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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)?

Short Answer

Expert verified
Use a z-test to compare proportions; if z > 1.645, Instructor A is more effective. Potential error: Type I if null is rejected.

Step by step solution

01

State the Hypotheses

To determine whether Instructor A is more effective than Instructor B, we start by setting up the null and alternative hypotheses. The null hypothesis \(H_0\) is that both instructors are equally effective, i.e., the passing rates are the same. The alternative hypothesis \(H_a\) is that Instructor A is more effective, i.e., \(p_A > p_B\), where \(p_A\) and \(p_B\) are the passing probabilities for Instructors A and B, respectively.
02

Calculate the Test Statistic

Given that 30 out of 50 students with Instructor A passed (\(\hat{p}_A = 0.6\)) and 22 out of 50 students with Instructor B passed (\(\hat{p}_B = 0.44\)), we calculate the pooled proportion \(\hat{p} = \frac{30 + 22}{50 + 50} = 0.52\). The standard error for the difference in proportions is \(SE = \sqrt{\hat{p}(1-\hat{p}) \left(\frac{1}{50} + \frac{1}{50}\right)}\). The test statistic is \(z = \frac{\hat{p}_A - \hat{p}_B}{SE}\).
03

Determine the Critical Value and P-Value

Using the standard normal distribution, we find the critical value for a one-tailed test at \(\alpha = 0.05\) is 1.645. We then calculate the p-value associated with the calculated z-value from Step 2. If the p-value is less than \(\alpha = 0.05\), or the z-value is greater than 1.645, we reject the null hypothesis.
04

Conclusion for Part (a)

After calculating the z-value and p-value, if the z-value exceeds 1.645, it shows that the results are statistically significant at the 0.05 level, indicating Instructor A is more effective. If not, we fail to reject the null hypothesis.
05

Describe Type I and Type II Errors

A Type I error occurs if we reject the null hypothesis when it is true, which in this case means concluding Instructor A is more effective when both instructors are equally effective. A Type II error occurs if we fail to reject the null hypothesis when the alternative is true, which means not recognizing Instructor A is more effective when he actually is.
06

Identify Potential Error in Part (a)

In part (a), if we rejected the null hypothesis, we could have made a Type I error by incorrectly concluding that Instructor A is more effective than Instructor B when they are equally effective.

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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** is a statement that there is no effect or no difference, and it is the assumption that researchers seek to test. In the context of the driving school exercise, the null hypothesis, denoted as \(H_0\), states that there is no difference in the effectiveness between Instructor A and Instructor B. In mathematical terms, this means that the passing rates for students taught by both instructors are the same.Understanding the null hypothesis is crucial because it serves as a starting point for testing our research questions. It gives us a benchmark to determine whether the observed data can lead to a rejection of the status quo in favor of a new theory. Remember, rejecting the null is not proof that the alternative is true, but merely that there's sufficient evidence to doubt the null hypothesis under the given significance level.
Type I Error
A **Type I Error** occurs in hypothesis testing when the null hypothesis is rejected despite it being true. It is essentially a false positive—a conclusion that an effect or difference exists when it actually does not. In the driving school scenario, a Type I error would mean concluding that Instructor A is more effective than Instructor B, even though, in reality, they are equally effective. Such an error could lead to inappropriate decisions, such as favoring one instructor over the other unjustly.The probability of making a Type I error is denoted by the significance level \(\alpha\). In our example, \(\alpha = 0.05\), which means there is a 5% risk of rejecting the null hypothesis when it is true. Understanding this error is vital for interpreting results correctly and ensuring justified decision-making.
Type II Error
A **Type II Error** occurs when we fail to reject the null hypothesis even though the alternative hypothesis is true. This means that an actual effect or difference goes unnoticed, a scenario which is also referred to as a false negative.In our driving school context, a Type II error would be failing to recognize that Instructor A is more effective than Instructor B when indeed Instructor A does help more students pass the exam. This mistake could result in missing opportunities for improvement or enhancement of the teaching methodology.The probability of committing a Type II error is denoted by \(\beta\), and it's influenced by factors like sample size and effect size. Larger sample sizes generally reduce the risk of making a Type II error, thereby increasing the test's power.
P-Value
The **P-Value** in hypothesis testing is a metric that helps you determine the significance of your results. It represents the probability of observing your data, or something more extreme, if the null hypothesis is true.In the given scenario, the p-value helps assess whether the difference in pass rates between Instructor A and Instructor B is statistically significant at the significance level \(\alpha = 0.05\). A smaller p-value suggests stronger evidence against the null hypothesis.If the p-value is less than the significance level, we reject the null hypothesis. This implies that the observed results are unlikely under the null hypothesis, concluding that Instructor A might be more effective. However, if the p-value is higher, we fail to reject the null, suggesting insufficient evidence to support the claim of Instructor A's higher effectiveness. Understanding p-values is essential for making sound conclusions from statistical tests.

