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Chapter 10: Problem 2

Binge drinking Who is more likely to binge drink鈥攎ale or 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. (a) Is this a problem about comparing means or comparing proportions? Explain. (b) What type of study design is being used to produce data?

Short Answer

Expert verified
(a) Comparing proportions. (b) Observational study.

Step by step solution

01

Understanding the Problem

Before diving into the specific questions, it's important to understand the data collected. This problem involves data from male and female undergraduates about binge drinking habits. They are surveyed from various colleges and universities.
02

Analyze Part (a) - Comparing Means or Proportions

In statistical problems, comparing means typically applies to numerical data, such as heights or GPA. Comparing proportions applies to categorical data, such as the percentage of people engaging in certain activities. Here, the data pertains to whether students engage in binge drinking (yes or no), which is categorical, so we compare proportions.
03

Analyze Part (b) - Study Design Identification

The study involves surveying random samples from a population, which indicates it is an observational study since the researchers are not assigning treatments or conditions to the subjects. They are merely observing the natural behavior (binge drinking) in different groups (male and female students).

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Key Concepts

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

Observational Study
An observational study is a research method widely used to collect information about subjects without interfering with their natural behavior or environment. In the context of studying binge drinking among college students, researchers collected data by simply observing and recording the incidence of binge drinking among male and female students.
The researchers did not influence or alter the students' drinking habits in any way.

Characteristics of Observational Studies

  • They do not manipulate the variables being studied, meaning researchers solely collect and analyze data based on natural occurrences.
  • Often used in social sciences and medicine to study behavioral trends or medical conditions.
  • Provide a real-world perspective, often leading to high external validity, which refers to the applicability of study results to real-life settings.
Observational studies, such as the one analyzing binge drinking, offer valuable insights into behaviors and trends without the complexities introduced by experimental manipulation.
Categorical Data Analysis
Categorical data analysis deals with analyzing variables that can be divided into categories, which are often labels or names. In the given exercise, the presence of binge drinking among students is an example of categorical data.

Understanding Proportions and Categorical Data

In this kind of data, you may encounter questions like "Do college students binge drink? Yes or No."
Categories in this example are essentially the responses ("yes" or "no") each student gives.
Since these responses are categories and not numbers, the process involves comparing the proportions of students in each category.
  • Proportions are expressed as fractions or percentages that indicate how often an event (like binge drinking) occurs within a group.
  • In analyzing categorical data, sample sizes and the number of occurrences in each category play critical roles.
To compare proportions effectively, statistical tests like the Chi-square test can be applied. This helps determine if observed differences between groups (e.g., male vs. female binge drinking rates) are statistically significant.
Random Sampling
Random sampling is a fundamental technique in statistics used to ensure that a sample is representative of the overall population. In the binge drinking essay scenario, random sampling was crucial to ensure that the conclusions made about the behavior of all male and female college students were reliable and unbiased.

Importance of Random Sampling

  • It allows each member of the population an equal chance of being included in the sample, which reduces selection bias.
  • Helps achieve a more accurate representation of the population, allowing the findings to be generalized.
  • Ensures the validity and reliability of statistical inference, providing trustworthy conclusions.
By randomly selecting male and female students from four-year colleges, researchers minimize disparities that could skew results. This randomness is the backbone of accurate, dependable data collection, crucial for making informed comparisons of student behaviors.

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Most popular questions from this chapter

Preventing strokes Aspirin prevents blood from clotting and so helps prevent strokes. The Second European Stroke Prevention Study asked whether adding another anticlotting drug, named dipyridamole, would be more effective for patients who had already had a stroke. Here are the data on strokes and deaths during the two years of the study:\(^{16}\) $$\begin{array}{ll} &{\text { Number of }} & {\text { Number of }} \\ & {\text { patients }} & {\text { strokes }} \\ \hline \text {Aspirin alone } & 1649 & {206} \\ \text {Aspirin + dipyridamole }& {1650} & {157}\end{array}$$ The study was a randomized comparative experiment. (a) Is there a significant difference in the proportion of strokes between these two treatments? Carry out an appropriate test to help answer this question. (b) Describe a Type I and a Type II error in this setting. Which is more serious? Explain.

What鈥檚 wrong? 鈥淲ould you marry a person from a lower social class than your own?鈥� Researchers asked this question of a random sample of 385 black, never- married students at two historically black colleges in the South. Of the 149 men in the sample, 91 said 鈥淵es.鈥� Among the 236 women, 117 said 鈥淵es.鈥漒(^{14}\) Is there reason to think that different proportions of men and women in this student population would be willing to marry beneath their class? Holly carried out the significance test shown below to answer this question. Unfortunately, she 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 $$H_{0} : p_{1}=\rho_{2}$$ $$H_{a} : p_{1} \neq p_{2}$$ at the 95% confidence level. Plan: If conditions are met, I鈥檒l do a one-sample \(z\) test for comparing two proportions. \(\cdot\) Random The data came from a random sample of 385 black, never-married students. \(\cdot\) Normal One student's answer to the question should have no relationship to another student's answer. \(\cdot\) Independent The counts of successes and falures in the two groups - \(91,58,117,\) and \(119-\) are all at least 10 . Do: From the data, \(\hat{p}_{1}=\frac{91}{149}=0.61\) and \(\hat{p}_{2}=\frac{117}{236}=0.46\) \(\bullet\) Test statistic $$z=\frac{(0.61-0.46)-0}{\sqrt{\frac{0.61(0.39)}{149}+\frac{0.46(0.54)}{236}}}=2.91$$ \(\cdot P\) value From Table \(A, P(z \geq 2.91)=1-0.9982=\) 0.0018 . Conclude: The P-value, \(0.0018,\) is less than \(0.05,\) so I'll reject the null hypothesis. This proves that a higher proportion of men than women are willing to marry someone from a social class lower than their own.

How tall? 6.2 ) The heights of young men follow a Normal distribution with mean 69.3 inches and standard deviation 2.8 inches. The heights of young women follow a Normal distribution with mean 64.5 inches and standard deviation 2.5 inches. (a) Let \(M=\) the height of a randomly selected young man and \(W=\) the height of a randomly selected young woman. Describe the shape, center, and spread of the distribution of \(M-W\) (b) Find the probability that a randomly selected young man is at least 2 inches taller than a randomly selected young woman. Show your work.

Credit cards and incentives A bank wants to know which of two incentive plans will most increase the use of its credit cards. 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. (a) Is this a problem about comparing means or comparing proportions? Explain. (b) What type of study design is being used to produce data?

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.
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