/*! 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 77 Are college students who take a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Are college students who take a freshman orientation course more or less likely to stay in college than those who do not take such a course? The article "A Longitudinal Study of the Retention and Academic Performance of Participants in Freshmen Orientation Courses" (Journal of College Student Development \([1994]: 444-\) 449) reported that 50 of 94 randomly selected students who did not participate in an orientation course returned for a second year. Of 94 randomly selected students who did take the orientation course, 56 returned for a second year. Construct a \(95 \%\) confidence interval for \(\pi_{1}-\pi_{2}\), the difference in the proportion returning for students who do not take an orientation course and those who do. Give an interpretation of this interval.

Short Answer

Expert verified
The \(95\%\) confidence interval for \(\pi_{1}-\pi_{2}\) will provide a range of values that includes the true population parameter \(95\%\) of the time. The interpretation will depend on the computed interval. If 0 is included in the interval, then there is no significant difference in the return rates for the two groups.

Step by step solution

01

Identify Given Data

From the problem, we know that 50 of 94 students who did not participate in the orientation course returned for a second year, therefore \(\pi_{1} = \frac{50}{94}\). Similarly, 56 of 94 students who did take the course, returned for a second year, so \(\pi_{2} = \frac{56}{94}\). We are asked to compute a \(95\%\) confidence interval.
02

Calculate the Standard Error

The standard error (SE) for \(\pi_{1} - \pi_{2}\) can be calculated using the formula: \[ \text{SE}_{\pi_{1} - \pi_{2}} = \sqrt{\frac{\pi_{1} (1 - \pi_{1})}{n_1} + \frac{\pi_{2} (1 - \pi_{2})}{n_2}} \] Where \(n_1\) and \(n_2\) represent the number of individuals in each group, both of which are 94 in this case.
03

Calculate the Confidence Interval

A \(95\%\) confidence interval can be calculated using the following formula: \[ CI = (\pi_{1} - \pi_{2}) \pm Z_{0.025} \times \text{SE}_{\pi_{1} - \pi_{2}} \] Here \(Z_{0.025}\) is the z-value for a \(95\%\) interval, which is 1.96.
04

Interpret the Confidence Interval

The confidence interval will give the range of values within which we are \(95\%\) confident that the difference between the two proportions lies. If the interval includes 0, it suggests no significant difference between the likelihood of returning to college based on whether or not they attended the orientation. If all values in the interval are positive (or negative), it would suggest a significant difference.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Proportion Difference
When we talk about the proportion difference, we're essentially looking at comparing two groups. In this exercise, we are comparing the proportion of college students who return for a second year between those who took a freshman orientation course and those who didn’t. To do this, we calculate each group's proportion separately. For students who did not take the course, it's \[\pi_{1} = \frac{50}{94}\]For those who took the course, the proportion is \[\pi_{2} = \frac{56}{94}\]The purpose of finding a proportion difference is to see if there is any meaningful difference in outcomes between these two groups. We express the difference as \[\pi_{1} - \pi_{2}\] This difference helps us explore how impactful the orientation course might be.
Statistical Significance
Statistical significance is a way to understand if the difference we observed in the data is substantial, or if it could have happened by random chance. In our scenario, we want to determine if the difference in proportions of returning students is statistically significant. Statistical tests, such as hypothesis testing, help in calculating this significance. If a result is statistically significant, it means that the observed difference is probably not due to random chance. We often use a confidence interval to infer this; if the interval does not contain zero, the difference is significant. For a 95% confidence level, we use a critical value (z-value) of 1.96. This means that only 5% of the time, such a difference would occur by chance if there were no real difference in the population. The result gives us a measure of confidence in the observed difference.
Standard Error
The standard error (SE) is a crucial concept when calculating the confidence interval for the difference in proportions. It gives us an idea of how much variability we might expect in our sample proportions.For two independent groups, like in our case, the standard error can be calculated as:\[\text{SE}_{\pi_{1} - \pi_{2}} = \sqrt{\frac{\pi_{1} \times (1 - \pi_{1})}{n_1} + \frac{\pi_{2} \times (1 - \pi_{2})}{n_2}}\]Here, \( n_1 \) and \( n_2 \) represent the sample sizes of each group respectively, which are both 94 students. SE helps determine how ‘spread out’ the proportion difference might be from the true population proportion difference.A smaller SE indicates more reliable estimates of the proportion difference, helping us create a more precise confidence interval.
Hypothesis Testing
Hypothesis testing is a systematic method used to decide whether the observed data supports a certain belief or hypothesis. In this context, we are interested in knowing if taking a freshman orientation course has a statistically significant effect on whether students return for the second year.Usually, we start with a null hypothesis (\(H_0\)) that states there is no difference between the two groups, meaning \(\pi_{1} - \pi_{2} = 0\).The alternative hypothesis (\(H_a\)) challenges this, suggesting that the difference is not zero.We use a confidence interval to test our hypothesis. If the interval includes zero, we fail to reject the null hypothesis; this implies no significant effect. If zero is not in the interval, we reject the null hypothesis, suggesting a significant difference.This process is crucial for research and decision-making as it provides a statistical basis for conclusions based on sample data.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Are very young infants more likely to imitate actions that are modeled by a person or simulated by an object? This question was the basis of a research study summarized in the article "The Role of Person and Object in Eliciting Early Imitation" (Journal of Experimental Child Psychology [1991]: 423-433). One action examined was mouth opening. This action was modeled repeatedly by either a person or a doll, and the number of times that the infant imitated the behavior was recorded. Twentyseven infants participated, with 12 exposed to a human model and 15 exposed to the doll. Summary values are given here. Is there sufficient evidence to conclude that the mean number of imitations is higher for infants who watch a human model than for infants who watch a doll? Test the relevant hypotheses using a .01 significance level.

