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Chapter 10: Problem 70

Multiple choice: Select the best answer for Exercises 67 to 70. Exercises 69 and 70 refer to the following setting. A study of road rage asked samples of 596 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 telephone numbers. The two-sample t statistic for the road rage study (male mean minus female mean) is \(t=3.18\). The \(P\)-value for testing the hypotheses from the previous exercise satisfies (a) \(0.001 < P < 0.005 . \quad\) (d) \(0.002 < P < 0.01\) (b) \(0.0005 < P < 0.001 . \quad(\mathrm{e}) P > 0.01\) (c) \(0.001 < P < 0.002\)

Short Answer

Expert verified
The answer is (a) \(0.001 < P < 0.005\).

Step by step solution

01

Understand the Hypothesis Test

We have a two-sample t-test where we compare the means of two groups (men and women) based on their road rage scores. The hypotheses are likely: \( H_0: \mu_{men} = \mu_{women} \) and \( H_a: \mu_{men} eq \mu_{women} \). The test statistic is given as \( t = 3.18 \).
02

Use the t-Statistic to Find the P-Value

The t-statistic of 3.18 is compared against a t-distribution. Given the large sample sizes, \( n_{men} = 596 \) and \( n_{women} = 523 \), the degrees of freedom (approximate) are high. The high degrees of freedom imply the distribution is close to a standard normal distribution.
03

Determine the P-Value Range

For a t-statistic of 3.18 in a two-tailed test with high degrees of freedom, the P-value is quite small. From statistical tables or software, we find that this P-value falls between 0.001 and 0.005.
04

Select the Best Answer

Review the options given. Align the P-value range determined with the answer choices. The range \( 0.001 < P < 0.005 \) matches option (a).

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Key Concepts

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

road rage study
The road rage study featured in this problem explores the differences in driving behavior between men and women. Researchers assigned road rage scores on a scale from 0 to 20, generating quantitative data for analysis. Participants were selected through random digit dialing, ensuring a sample representative of the general population.

By randomly choosing 596 men and 523 women, the study set up a robust basis for comparing the road rage levels between these groups. The random sampling method is crucial because it helps reduce bias, making the results more reliable.

This study serves as a practical application of statistical methods, such as the two-sample t-test, which we use to determine if there are significant differences in average scores between the two genders.
p-value interpretation
In hypothesis testing, the p-value plays a pivotal role in decision-making. It quantifies the evidence against a null hypothesis, helping us determine if observed results are statistically significant.

For the road rage study, the null hypothesis (\(H_0\)) posits that there is no difference in road rage scores between men and women. On the other hand, the alternative hypothesis (\(H_a\)) suggests a significant difference exists.

A small p-value indicates strong evidence against the null hypothesis, leading us to reject it. Here, the p-value range of 0.001 to 0.005 suggests there's significant evidence that road rage scores differ by gender. We use specified significance levels (usually 0.05 or 0.01) to guide our interpretation, where a p-value less than the significance level signals a statistically significant result.
hypothesis testing
Hypothesis testing is a statistical method used to decide whether evidence exists to reject a null hypothesis. In this study, we test the hypothesis that men and women have equal road rage scores.

The framework involves two hypotheses:
  • Null hypothesis (\(H_{0}\)): Men and women have equal road rage scores (\(\mu_{men} = \mu_{women}\)).
  • Alternative hypothesis (\(H_{a}\)): Men and women have different road rage scores (\(\mu_{men} eq \mu_{women}\)).
We use a two-sample t-test to evaluate these hypotheses. By calculating the test statistic and comparing it to a threshold based on degrees of freedom, we assess the likelihood of the observed data under the null hypothesis. The results of this test help determine whether to accept or reject the null hypothesis, using evidence from the p-value.
t-statistic calculation
Calculating the t-statistic is a vital part of performing a two-sample t-test. It determines the standardized difference between two group means, allowing us to assess if the means are statistically different. In the road rage study, this calculation involved the means of male and female road rage scores.

The formula for the t-statistic in a two-sample t-test is:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where:
  • \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means of the two groups.
  • \(s_1^2\) and \(s_2^2\) are the variances of the two groups.
  • \(n_1\) and \(n_2\) are the sample sizes of the two groups.
For the given study, the t-statistic was calculated as 3.18. This statistic is then used to find the p-value, which helps in interpreting the significance of the difference between the groups.