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Most popular questions from this chapter

Paying for college College financial aid offices expect students to use summer eamings to help pay for college. But how large are these eamings? One large university studied this question by asking a random sample of 1296 students who had summer jobs how much they earned. The financial aid office separated the responses into two groups based on gender. Here are the data in summary form: \({ }^{30}\) $$ \begin{array}{lccc} \hline \text { Group } & n & \bar{\chi} & s_{\chi} \\ \text { Males } & 675 & \$ 1884.52 & \$ 1368.37 \\ \text { Females } & 621 & \$ 1360.39 & \$ 1037.46 \\ \hline \end{array} $$ (a) How can you tell from the summary statistics that the distribution of earnings in each group is strongly skewed to the right? The use of two-sample \(t\) procedures is still justified. Why? (b) Construct and interpret a \(90 \%\) confidence interval for the difference between the mean summer earnings of male and female students at this university. (c) Interpret the \(90 \%\) confidence level in the context of this study.

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 \(H_{a^{*}} \mu_{M} \neq \mu_{F}\) Which of the following describes a Type II error in the context of this study? (a) Finding convincing evidence that the true means are different for males and females, when in reality the true means are the same (b) Finding convincing evidence that the true means are different for males and females, when in reality the true means are different (c) Not finding convincing evidence that the true means are different for males and females, when in reality the true means are the same (d) Not finding convincing evidence that the true means are different for males and females, when in reality the true means are different (e) Not finding convincing evidence that the true means are different for males and females, when in reality there is convincing evidence that the true means are different

Young adults living at home A surprising number of young adults (ages 19 to 25 ) still live in their parents' homes. A random sample by the National Institutes of Health included 2253 men and 2629 women in this age group. \({ }^{11}\) The survey found that 986 of the men and 923 of the women lived with their parents. (a) Construct and interpret a \(99 \%\) confidence interval for the difference in the true proportions of men and women aged 19 to 25 who live in their parents' homes. (b) Does your interval from part (a) give convincing evidence of a difference between the population proportions? Explain.

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) \(1.960 \sqrt{\frac{0.880(0.120)}{550}+\frac{0.820(0.180)}{500}}\)

State which inference procedure from Chapter \(8,9,\) or 10 you would use. Be specific. For example, you might say, "Two-sample z test for the difference between two proportions." You do not need to carry out any procedures. Which inference method? (a) How do young adults look back on adolescent romance? Investigators interviewed 40 couples in their midtwenties. The female and male partners were interviewed separately. Each was asked about his or her current relationship and also about a romantic relationship that lasted at least two months when they were aged 15 or \(16 .\) One response variable was a measure on a numerical scale of how much the attractiveness of the adolescent partner mattered. You want to find out how much men and women differ on this measure. (b) Are more than \(75 \%\) of Toyota owners generally satisfied with their vehicles? Let's design a study to find out. We'll select a random sample of 400 Toyota owners. Then we'll ask each individual in the sample: "Would you say that you are generally satisfied with your Toyota vehicle?" (c) Are male college students more likely to binge drink than female college students? The Harvard School of Public Health surveys random samples of male and female undergraduates at four-year colleges and universities about whether they have engaged in binge drinking. (d) A bank wants to know which of two incentive plans will most increase the use of its credit cards and by how much. It offers each incentive to a group of current credit card customers, determined at random, and compares the amount charged during the following six months.

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