/*! 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 83 The Sheboygan (Wisconsin) Fire D... [FREE SOLUTION] | 91影视

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The Sheboygan (Wisconsin) Fire Department received a report on the potential effects of reductions in the number of firefighters it employs ("Study of Fire Department Causes Controversy," USA TODAY NETWORK-Wisconsin, December \(22,\) 2016) In one section of the report, the average working heart rate percentage (the percentage value of the observed maximum heart rate during firefighting drills divided by an age-adjusted maximum heart rate for each firefighter) was reported for the driver of the first-arriving fire engine when only two firefighters (including the driver) were present, and for the driver of the first-arriving fire engine when more than two firefighters (up to five) were present. The average working heart rate percentages were based on an earlier study, which included data from a sample of six drills using only two firefighters and from a sample of 18 drills using more than two firefighters. For purposes of this exercise, you can assume that these samples are representative of all drills with two firefighters and all drills with more than two firefighters. The following data values are consistent with summary statistics given in the paper. Do these data support the claim that the mean average working heart rate percentage for the driver in a two-firclighter team is greater than the mean average working heart rate percentage for the driver in teams containing from three to five firefighters? Use a 0.05 significance level to carry out a randomization test of the given claim. You can use make use of the Shiny apps in the collection at statistics.cengage.com/Peck2e/Apps.html.

Short Answer

Expert verified
In conclusion, after performing the randomization test using the Shiny app and comparing the obtained p-value with the significance level of 0.05, we can either reject or fail to reject the null hypothesis. If the p-value is less than 0.05, we reject the null hypothesis and conclude that the data supports the claim that the mean average working heart rate percentage for the driver in a two-firefighter team is greater than the mean average working heart rate percentage for the driver in teams containing from three to five firefighters. If the p-value is greater than or equal to 0.05, we fail to reject the null hypothesis, suggesting that the data does not support the claim.

Step by step solution

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1. Hypotheses Setting

First, we need to state our null hypothesis (H鈧) and the alternative hypothesis (H鈧): H鈧: The mean average working heart rate percentage for the driver in a two-firefighter team is equal to the mean average working heart rate percentage for the driver in teams containing three to five firefighters (渭鈧 = 渭鈧). H鈧: The mean average working heart rate percentage for the driver in a two-firefighter team is greater than the mean average working heart rate percentage for the driver in teams containing three to five firefighters (渭鈧 > 渭鈧).
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2. Check Assumptions and Requirements

Because we are given the sample sizes for each group (n鈧=6 drills for two-firefighter teams and n鈧=18 drills for three-to-five firefighter teams) and we have assumed that these samples are representative of all drills, we can proceed with the analysis without worrying about the assumptions or requirements of our test.
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3. Perform the Randomization Test

Using the Shiny apps provided at statistics.cengage.com/Peck2e/Apps.html, we can perform a randomization test using our data and a significance level of 0.05.
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4. Analyze Results and Make a Decision

After inputting the data into the randomization test app and observing the test results, compare the p-value obtained from the app with the significance level (0.05). If the p-value is less than the significance level, 0.05, then we reject the null hypothesis (H鈧) in favor of the alternative hypothesis (H鈧). If the p-value is greater than or equal to the significance level, we fail to reject the null hypothesis.
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5. Conclusion

Based on the results of the randomization test and the comparison of the p-value with the significance level, make a conclusion about whether or not the data supports the claim that the mean average working heart rate percentage for the driver in a two-firefighter team is greater than the mean average working heart rate percentage for the driver in teams containing from three to five firefighters.

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Key Concepts

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

Understanding the Null Hypothesis
When conducting a randomization test, one of the first steps is to define the **null hypothesis**. In essence, the null hypothesis represents the default assumption that there is no significant effect or difference. In the context of our exercise, the null hypothesis ( H鈧 ) states that the mean average working heart rate percentage for the driver of a two-firefighter team is equal to that for the driver in teams with three to five firefighters.
This hypothesis is our starting point, and we assume it to be true until we have enough evidence to the contrary. The null hypothesis is critical because it puts a concrete statement into what 'no effect' or 'no difference' would mean in our scenario. By clearly defining this, we're set up to test if the observed differences are significant enough to say otherwise.
Exploring the P-Value
The **p-value** is a crucial component in statistical hypothesis testing, including randomization tests. It tells us how likely we are to observe a result as extreme as, or more extreme than, the result observed, assuming that the null hypothesis is true. In simple terms, it's a measure of the strength of the evidence against the null hypothesis.
After performing the randomization test and calculating the p-value, we compare it against our predetermined significance level (often denoted as 伪 ). A smaller p-value indicates stronger evidence against the null hypothesis, suggesting that the observed effect may be real. For instance, if we get a p-value of 0.03 and our significance level is 0.05, this would mean there is only a 3% chance that the observed results or more extreme ones could occur if the null hypothesis were true.
  • Key Insight: If the p-value is less than your significance level, you would reject the null hypothesis.
  • Interpretation: In our case, if the p-value is lower than 0.05, it suggests there's a significant difference in heart rate percentages between the two firefighting team configurations.
Alternative Hypothesis Demystified
The **alternative hypothesis** is what we consider if the null hypothesis is rejected. It represents the presence of an actual effect or difference. In our case, the alternative hypothesis ( H鈧 ) posits that the mean average working heart rate percentage for the driver of a two-firefighter team is greater than that of a team with three to five firefighters.
This hypothesis aligns with the claim under investigation. The alternative hypothesis is crucial because it gives direction to our test, telling us exactly what difference we are looking for. It's essentially the claim that we aim to support with statistical evidence.
Understanding Significance Level
The **significance level**, often denoted by 伪 , determines the threshold at which we decide whether to reject the null hypothesis. It represents the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true. Commonly used significance levels are 0.05, 0.01, and 0.10. In our exercise, a significance level ( 伪 ) of 0.05 means that we are willing to accept a 5% risk of concluding that a difference exists when there is none.
The choice of significance level can affect the outcome of the test, as a lower significance level requires stronger evidence to reject the null hypothesis. This trade-off is important as it balances the risk of false positives with the ability to detect a true effect.
  • Conclusion: If the computed p-value is lower than 0.05, it suggests a statistically significant difference, thus leading us to reject the null hypothesis in favor of the alternative hypothesis.

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