
Conducting a Two-Sample Z-Test
When comparing two independent groups to determine if there is a statistically significant difference between their proportions, as in the case of the cleat study, the two-sample Z-test is a commonly used statistical method.

The two-sample Z-test works under the assumption that the samples are independent and come from normally distributed populations with known variances, or that the sample sizes are large enough that the distribution of the test statistic will be approximately normally distributed, according to the Central Limit Theorem.

To perform this test, researchers calculate the test statistic, which measures the difference between the sample proportions relative to the variability expected under the null hypothesis. The null hypothesis typically states that there is no difference in proportions between the two groups. In contrast, the alternative hypothesis would suggest a difference鈥攊n this exercise, that the injury proportion is smaller for the swivel disc shoe.

The calculated Z-value from the data then gets compared to a critical value from the Z-distribution for a given significance level, which is 0.05 in this example. If the calculated Z-value is higher than the critical value, the null hypothesis is rejected, indicating that there is a statistically significant difference between the groups.
Understanding Proportion Hypothesis Testing
The proportion hypothesis test is a specific type of hypothesis testing used when the goal is to compare proportions鈥攆or instance, comparing the proportion of injuries between two different types of footwear, as seen in our example. This test is particularly useful when dealing with binary data, such as 'injured' or 'not injured'.

The proportion hypothesis test starts by defining the null and alternative hypotheses. The null hypothesis (\(H_0\)) often states that there is no difference in proportions between the groups, while the alternative hypothesis (\(H_1\)) proposes what the researcher suspects might be true鈥攊n our case, that the swivel disc shoe has a lower injury proportion.

Using the sample data, the researcher calculates the sample proportions and the pooled proportion, which is an estimate of the common proportion assuming the null hypothesis is true. The Z-statistic is then calculated, taking into account the pooled proportion and the sizes of both samples. If the p-value associated with the calculated Z-statistic is less than the chosen significance level, the null hypothesis is rejected, and it is concluded that there is a significant difference in proportions. This kind of test provides a framework for making inferences about population proportions based on sample data, enabling researchers to draw conclusions about an entire population of interest.

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Most popular questions from this chapter

A Harris Poll press release dated November 1, 2016 summarized results of a survey of 2463 adults and 510 teens age 13 to 17 ("American Teens No Longer More Likely Than Adults to Believe in God, Miracles, Heaven, Jesus, Angels, or the Devil," www.theharrispoll.com, retrieved December 12 , 2016). It was reported that \(19 \%\) of the teens surveyed and \(26 \%\) of the adults surveyed indicated that they believe in reincarnation. The samples were selected to be representative of American adults and teens. Use the data from this survey to estimate the difference in the proportion of teens who believe in reincarnation and the proportion of adults who believe in reincarnation. Be sure to interpret your interval in context.

Choice blindness is the term that psychologists use to describe a situation in which a person expresses a preference and then doesn't notice when they receive something different than what they asked for. The authors of the paper "Can Chocolate Cure Blindness? Investigating the Effect of Preference Strength and Incentives on the Incidence of Choice Blindness" Uournal of Behavioral and Experimental Economics [2016]: 1-11) wondered if choice blindness would occur more often if people made their initial selection by looking at pictures of different kinds of chocolate compared with if they made their initial selection by looking as the actual different chocolate candies. Suppose that 200 people were randomly assigned to one of two groups. The 100 people in the first group are shown a picture of eight different kinds of chocolate candy and asked which one they would like to have. After they selected, the picture is removed and they are given a chocolate candy, but not the one they actually selected. The 100 people in the second group are shown a tray with the eight different kinds of candy and asked which one they would like to receive. Then the tray is removed and they are given a chocolate candy, but not the one they selected. If 20 of the people in the picture group and 12 of the people in the actual candy group failed to detect the switch, would you conclude that there is convincing evidence that the proportion who experience choice blindness is different for the two treatments (choice based on a picture and choice based on seeing the actual candy)? Test the relevant hypotheses using a 0.01 significance level.

The paper "Passenger and Cell Phone Conversations in Simulated Driving" (Journal of Experimental Psychology: Applied [2008]: 392-400) describes an experiment that investigated if talking on a cell phone while driving is more distracting than talking with a passenger. Drivers were randomly assigned to one of two groups. The 40 drivers in the cell phone group talked on a cell phone while driving in a simulator. The 40 drivers in the passenger group talked with a passenger in the car while driving in the simulator. The drivers were instructed to exit the highway when they came to a rest stop. Of the drivers talking to a passenger, 21 noticed the rest stop and exited. For the drivers talking on a cell phone, 11 noticed the rest stop and exited. a. Use the given information to construct and interpret a \(95 \%\) confidence interval for the difference in the proportions of drivers who would exit at the rest stop. b. Does the interval from Part (a) support the conclusion that drivers using a cell phone are more likely to miss the exit than drivers talking with a passenger? Explain how you used the confidence interval to answer this question.

The Insurance Institute for Highway Safety issued a news release titled "Teen Drivers Often Ignoring Bans on Using Cell Phones" (June 9,2008 ). The following quote is from the news release: Just \(1-2\) months prior to the ban's Dec. \(1,2006,\) start, \(11 \%\) of teen drivers were observed using cell phones as they left school in the afternoon. About 5 months after the ban took effect, \(12 \%\) of teen drivers were observed using cell phones. Suppose that the two samples of teen drivers (before the ban, after the ban) are representative of these populations of teen drivers. Suppose also that 200 teen drivers were observed before the ban (so \(n_{1}=200\) and \(\hat{p}_{1}=0.11\) ) and that 150 teen drivers were observed after the ban. a. Construct and interpret a \(95 \%\) large-sample confidence interval for the difference in the proportion using a cell phone while driving before the ban and the proportion after the ban. b. Is 0 included in the confidence interval of Part (a)? What does this imply about the difference in the population proportions?

In the experiment described in the article "Study Points to Benefits of Knee Replacement Surgery Over Therapy Alone" (The New York Times, October 21,2015 ), adults who were considered candidates for knee replacement were followed for one year. Suppose that 200 patients were randomly assigned to one of two groups. One hundred were assigned to a group that had knee replacement surgery followed by therapy and the other half were assigned to a group that did not have surgery but did receive therapy. After one year, \(86 \%\) of the patients in the surgery group and \(68 \%\) of the patients in the therapy only group reported pain relief. Is there convincing evidence that the proportion experiencing pain relief is greater for the surgery treatment than for the therapy treatment? Use a significance level of 0.05
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