/*! 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 25 Tongue Piercing May Speed Tooth ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Tongue Piercing May Speed Tooth Loss, }}\( Researchers Say" is the headline of an article that appeared in the San Luis Obispo Tribune (June 5,2002 ). The article describes a study of 52 young adults with pierced tongues. The researchers found receding gums, which can lead to tooth loss, in 18 of the participants. Construct a \)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 95% confidence interval fluctuates based on the calculated values. The assumptions that must be made for the z interval to be appropriate are that observations are independent and come from a random sample. Additionally, the Central Limit Theorem states if the sample size is large (n > 30), the distribution is approximately normal.

Step by step solution

01

Interpret the problem and data

From the exercise, we have a study of 52 young adults with pierced tongues, and among them, 18 have receding gums.
02

Calculate the sample proportion

The sample proportion (p) is calculated by dividing the number of young adults with receding gums by the total number of young adults in the study. In this case, the sample proportion is \(p = 18/52\).
03

Calculate the standard error

The standard error (SE) is calculated using the formula \(SE = \sqrt{(p(1-p)/n)}\), where p is the sample proportion, and n is the sample size. Plugging the values will allow us to get the standard error.
04

Find the critical value

Using the Z-table, the critical value for a 95% confidence interval is approximately 1.96.
05

Calculate the 95% confidence interval

The 95% confidence interval is calculated using the formula: \(p \pm Z*SE\), where p is the sample proportion, Z is the critical value, and SE is the standard error. Calculate this to get the confidence interval.
06

State the assumptions

The assumptions for this interval are that the samples are independent and are randomly sampled. Also, the Central Limit Theorem states that the distribution tends to be normal if the sample size is large (n > 30), which is the case here. A point to note is that these results are dependant on the validity of the study's methodology.

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.

Statistics
In the world of data and decision-making, statistics serves as a crucial tool for understanding and interpreting the information gathered from the world around us. It's the science of collecting, analyzing, interpreting, presenting, and organizing data to find patterns, trends, and probabilities. When a study, such as the investigation into receding gums among young adults with pierced tongues, seeks to understand a phenomenon in a larger population, statistics provides the methodology to make informed conclusions based on a sample.

Understanding the results from such studies involves familiarizing oneself with several statistical concepts, including proportion, standard error, and confidence intervals. These allow researchers to infer characteristics of a larger population without having to survey every individual, an often impractical or impossible task.
Sample Proportion
The sample proportion is a snapshot of what you're studying — in this case, the fraction of young adults with pierced tongues who are experiencing receding gums. It's represented by the symbol \(p\) and is calculated by dividing the number of 'successes' by the total number of observations in the sample. If you have 18 young adults with receding gums out of 52 surveyed, then the sample proportion \(p\) is \(18/52\).

Understanding sample proportions is essential to making predictions about the broader population. As you work with this kind of data, remember that sample proportions can vary from sample to sample; hence, statisticians use them alongside other statistical measures, such as the standard error and confidence intervals, to create a more complete, reliable picture.
Standard Error
The standard error (SE) measures how much the sample proportion (\(p\)) might differ from the true population proportion due to random sampling variability. It's the standard deviation of the sampling distribution of the sample proportion and is calculated using the formula \(SE = \sqrt{(p(1-p)/n)}\), where \(n\) is the sample size. In our example, with the sample proportion of receding gums calculated at \(18/52\), and with a sample size of 52, you can insert these values into the formula to determine the standard error.

This metric is vital because it allows researchers to create a range (known as a confidence interval) around the sample proportion to estimate where the true population proportion is likely to fall. The lower the standard error, the more precise the estimation is.
Central Limit Theorem
The Central Limit Theorem is a cornerstone in the field of statistics. This theorem states that, given a sufficiently large sample size, the sampling distribution of the sample mean will be normally distributed, regardless of the original distribution of the population. This principle is what underlies the approach to calculate confidence intervals using Z-scores or T-scores.

In the context of the pierced tongue study, considering the sample size of 52 is greater than the common threshold of 30, the central limit theorem assures us that the sampling distribution of our sample proportion will tend to be normal. This normality allows us to apply the Z-score to our calculations to determine our confidence interval. When we assume a normal distribution, we can assert with some degree of 'confidence' that the true population proportion lies within this interval, giving us a practical and statistically valid estimation.

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

Suppose that a random sample of 50 bottles of a par. ticular brand of cough medicine is selected and the alcohol content of each bottle is determined. Let \(\mu\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the sample of 50 results in a \(95 \%\) confidence interval for \(\mu\) of \((7.8,9.4)\). a. Would a \(90 \%\) confidence interval have been narrower or wider than the given interval? Explain your answer. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between \(7.8\) and \(9.4\). Is this statement correct? Why or why not? c. Consider the following statement: If the process of selecting a sample of size 50 and then computing the corresponding \(95 \%\) confidence interval is repeated 100 times, 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

The article "CSI Effect Has Juries Wanting More Evidence" (USA Today, August 5,2004 ) examines how the popularity of crime-scene investigation television shows is influencing jurors' expectations of what evidence should be produced at a trial. In a survey of 500 potential jurors, one study found that 350 were regular watchers of at least one crime-scene forensics television series. a. Assuming that it is reasonable to regard this sample of 500 potential jurors as representative of potential jurors in the United States, use the given information to construct and interpret a \(95 \%\) confidence interval for the true proportion of potential jurors who regularly watch at least one crime-scene investigation series. b. Would a \(99 \%\) confidence interval be wider or narrower than the \(95 \%\) confidence interval from Part (a)?

In a study of 1710 schoolchildren in Australia (Herald Sun, October 27,1994 ), 1060 children indicated that they normally watch TV before school in the morning. (Interestingly, only \(35 \%\) of the parents said their children watched TV before school!) Construct a \(95 \%\) confidence interval for the true proportion of Australian children who say they watch TV before school. What assumption about the sample must be true for the method used to construct the interval to be valid?

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.

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 assume that breaking force under these conditions has a distribution that is approximately normal, estimate the true average breaking force for acrylic bone cement under the specified conditions.

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.