/*! 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 57 Tongue Piercing May Speed Tooth ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 57

Tongue Piercing May Speed Tooth Loss. Researchers Say" is the headline of an article that appeared in the San Luis Obispo Tribune (June 5,2002 ). The article describes a study of 52 young adults with pierced tongues. The rescarchers found receding gums, which can lead to tooth loss, in 18 of the participants. Construct a \(95 \%\) confidence interval for the proportion of young adults with pierced tongues who have receding gums. What assumptions must be made for use of the \(z\) confidence interval to be appropriate?

Short Answer

Expert verified
The sample proportion is given by \(\hat{p} = \frac{18}{52}\). The 95% confidence interval is found using the formula \(\hat{p} \pm E\) where E is defined as \(1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{52}}\). The assumptions for the validity of the \(z\) interval are a random sample and an approximately normal distribution.

Step by step solution

01

Calculate the Sample Proportion

First calculate the sample proportion \(\hat{p}\), which is the proportion of young adults in the sample with receding gums. This is calculated as \(\hat{p} = \frac{x}{n} = \frac{18}{52}\), where \(x\) is the number of successful outcomes and \(n\) is the size of the sample.
02

Calculate the Margin of Error

Next, calculate the margin of error (E) using the formula \(E = z \sqrt{(\frac{\hat{p}(1-\hat{p})}{n})}\) where z is the z-score corresponding to the desired degree of confidence (for 95% confidence interval, z-score is 1.96).
03

Calculate the Confidence Interval

Use the margin of error to calculate the confidence interval. The interval is \(\hat{p} \pm E\). This will give an interval within which the proportion of all young adults with pierced tongues who have receding gums lies with \(95 \%\) confidence.
04

State the Assumptions for \(z\) Interval

Assumptions for use of \(z\) interval: 1) The sample should be random. 2) The distribution should be approximately normal. This is usually the case when \(n\hat{p} > 5\) and\(n(1-\hat{p}) > 5\) where \(n\) is the sample size and \(\hat{p}\) is the sample proportion.

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 91影视!

Key Concepts

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

Understanding Sample Proportion
The sample proportion is an essential concept when dealing with statistics and surveys. It represents the fraction of individuals in a sample who exhibit a particular characteristic. In our exercise, the sample proportion (\(\hat{p}\)) is the ratio of young adults with receding gums to the total number of participants with pierced tongues. To find this, we divide the number of people with receding gums (18) by the total sample size (52). Thus, \(\hat{p} = \frac{18}{52}\). This proportion gives us a snapshot of the larger population's characteristics based on our sample.
What is Margin of Error?
The margin of error is a critical value in statistics that provides an estimate of the likely range within which the true population measure lies. It accounts for sampling variability and expresses how much the sample proportion may vary from the actual population proportion. For a confidence interval, the margin of error (E) is calculated with the formula \[E = z \sqrt{\left( \frac{\hat{p}(1-\hat{p})}{n} \right)}\]where:
  • \( z \) is the z-score associated with the confidence level (1.96 for 95% confidence),
  • \( \hat{p} \) is the sample proportion, and
  • \( n \) is the sample size.
The margin of error helps define the range of the confidence interval, which indicates the uncertainty around the sample proportion estimate.
The Role of Z-score
The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. In the context of a confidence interval, it's used to determine the number of standard deviations a data point is from the mean. For a 95% confidence level, the z-score is 1.96. This means we are 95% confident that the actual proportion lies within the calculated confidence interval. The higher the confidence level, the larger the z-score, which widens the confidence interval, indicating greater uncertainty. By using the z-score, statisticians can ensure that the calculated interval is likely to contain the true population proportion.
Understanding Normal Distribution
Normal distribution is a fundamental concept in statistics, often referred to as the bell curve due to its shape. It's characterized by its symmetry around the mean, with most of the data points clustering around a central high point. In the context of confidence intervals, normal distribution is vital because many statistical methods, including the use of z-scores, presuppose normality. For the confidence interval to be valid:
  • The sample data should ideally follow a normal distribution.
  • This becomes approximately true if conditions like \( n\hat{p} > 5 \) and \( n(1-\hat{p}) > 5 \) are satisfied.
Importance of Random Sample
A random sample serves as the foundation for reliable statistical analysis. It ensures that each individual in a population has an equal probability of being chosen, eliminating bias and yielding a sample that accurately reflects the population. In this exercise, one of the critical assumptions for constructing a \( z \) confidence interval is that the sample is random. Without this, our findings might not be generalizable to the broader population. Random sampling enhances the validity and reliability of statistical conclusions, making it a crucial component in the study design.

