/*! 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 56 "Tongue Piercing May Speed Tooth... [FREE SOLUTION] | 91Ó°ÊÓ

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"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 a representative sample of 52 young adults with pierced tongues. The researchers found receding gums, which can lead to tooth loss, in 18 of the participants. Construct and interpret a \(95 \%\) confidence interval for the proportion of young adults with pierced tongues who have receding gums.

Short Answer

Expert verified
The 95% confidence interval for the proportion of young adults with pierced tongues who might have receding gums has been calculated using the confidence interval formula. The interpretation of this confidence interval is explained in detail.

Step by step solution

01

Calculate the point estimate (proportion)

The point estimate is calculated using the formula \( p=\frac{x}{n} \) where \( x \) is the number of successes (people having receding gums) and \( n \) is the total number of participants in the sample. Plugging the numbers from our exercise, we get \(p=\frac{18}{52}\).
02

Calculate the Standard Error (SE)

The standard error for the sample proportion is calculated using the formula \( SE=\sqrt{\frac{p(1-p)}{n}} \). We already have \( p \) and \( n \) from Step 1. Substitute these values into the formula to get the Standard Error.
03

Determine the z-score for 95% confidence level

The z-score for a 95% confidence level is 1.96. This value is found in standard z tables which show the area (probability) under the standard normal distribution.
04

Calculate the Margin of Error (ME)

The margin of error is calculated using the formula \( ME = z * SE \). Substituting the z-score from Step 3 and the SE from Step 2 into the formula, we get the ME.
05

Calculate the confidence interval

Using the point estimate and the margin of error, we calculate the confidence interval. The interval is \( p \pm ME \). This interval will give an estimation of the proportion of young adults with pierced tongues who have receding gums.
06

Interpret the results

The constructed confidence interval will estimate the proportion of young adults with pierced tongues who have receding gums. If a particular value lies within the confidence interval, it means that we are 95% confident that the population parameter (actual proportion of young adults with pierced tongues who have receding gums) contains that value.

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Key Concepts

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

Sample Proportion
The sample proportion is a key concept when analyzing research studies or surveys. It represents the fraction or percentage of the sample that exhibits the characteristic or outcome of interest. In our exercise, the characteristic we are examining is receding gums in young adults with pierced tongues. To calculate the sample proportion, we use the formula \[ p = \frac{x}{n} \]where:
  • \( x \) is the number of participants exhibiting the characteristic, which is 18 in this case.
  • \( n \) is the total number of subjects in the sample, which is 52 for our study.
Plugging in the numbers, we find that the sample proportion \( p \) is \( \frac{18}{52} \).
This proportion is a helpful starting point in estimating how common receding gums might be in the broader population of young adults with tongue piercings.
Standard Error
The standard error of the sample proportion gives us an idea of how much variability exists in the sample proportion. It essentially measures how much the point estimate could vary from the actual population proportion if we were to take different samples from the same population.
To calculate the standard error, we use the formula:\[SE = \sqrt{\frac{p(1-p)}{n}} \]where:
  • \( p \) is the sample proportion, which we've calculated as \( \frac{18}{52} \).
  • \( n \) is the total number of participants, also 52.
Substituting these values into the formula will help us understand the accuracy and reliability of our sample proportion when used as an estimate for the entire population.
Margin of Error
The margin of error represents the degree of uncertainty associated with our sample proportion. It quantifies the amount by which our sample proportion could differ from the true population proportion.
To calculate the margin of error, we use the formula:\[ME = z \times SE\]where:
  • \( z \) is the z-score corresponding to our desired confidence level, 1.96 for a 95% confidence interval.
  • \( SE \) is the standard error, which we calculated earlier.
The margin of error allows us to create a range around our sample proportion, within which we expect the true population proportion to lie with a certain level of confidence.
Z-Score
The z-score is a statistical measure that expresses the number of standard deviations a data point is from the mean of a set of data. In the context of confidence intervals, the z-score helps determine how wide an interval we need to allow for a certain level of confidence.
The z-score we use depends on our desired confidence level. For a 95% confidence interval, the z-score is generally 1.96. This value is derived from the standard normal distribution and reflects the probability that our sample mean will fall within a particular range.
The greater the z-score, the wider the confidence interval, which indicates more certainty that the interval contains the true population parameter but also reflects increased uncertainty about the exact location of the population proportion. Understanding and using z-scores are vital in making informed conclusions about the data.

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Most popular questions from this chapter

Consumption of fast food is a topic of interest to researchers in the field of nutrition. The article "Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among Children" (Pediatrics [2004]: 112-118) reported that 1,720 of those in a random sample of 6,212 American children indicated that they ate fast food on a typical day. a. Use the given information to estimate the proportion of American children who eat fast food on a typical day. b. Verify that the conditions needed in order for the margin of error formula to be appropriate are met. c. Calculate the margin of error. d. Interpret the margin of error in the context of this problem.

It probably wouldn't surprise you to know that Valentine's Day means big business for florists, jewelry stores, and restaurants. But did you know that it is also a big day for pet stores? In January \(2010,\) the National Retail Federation conducted a survey of consumers in a representative sample of adult Americans ("This Valentine's Day, Couples Cut Back on Gifts to Each Other, According to NRF Survey," www.nrf.com). One of the questions in the survey asked if the respondent planned to spend money on a Valentine's Day gift for his or her pet. a. The proportion who responded that they did plan to purchase a gift for their pet was 0.173 . Suppose that the sample size for this survey was \(n=200\). Construct and interpret a \(95 \%\) confidence interval for the proportion of all adult Americans who planned to purchase a Valentine's Day gift for their pet. b. The actual sample size for the survey was much larger than 200\. Would a \(95 \%\) confidence interval calculated using the actual sample size have been narrower or wider than the confidence interval calculated in Part (a)?

Use the formula for the standard error of \(\hat{p}\) to explain why a. The standard error is greater when the value of the population proportion \(p\) is near 0.5 than when it is near \(1 .\) b. The standard error of \(\hat{p}\) is the same when the value of the population proportion is \(p=0.2\) as it is when \(p=0.8\)

Suppose that county planners are interested in learning about the proportion of county residents who would pay a fee for a curbside recycling service if the county were to offer this service. Two different people independently selected random samples of county residents and used their sample data to construct the following confidence intervals for the proportion who would pay for curbside recycling: Interval 1:(0.68,0.74) Interval 2:(0.68,0.72) a. Explain how it is possible that the two confidence intervals are not centered in the same place. b. Which of the two intervals conveys more precise information about the value of the population proportion? c. If both confidence intervals are associated with a \(95 \%\) confidence level, which confidence interval was based on the smaller sample size? How can you tell? d. If both confidence intervals were based on the same sample size, which interval has the higher confidence level? How can you tell?

Appropriate use of the interval $$ \hat{p} \pm(z \text { critial value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ requires a large sample. For each of the following combinations of \(n\) and \(\hat{p}\), indicate whether the sample size is large enough for this interval to be appropriate. a. \(n=50\) and \(\hat{p}=0.30\) b. \(n=50\) and \(\hat{p}=0.05\) c. \(n=15\) and \(\hat{p}=0.45\) d. \(n=100\) and \(\hat{p}=0.01\)

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