/*! 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 59 Exercises 58 to 60 refer to the ... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercises 58 to 60 refer to the following setting. A study of road rage asked random samples of \(596 \mathrm{men}\) and 523 women about their behavior while driving. Based on their answers, each person was assigned a road rage score on a scale of 0 to \(20 .\) The participants were chosen by random digit dialing of phone numbers. The researchers performed a test of the following hypotheses: \(H_{0}: \mu_{M}=\mu_{F}\) versus \(H_{a^{*}} \mu_{M} \neq \mu_{F}\) The \(P\) -value for the stated hypotheses is 0.002 . Interpret this value in the context of this study. (a) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting a difference in sample means. (b) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting an observed difference at least as extreme as the observed difference. (c) Assuming that the true mean road rage score is different for males and females, there is a 0.002 probability of getting an observed difference at least as extreme as the observed difference. (d) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the null hypothesis is true. (e) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the alternative hypothesis is true.

Short Answer

Expert verified
The correct interpretation is option (b).

Step by step solution

01

Understanding the Hypothesis Test

We are testing the null hypothesis \(H_{0}: \mu_{M} = \mu_{F}\) against the alternative hypothesis \(H_{a}: \mu_{M} eq \mu_{F}\), where \(\mu_M\) is the mean road rage score for men and \(\mu_F\) is the mean road rage score for women. The null hypothesis asserts there is no difference between the mean scores, while the alternative suggests a difference exists.
02

Understanding the P-Value

The \(P\)-value, given as 0.002, quantifies the probability of observing a test statistic as extreme as, or more extreme than, the one observed in the sample, under the assumption that the null hypothesis is true. A small \(P\)-value indicates that the observed data is unlikely under the null hypothesis.
03

Identify the Correct Interpretation

Out of the given options, the interpretation closest to the definition of a \(P\)-value is option (b): "Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting an observed difference at least as extreme as the observed difference." This aligns with the concept of a \(P\)-value testing the hypothesis assuming \(H_0\) is true.

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Key Concepts

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

P-value
The P-value in hypothesis testing is a crucial concept to understand. It tells us how likely the observed data would occur if the null hypothesis were true. In simpler terms, it measures how well the sample data supports the null hypothesis. For our study, the P-value is given as 0.002. This low P-value suggests that the observed differences in road rage scores between men and women are unlikely if we assume the null hypothesis to be correct.
When interpreting a P-value, remember that:
  • A small P-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, so we reject the null hypothesis.
  • A large P-value (> 0.05) suggests weak evidence against the null hypothesis, so we fail to reject it.
  • The threshold (like 0.05) is called the significance level, and it is chosen based on how much uncertainty we are willing to accept in our conclusions.
Therefore, the 0.002 P-value in this study strongly suggests a meaningful difference in road rage scores between the genders, prompting researchers to likely reject the null hypothesis.
Null Hypothesis
The null hypothesis (often denoted as \(H_0\)) is the starting assumption for statistical tests. It suggests no effect or no difference exists between groups or conditions being compared. In the road rage study, the null hypothesis \(H_0: \mu_{M} = \mu_{F}\) claims that the average road rage scores are the same for men and women.
The null hypothesis serves several purposes:
  • It acts as a baseline that we test against using the sample data.
  • It provides a specific scenario under which we calculate the P-value.
  • In many tests, it is favored or assumed true unless evidence strongly suggests otherwise.
Rejecting the null hypothesis does not prove the null hypothesis false; it just suggests that the observed data are not consistent with the assumption of no difference.
Alternative Hypothesis
The alternative hypothesis (denoted as \(H_a\)) is what researchers typically want to prove. It suggests that there is an effect or a difference. In our example, the alternative hypothesis \(H_a: \mu_M eq \mu_F\) asserts that there is a difference in mean road rage scores between men and women.
Here are key points about the alternative hypothesis:
  • It is what we accept when the null hypothesis is rejected based on the P-value and significance level.
  • It provides the basis for concluding that the observed effect is statistically significant.
  • Unlike the null hypothesis, the alternative hypothesis is usually not a statement of no effect.
In this exercise, the extremely low P-value of 0.002 provides enough evidence for the researchers to consider accepting the alternative hypothesis, indicating a difference in road rage scores based on gender.

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Most popular questions from this chapter

Driving school A driving school owner believes that Instructor \(\mathrm{A}\) is more effective than Instructor \(\mathrm{B}\) at preparing students to pass the state's driver's license exam. An incoming class of 100 students is randomly assigned to two groups, each of size \(50 .\) One group is taught by Instructor \(A ;\) the other is taught by Instructor B. At the end of the course, 30 of Instructor A's students and 22 of Instructor \(\mathrm{B}\) 's students pass the state exam. (a) Do these results give convincing evidence at the \(\alpha=0.05\) level that Instructor \(\mathrm{A}\) is more effective? (b) Describe a Type I and a Type II error in this setting. Which error could you have made in part (a)?

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Which of the following is the correct margin of error for a \(99 \%\) confidence interval for the difference in the proportion of male and female college students who worked for pay last summer? (a) \(2.576 \sqrt{\frac{0.851(0.149)}{550}+\frac{0.851(0.149)}{500}}\) (b) \(2.576 \sqrt{\frac{0.851(0.149)}{1050}}\) (c) \(2.576 \sqrt{\frac{0.880(0.120)}{550}+\frac{0.820(0.180)}{500}}\) (d) \(1.960 \sqrt{\frac{0.851(0.149)}{550}+\frac{0.851(0.149)}{500}}\) (e) \(1.960 \sqrt{\frac{0.880(0.120)}{550}+\frac{0.820(0.180)}{500}}\)

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