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Most popular questions from this chapter

Explain why the conditions for using two-sample z procedures to perform inference about \(p_{1}-p_{2}\) are not met in the settings of Exercises 7 through 10 . Broken crackers We don鈥檛 like to find broken crackers when we open the package. How can makers reduce breaking? One idea is to microwave the crackers for 30 seconds right after baking them. Breaks start as hairline cracks called 鈥渃hecking.鈥� Assign 65 newly baked crackers to the microwave and another 65 to a control group that is not microwaved. After one day, none of the microwave group and 16 of the control group show checking.\(^{8}\)

What鈥檚 wrong? A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state鈥檚 driver鈥檚 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鈥檚 students and 22 of Instructor B鈥檚 students pass the state exam. Do these results give convincing evidence that Instructor A is more effective? Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one. State: I want to perform a test of $$H_{0} : p_{1}-p_{2}=0$$ $$H_{a} : p_{1}-p_{2}>0$$ where \(p_{1}=\) the proportion of Instructor A's students that passed the state exam and \(p_{2}=\) the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use \(\sigma=0.05\) Plan: If conditions are met, I'll do a two-sample \(z\) test for comparing two proportions. \(\bullet\) Random The data came from two random samples of 50 students. \(\bullet\) Normal The counts of successes and failures in the two groups - \(30,20,22\) , and \(28-\) are all at least \(10 .\) \(\bullet\) Independent There are at least 1000 students who take this driving school's class. Do: From the data, \(\hat{p}_{1}=\frac{20}{50}=0.40\) and \(\hat{p}_{2}=\frac{30}{50}=0.60 .\) So the pooled proportion of successes is $$\hat{p}_{C}=\frac{22+30}{50+50}=0.52$$ \(\bullet\) Test statistic $$z=\frac{(0.40-0.60)-0}{\sqrt{\frac{0.52(0.48)}{100}+\frac{0.52(0.48)}{100}}}=-2.83$$ Conclude: The P-value, \(0.9977,\) is greater than \(\alpha=\) \(0.05,\) so we fail to reject the null hypothesis. There is not convincing evidence that Instructor A's pass rate is higher than Instructor B's.

Multiple choice: Select the best answer for Exercises 29 to 32. A sample survey interviews SRSs of 500 female college students and 550 male college students. Each student is asked whether he or she worked for pay last summer. In all, 410 of the women and 484 of the men say 鈥淵es.鈥� Exercises 29 to 31 are based on this survey. The 95\(\%\) confidence interval for the difference \(p_{M}-p_{F}\) in the proportions of college men and women who worked last summer is about (a) \(0.06 \pm 0.00095\) (b) \(0.06 \pm 0.043\) (c) \(0.06 \pm 0.036\) (d) \(-0.06 \pm 0.043\) (e) \(-0.06 \pm 0.036\)

Listening to rap Is rap music more popular among young blacks than among young whites? A sample survey compared 634 randomly chosen blacks aged 15 to 25 with 567 randomly selected whites in the same age group. It found that 368 of the blacks and 130 of the whites listened to rap music every day.\(^{10}\) Construct and interpret a 95% confidence interval for the difference between the proportions of black and white young people who listen to rap every day.

In Exercises 39 to 42, determine whether or not the conditions for using two- sample t procedures are met. Literacy rates Do males have higher average literacy rates than females in Islamic countries? The table below shows the percent of men and women at least 15 years old who were literate in 2008 in the major Islamic nations. (We omitted countries with populations of less than 3 million.) Data for a few nations, such as Afghanistan and Iraq, were not available.\(^{30}\) $$\begin{array}{lll}{\text { Country }} & {\text { Female percent }} & {\text { Male percent }} \\ {\text { Algeria }} & {66} & {94} \\ {\text { Bangladesh }} & {48} & {71} \\ {\text { Egypt }} & {58} & {88} \\ {\text { lran }} & {77}& {97} \\ {\text { Jordan }} & {87} & {99} \\ {\text { Kazakhstan }} & {100} & {100}\\\\{\text { Lebanon }} & {86} & {98} \\ {\text { Libya }} & {78} & {100}\\\\{\text { Libya }} & {78} & {100} \\ {\text { Malaysia }} & {90} & {98}\\\ {\text { Morocoo }} & {43}& {84} \\ {\text { Saudi Arabia }} & {79} & {98}\\\\{\text { Syria }} & {77} & {95} \\ {\text { Taijkistan }} & {100} & {100} \\ {\text { Tunisia }} & {69} & {97}\\\\{\text { Turkey }} & {81}& {99} \\ {\text { Uzbekistan }} & {96} & {99} \\ {\text { Yemen }} & {41} & {93}\end{array}$$
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