Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The Chronicle of Higher Education (January 13, 1993) reported that \(72.1 \%\) of those responding to a national survey of college freshmen were attending the college of their first choice. Suppose that \(n=500\) students responded to the survey (the actual sample size was much larger). a. Using the sample size \(n=500\), calculate a \(99 \%\) confidence interval for the proportion of college students who are attending their first choice of college. b. Compute and interpret a \(95 \%\) confidence interval for the proportion of students who are not attending their first choice of college. c. The actual sample size for this survey was much larger than 500 . Would a confidence interval based on the actual sample size have been narrower or wider than the one computed in Part (a)?

Short Answer

Expert verified
a. The 99% confidence interval for students attending their first choice of college is from 0.692 to 0.750. b. The 95% confidence interval for students not attending their first choice of college is from 0.245 to 0.313. c. A confidence interval based on the actual sample size would have been narrower than the one computed in part (a).

Step by step solution

01

Calculate 99% confidence interval for first choice

The formula for confidence interval is \(p \pm Z_{\frac{a}{2}} \cdot \sqrt{\frac{p(1-p)}{n}}\), where \(p\) is proportion, \(n\) is the sample size and \(Z_{\frac{a}{2}}\) is the respective Z-score for a given confidence level. For 99% confidence, \(Z_{\frac{a}{2}} = 2.576\). Substitute \(p = 0.721\) and \(n = 500\) into the formula to receive the confidence interval.
02

Calculate 95% confidence interval for not first choice

The proportion of students not attending their first choice of college is \(1 - p = 1 - 0.721 = 0.279\). For 95% confidence, \(Z_{\frac{a}{2}} = 1.96\). Substitute these values into the formula from Step 1 to obtain the confidence interval.
03

Compare the width of confidence intervals

A confidence interval becomes narrower with a larger sample size. In part (c) it specifies the actual sample size was much larger than 500. An interval based on a larger sample size than 500, therefore, would have been narrower than the one computed in part (a).

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 Vaia!

Key Concepts

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

Statistics
Statistics is the science of collecting, analyzing, and interpreting data to make informed decisions. It uses a combination of mathematics, probability, and computational techniques to convert raw data into useful information. In the context of our exercise, statistics helps us understand aspects of college freshmen's choices regarding their education through the construction of confidence intervals, which offer a likely range of values based on their responses to a survey.

For students grappling with statistical concepts, remember that every dataset provides a story about a particular population. In this case, it's about college freshmen and their college preferences. By delving into the statistics, they can not only get numerical results but can also paint a broader picture of the population's behavior or preferences, leading to valuable insights.
Data Analysis
Data analysis involves the systematic application of statistical and logical techniques to describe and illustrate, condense and recap, and evaluate data. The primary purpose is to extract useful information and develop conclusions. In our exercise, we examine survey data to better understand a trend within the student population. Through the calculated confidence intervals, we can analyze how confident we can be in our estimates of the entire college freshmen population's preferences.

Through a relatable example, such as students attending their first choice of college, data analysis becomes less daunting. It's akin to surveying a handful of friends about their favorite ice cream flavors to predict the next flavor the group might enjoy.
Confidence Level
The confidence level represents the degree of certainty that the calculated confidence interval includes the true population parameter. For example, a 99% confidence level means that if we were to take 100 different random samples and compute a confidence interval for each sample, we would expect 99 of the confidence intervals to contain the true population proportion.

A higher confidence level indicates a larger margin of error and a broader confidence interval. It's important to note that while a higher confidence level may seem like it would yield better results, it also means the interval will be less precise. This is the trade-off when choosing a confidence level.
Sample Size
Sample size, denoted by 'n' in our exercise, significantly affects the calculation of confidence intervals. Larger sample sizes tend to yield more precise estimates of the population parameter, resulting in narrower confidence intervals. This is because they provide more data points, thus reflecting the population more accurately and reducing the margin of error.

In plain terms, if you want to know what kind of pizza toppings a city prefers, asking 500 people will provide a clearer picture than asking just 5. Therefore, when we discuss the confidence interval becoming narrower with a larger sample size, it's akin to sharpening the focus on our pizza preference picture; details become clearer and more reliable.
Proportion
Proportion in our exercise refers to the part of the surveyed sample that represents a particular characteristic—like the proportion of students attending their first choice of college. A proportion is a type of ratio, which can be expressed as a fraction, percentage, or decimal, and is fundamental in calculating the confidence interval.

Imagery can sometimes simplify the understanding of a proportion. Think of it as slicing a pie in accordance with who wants a piece. In our case, 72.1% is akin to cutting the pie so that seven out of ten friends ideally get the exact subpart they want. When you work with proportions in statistics, you are allocating parts of a whole based on certain criteria or responses, just as you would divide a pie to satisfy varying appetites.

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

The article "Doctors Cite Burnout in Mistakes" (San Luis Obispo Tribune, March 5,2002 ) reported that many doctors who are completing their residency have financial struggles that could interfere with training. In a sample of 115 residents, 38 reported that they worked moonlighting jobs and 22 reported a credit card debt of more than \(\$ 3000\). Suppose that it is reasonable to consider this sample of 115 as a random sample of all medical residents in the United States. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. medical residents who work moonlighting jobs. b. Construct and interpret a \(90 \%\) confidence interval for the proportion of U.S. medical residents who have a credit card debt of more than \(\$ 3000\). c. Give two reasons why the confidence interval in Part (a) is wider than the confidence interval in Part (b).

In the article "Fluoridation Brushed Off by Utah" (Associated Press, August 24,1998 ), it was reported that a small but vocal minority in Utah has been successful in keeping fluoride out of Utah water supplies despite evidence that fluoridation reduces tooth decay and despite the fact that a clear majority of Utah residents favor fluoridation. To support this statement, the article included the result of a survey of Utah residents that found \(65 \%\) to be in favor of fluoridation. Suppose that this result was based on a random sample of 150 Utah residents. Construct and interpret a \(90 \%\) confidence interval for \(\pi\), the true proportion of Utah residents who favor fluoridation. Is this interval consistent with the statement that fluoridation is favored by a clear majority of residents?

Seventy-seven students at the University of Virginia were asked to keep a diary of a conversation with their mothers, recording any lies they told during these conversations (San Luis Obispo Telegram-Tribune, August 16 , 1995 ). It was reported that the mean number of lies per conversation was \(0.5 .\) Suppose that the standard deviation (which was not reported) was \(0.4\). a. Suppose that this group of 77 is a random sample from the population of students at this university. Construct a \(95 \%\) confidence interval for the mean number of lies per conversation for this population. b. The interval in Part (a) does not include 0 . Does this imply that all students lie to their mothers? Explain.

Five hundred randomly selected working adults living in Calgary, Canada were asked how long, in minutes, their typical daily commute was (Calgary Herald Traffic Study, Ipsos, September 17,2005 ). The resulting sample mean and standard deviation of commute time were \(28.5\) minutes and \(24.2\) minutes, respectively. Construct and interpret a \(90 \%\) confidence interval for the mean commute time of working adult Calgary residents.

A study reported in Newsweek (December 23, 1991) involved a sample of 935 smokers. Each individual received a nicotine patch, which delivers nicotine to the bloodstream but at a much slower rate than cigarettes do. Dosage was decreased to 0 over a 12-week period. Suppose that 245 of the subjects were still not smoking 6 months after treatment (this figure is consistent with information given in the article). Estimate the percentage of all smokers who, when given this treatment, would refrain from smoking for at least 6 months.

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.

Sign-up for free