/*! 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 61 In the article "Fluoridation Bru... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(p\), 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?

Short Answer

Expert verified
The \(90\%\) confidence interval for the proportion of Utah residents who favor fluoridation is \(58.6\%\) to \(71.4\%\), implying a clear majority favors fluoridation.

Step by step solution

01

Calculate the sample proportion

The sample proportion, \(\hat{p}\), of Utah residents who favor fluoridation is \(65\%\) or \(0.65\). The sample size, \(n\), is \(150\).
02

Identify the confidence level

The desired level of confidence is \(90\%\), meaning the significance level (\(\alpha\)) is \(1 - 0.90 = 0.10\). Half of this value goes into each tail of the distribution, hence \(\alpha/2 = 0.05\).
03

Compute the Z-Score for the Sample Proportion

From the normal distribution table, for \(\alpha/2 = 0.05\), we find the Z-score is \(1.645\).
04

Compute the standard error of the proportion

Standard error (SE) is given by the formula \(SE = \sqrt{\hat{p}(1-\hat{p})/n}\) = \(\sqrt{0.65(1-0.65)/150} = 0.039\).
05

Compute the Margin of Error

The Margin of Error (E) is computed as \(E = Z \times SE = 1.645 \times 0.039 = 0.064\).
06

Construct the Confidence Interval

The \(90\%\) confidence interval for \(p\) is given by the formula \(\hat{p} \pm E\), which is \(0.65 \pm 0.064\). This gives the interval \(0.586\) to \(0.714\) or \(58.6\%\) to \(71.4\%\).
07

Interpret the Confidence Interval

With \(90\%\) confidence, the true proportion of Utah residents who favor fluoridation lies between \(58.6\%\) and \(71.4\%\). This suggests a clear majority of residents favor fluoridation.

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.

Statistical Significance
Understanding statistical significance helps us determine whether a result is likely due to chance or a specific factor such as a medical treatment, educational program, or, in this case, public opinion on water fluoridation. A result is statistically significant if the likelihood of it occurring by random chance is low. This is typically assessed using a p-value, which is compared to the significance level, commonly denoted as \(\alpha\). If the p-value is less than \(\alpha\), the result is considered statistically significant.

In the context of our exercise, statistical significance plays a part in building a confidence interval, which indirectly tells us if a majority favoring fluoridation could be significant or not. Since the sample's favor for fluoridation is captured within our confidence interval, we may infer the support is indeed significant and not random.
Sample Proportion
The sample proportion, represented as \(\hat{p}\), measures the fraction of subjects in a sample that exhibit a certain characteristic - in this example, the percentage of Utah residents in the sample who favor fluoridation. Calculating \(\hat{p}\) is straightforward: It's the number of individuals with the characteristic divided by the total sample size.

With a sample of 150 individuals and 65% favoring fluoridation, \(\hat{p}\) is 0.65. Knowing \(\hat{p}\) is crucial as it's a key component in other computations, such as determining the margin of error and constructing the confidence interval, which provide more context about the entire population's stance on an issue.
Margin of Error
The margin of error (E) represents the range above and below the sample proportion within which we can expect to find the true population proportion with a specified level of confidence. It's a crucial concept in survey analysis because it acknowledges that our sample might not perfectly reflect the population.

The margin of error increases with the level of confidence we want to have in our results. It’s affected by the sample size and variability within the data. In our exercise, we calculate E using the Z-score for the desired confidence level and the standard error. A smaller margin of error suggests a more precise estimate of the population's true proportion.
Z-score
A Z-score is a statistical measure that describes a data point's relation to the mean of a group of values, measured in terms of standard deviations from the mean. In the context of confidence intervals, the Z-score corresponds to the selected confidence level and indicates how many standard deviations an element is from the mean of the distribution.

For common confidence levels, such as 90%, 95%, or 99%, there are standard Z-scores used to calculate the margin of error. For example, a 90% confidence level corresponds to a Z-score of 1.645. This means that we are looking at the point in our normal distribution that has 5% of the data beyond it on either side, encapsulating 90% of the data within those bounds.

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

Given below are the sodium contents (in mg) for seven brands of hot dogs rated as "very good" by Consumer Reports (www.consumerreports.org): $$\begin{array}{lllllll} 420 & 470 & 350 & 360 & 270 & 550 & 530 \end{array}$$ a. Use the given data to produce a point estimate of \(\mu\), the true mean sodium content for hot dogs. b. Use the given data to produce a point estimate of \(\sigma^{2}\), the variance of sodium content for hot dogs. c. Use the given data to produce an estimate of \(\sigma\), the standard deviation of sodium content. Is the statistic you used to produce your estimate unbiased?

One thousand randomly selected adult Americans participated in a survey conducted by the Assodated Press (June 2006 ). When asked "Do you think it is sometimes justified to lie or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes okay. a. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who think lying is never justified. b. Construct a \(90 \%\) confidence interval for the proportion of adult American who think that it is often or sometimes okay to lie to avoid hurting someone's feelings. c. Comment on the apparent inconsistency in the responses given by the individuals in this sample.

Samples of two different models of cars were selected, and the actual speed for each car was determined when the speedometer registered \(50 \mathrm{mph}\). The resulting \(95 \%\) confidence intervals for mean actual speed were \((51.3,52.7)\) and \((49.4,50.6)\). Assuming that the two sample standard deviations are equal, which confidence interval is based on the larger sample size? Explain your reasoning.

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 proportion of all 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)?

"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?

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.