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

Each person in a random sample of 20 students at a particular university was asked whether he or she is registered to vote. The responses \((\mathrm{R}=\) registered, \(\mathrm{N}=\) not registered) are given here: R \(R N R N N R R R N R R R R R N R R R N\) Use these data to estimate \(p\), the proportion of all students at the university who are registered to vote.

The paper "The Curious Promiscuity of Queen Honey Bees (Apis mellifera), Evolutionary and BehavIoral Mechanisms" (Annals of Zoology Fennic \(120011: 255-\) 265 ) describes a study of the mating behavior of queen honeybees. The following quote is from the paper: Queens flew for an average of \(24.2 \pm 9.21\) minutes on their mating flights, which is consistent with previous findings. On those flighrs, queens effectively mated with \(4.6 \pm 3.47\) males (mean \(\pm \mathrm{SD})\). a. The intervals reported in the quote from the paper were based on data from the mating flights of \(n=\) 30 queen honeybees. One of the two intervals reported is stated to be a confidence interval for a population mean. Which interval is this? Justify your choice. b. Use the given information to construct a \(95 \%\) confidence interval for the mean number of partners on a mating flight for queen honeybees. For purposes of this exercise, assume that it is reasonable to consider these 30 queen honeybees as representative of the population of queen honeybees.

The article "Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes" (Ipsos-Reid, October 24, 2005 ) summarized results from a survey of 1000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween during \(2005 .\) The resulting sample mean and standard deviation were \(\$ 46.65\) and \(\$ 83.70,\) respectively. a. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be larger than the mean anticipated expense. b. Is it reasonable to think that the distribution of the yariable anticipated Halloween expense is approximately normal? Explain why or why not. c. Is it appropriate to use the \(t\) confidence interval to estimate the mean anticipated Halloween expense for Canadian residents? Explain why or why not. d. If appropriate, construct and interpret a \(99 \%\) confidence interval for the mean anticipated Halloween exnense for Canadian residents.

Acrylic bone cement is sometimes used in hip and knee replacements to fix an artificial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens under specified conditions, and the resulting mean and standard deviation were 306.09 Newtons and 41.97 Newtons, respectively. Assuming that it is reasonable to believe that breaking force under these conditions has a distribution that is approximately normal, estimate the mean breaking force for acrylic bone cement under the specified conditions using a \(95 \%\) confidence interval.

In a study of academic procrastination, the authors of the paper "Correlates and Consequences of Behavioral Procrastination" (Procrastination, Current Issues and New Directions [20001) reported that for a sample of 411 undergraduate students at a midsize public university preparing for a final exam in an introductory psychology course, the mean time spent studying for the exam was 7.74 hours and the standard deviation of study times was 3.40 hours. For purposes of this exercise, assume thar it is reasonable to regard this sample as represcntative of students taking introductory psychology at this university. a. Construct a \(95 \%\) confidence interval to estimate \(\mu\), the mean time spent studying for the final exam for students taking introductory psychology at this university. b. The paper also gave the following sample statistics for the percentage of study time that occurred in the 24 hours prior to the exam: \(n=411 \quad \bar{x}=43.18 \quad s=21.46\) Construct and interpret a \(90 \%\) confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the exam.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.