/*! 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

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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.

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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.

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Most popular questions from this chapter

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.

A random sample of \(n=12\) four-year-old red pine trees was selected, and the diameter (in inches) of each tree's main stem was measured. The resulting observations are as follows: \(\begin{array}{llllllll}11.3 & 10.7 & 12.4 & 15.2 & 10.1 & 12.1 & 16.2 & 10.5\end{array}\) \(\begin{array}{llll}11.4 & 11.0 & 10.7 & 12.0\end{array}\) a. Compute a point estimate of \(\sigma\), the population standard deviation of main stem diameter. What statistic did you use to obtain your estimate? b. Making no assumptions about the shape of the population distribution of diameters, give a point estimate for the population median diameter. What statistic did you use to obtain the estimate? c. Suppose that the population distribution of diameter is symmetric but with heavier tails than the normal distribution. Give a point estimate of the population mean diameter based on a statistic that gives some protection against the presence of outliers in the sample. What statistic did you use? d. Suppose that the diameter distribution is normal. Then the 90 th percentile of the diameter distribution is \(\mu+1.28 \sigma\) (so \(90 \%\) of all trees have diameters less than this value). Compute a point estimate

The formula used to compute a large-sample confidence interval for \(p\) is $$\hat{p} \pm(z \text { critical value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)

The paper "The Curious Promiscuity of Queen Honey Bees (Apis mellifera): Evolutionary and Behavioral Mechanisms" (Annals of Zoology Fennici [2001]: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 flights, 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.

In spite of the potential safety hazards, some people would like to have an Internet connection in their car. A preliminary survey of adult Americans has estimated this proportion to be somewhere around. 30 (USA Today. May 1,2009 ). a. Use the given preliminary estimate to determine the sample size required to estimate the proportion of adult Americans who would like an Internet connection in their car to within \(.02\) with \(95 \%\) confidence. b. The formula for determining sample size given in this section corresponds to a confidence level of \(95 \%\). How would you modify this formula if a \(99 \%\) confidence level was desired? c. Use the given preliminary estimate to determine the sample size required to estimate the proportion of adult Americans who would like an Internet connection in their car to within \(.02\) with \(99 \%\) confidence.

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