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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 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 for the proportion of young adults with pierced tongues who have gum recession is between 20.7% and 48.5%, assuming our sample is random, and the observations are independent and normally distributed.

Step by step solution

01

Identify the sample proportion

Here, we are given a sample of 52 young adults with pierced tongues, among which 18 are experiencing gum recession. So, the sample proportion \(\hat{p}\) can be calculated as \(\hat{p} = \frac{number \ of \ successes}{sample \ size} = \frac{18}{52} = 0.34615.\)
02

Calculate the standard error

The standard error of the proportion can be calculated by the formula \(SE = \sqrt{ \frac{ \hat{p}(1-\hat{p}) }{n}} =\sqrt{ \frac{0.34615 \times (1-0.34615) }{52}} =0.071.\)
03

Compute the z-score for a 95% confidence interval

Assuming a 95% confidence level, the z-score (given normal distribution) is 1.96.
04

Calculate the Confidence Interval

The 95% confidence interval can be computed like this: \(CI = \hat{p} \pm (Z*SE) = 0.34615 \pm (1.96*0.071) = (0.207, 0.485). \) So, we're 95% confident that the true population proportion of young adults with tongue piercings who have receding gums is between 20.7% and 48.5%.
05

Identify the assumptions

The assumptions required for this z-interval to be appropriate are: (1) the sample is random, (2) the sampling distribution of the proportion is approximately normal, which is generally the case given a large sample size (typically if np 鈮� 10 and n(1-p) 鈮� 10). It also assumes that (3) the observations are independent of each other, i.e. one individual's chance of having receding gums doesn't affect another's.

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Key Concepts

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

Sample Proportion
In statistics, the sample proportion is the fraction of the sample that shows the feature of interest. It is denoted as \(\textstyle\hat{p}\). For instance, if you have a sample of 100 individuals and 40 of them have a particular characteristic, the sample proportion \(\hat{p}\) is \(\frac{40}{100} = 0.4\).

In the context of the original problem about young adults with tongue piercings, the sample proportion was found by dividing the number of individuals with receding gums by the total number of individuals in the study. This gives a crucial point estimate for constructing confidence intervals and helps in understanding the likelihood of gum recession among the studied group.
Standard Error Computation
The standard error (SE) measures the variability of a sample proportion. Computation of SE is essential for forming confidence intervals and for hypothesis testing. The formula for the standard error of the proportion is \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where \(\textstyle\hat{p}\) is the sample proportion, and \(n\) is the sample size.

A smaller SE indicates that the sample proportion is a more precise estimate of the population proportion. As the sample size increases, the SE decreases, improving the estimate's reliability. In our piercing study, SE is calculated to encompass the variability of the proportion of individuals with potential gum issues.
Z-Score
A z-score is a numerical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. If a z-score is 0, it represents the score as identical to the mean score.

A z-score can also be used to calculate the probability of a score occurring within a normal distribution and to compare scores from different distributions. In the case of constructing a confidence interval, the z-score corresponds to the desired level of confidence. For a 95% confidence interval, the z-score is typically 1.96, which means that the interval extends from the mean to 1.96 standard deviations below and above the mean on a normal distribution.
Sampling Distribution
The sampling distribution is a probability distribution of a statistic obtained through a large number of samples drawn from a specific population. It is critical in the calculation of confidence intervals. This distribution tells us how the sample proportion varies from sample to sample.

For the z-confidence interval to be suitable, we assume the sampling distribution of the sample proportion follows a normal distribution, especially true when the sample size is large enough (according to the Central Limit Theorem). This is the case if the conditions \(np \geq 10\) and \(n(1-p) \geq 10\) are met, ensuring the sample proportion will be approximately normally distributed.

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Most popular questions from this chapter

USA Today (October 14, 2002) reported that \(36 \%\) of adult drivers admit that they often or sometimes talk on a cell phone when driving. This estimate was based on data from a sample of 1004 adult drivers, and a bound on the error of estimation of \(3.1 \%\) was reported. Assuming a \(95 \%\) confidence level, do you agree with the reported bound on the error? Explain.

The article "Consumers Show Increased Liking for Diesel Autos" (USA Today. January 29,2003 ) reported that \(27 \%\) of U.S. consumers would opt for a diesel car if it ran as cleanly and performed as well as a car with a gas engine. Suppose that you suspect that the proportion might be different in your area and that you want to conduct a survey to estimate this proportion for the adult residents of your city. What is the required sample size if you want to estimate this proportion to within \(.05\) with \(95 \%\) confidence? Compute the required sample size first using \(.27\) as a preliminary estimate of \(p\) and then using the conservative value of \(.5 .\) How do the two sample sizes compare? What sample size would you recommend for this study?

The study 鈥淒igital Footprints" (Pew Internet \& American Life Project, www.pewinternet.org, 2007) reported that \(47 \%\) of Internet users have searched for information about themselves online. The \(47 \%\) figure was based on a random sample of Internet users. For purposes of this exercise, suppose that the sample size was \(n=300\) (the actual sample size was much larger). Construct and interpret a \(90 \%\) confidence interval for the proportion of Internet users who have searched online for information about themselves.

Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were as follows: \(\begin{array}{ccccc}6 & 17 & 11 & 22 & 29\end{array}\) Assuming that these five students can be considered a random sample of all students participating in the free checkup program, construct a \(95 \%\) confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program.

The article "Nine Out of Ten Drivers Admit in Survey to Having Done Something Dangerous" (Knight Ridder Newspapers, July 8,2005 ) reported the results of a survey of 1100 drivers. Of those surveyed, 990 admitted to careless or aggressive driving during the previous 6 months. Assuming that it is reasonable to regard this sample of 1100 as representative of the population of drivers, use this information to construct a \(99 \%\) confidence interval to estimate \(p\), the proportion of all drivers who have engaged in careless or aggressive driving in the previous 6 months.
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