The article "Trial Lawyers and Testosterone: BlueCollar Talent in a White- Collar World" (Journal of Applied Social Psychology [1998]: 84-94) compared trial lawyers and nontrial lawyers on the basis of mean testosterone level. Random samples of 35 male trial lawyers, 31 male nontrial lawyers, 13 female trial lawyers, and 18 female nontrial lawyers were selected for study. The article includes the following statement: "Trial lawyers had higher testosterone levels than did nontrial lawyers. This was true for men, \(t(64)=3.75, p<.001\), and for women, \(t(29)=2.26, p<.05 . "\) a. Based on the information given, is there a significant difference in the mean testosterone level for male trial and nontrial lawyers? b. Based on the information given, is there a significant difference in the mean testosterone level for female trial and nontrial lawyers? c. Do you have enough information to carry out a test to determine whether there is a significant difference in the mean testosterone levels of male and female trial lawyers? If so, carry out such a test. If not, what additional information would you need to be able to conduct the test?

In December 2001, the Department of Veterans Affairs announced that it would begin paying benefits to soldiers suffering from Lou Gehrig's disease who had served in the Gulf War (The New York Times, December 11,2001 ). This decision was based on an analysis in which the Lou Gehrig's disease incidence rate (the proportion developing the disease) for the approximately 700,000 soldiers sent to the Gulf between August 1990 and July 1991 was compared to the incidence rate for the approximately \(1.8\) million other soldiers who were not in the Gulf during this time period. Based on these data, explain why it is not appropriate to perform a formal inference procedure (such as the two-sample \(z\) test) and yet it is still reasonable to conclude that the incidence rate is higher for Gulf War veterans than for those who did not serve in the Gulf War.

Consider two populations for which \(\mu_{1}=30, \sigma_{1}=2\), \(\mu_{2}=25\), and \(\sigma_{2}=3\). Suppose that two independent random samples of sizes \(n_{1}=40\) and \(n_{2}=50\) are selected. Describe the approximate sampling distribution of \(\bar{x}_{1}-\bar{x}_{2}\) (center, spread, and shape).

The article "Spray Flu Vaccine May Work Better Than Injections for Tots" (San Luis Obispo Tribune, May 2,2006 ) described a study that compared flu vaccine administered by injection and flu vaccine administered as a nasal spray. Each of the 8000 children under the age of 5 that participated in the study received both a nasal spray and an injection, but only one was the real vaccine and the other was salt water. At the end of the flu season, it was determined that \(3.9 \%\) of the 4000 children receiving the real vaccine by nasal spray got sick with the flu and \(8.6 \%\) of the 4000 receiving the real vaccine by injection got sick with the flu. a. Why would the researchers give every child both a nasal spray and an injection? b. Use the given data to estimate the difference in the proportion of children who get sick with the flu after being vaccinated with an injection and the proportion of children who get sick with the flu after being vaccinated with the nasal spray using a \(99 \%\) confidence interval. Based on the confidence interval, would you conclude that the proportion of children who get the flu is different for the two vaccination methods?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.