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Although most people are aware of minor dehydration symptoms such as dry skin and headaches, many are less knowledgeable about the causes of dehydration. According to a poll done for the Nutrition Information Center, the results of a random sample of 3003 American adults showed that \(20 \%\) did not know that caffeine dehydrates. The survey listed a margin of error of plus or minus \(1.8 \%\). a. Describe how this survey of 3003 American adults fits the properties of a binomial experiment. Specifically identify \(n,\) a trial, success, \(p,\) and \(x .\) b. What is the point estimate for the proportion of all Americans who did not know that caffeine dehydrates? Is it a parameter or a statistic? c. Calculate the \(95 \%\) confidence maximum error of estimate for a binomial experiment of 3003 trials that result in an observed proportion of \(0.20 .\) d. How is the maximum error, found in part c, related to the \(1.8 \%\) margin of error quoted in the survey report? e. Find the \(95 \%\) confidence interval for the true proportion \(p\) based on a binomial experiment of 3003 trials that results in an observed proportion of 0.20.

Short Answer

Expert verified
The random sample of American adults fits properties of a binomial experiment where \(n = 3003\), the trial is observing if an adult knew that caffeine dehydrates, and success is/means they did not. The observed proportion of success \(p = 0.20\) is a statistic as it is calculated from the sample. The maximum error estimate and confidence interval are calculated, allowing us to compare the statistic to a quoted survey margin of error to further understand the survey's accuracy.

Step by step solution

01

Identify the components of the binomial experiment

The number of trials, \(n\), is given as 3003. In context to this problem, a trial is the process of asking an American adult if they know caffeine dehydrates. A success is considered if the person did not know about it. The proportion of successes (\(p\)) is observed as 20%, converted to decimal equals 0.20. The random variable \(x\) in this case would represent the number of successes (people who did not know caffeine dehydrates).
02

Calculate the point estimate

The point estimate of the proportion (\(p\)) of all Americans who did not know that caffeine dehydrates is given directly in the problem as 0.20. It is a statistic (an estimate derived from the measure of a sample) and not a parameter (a value that describes a characteristic of the entire population).
03

Calculate the maximum error of estimate

The maximum error of estimate is found by taking the square-root of the calculated variance. Variance for a binomial experiment is calculated by formula \(np(1 - p)\). Simplify by substituting \(n = 3003\) and \(p = 0.20\). Multiply the squared result by \(1.96\) (i.e. z value for a \(95\%\) confidence), which gives us the maximum error estimate.
04

Comparing maximum error estimate to the given margin of error

Compare the maximum error estimate calculated in the previous step to the provided margin of error in the problem (1.8%). This would give an understanding of how the calculated error measure correlates with the quoted statistic in the survey.
05

Find the confidence interval

The \(95\%\) confidence interval is found by taking the point estimate and adding & subtracting the maximum error estimate: \(CI = 0.20 ± \text{(maximum error estimate)}\)

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Key Concepts

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

Confidence Interval
A fundamental concept in statistics is the confidence interval (CI). It's a range of values that's used to estimate an unknown population parameter. The CI provides a span within which we believe the true value lies with a certain level of confidence.

In the context of our exercise, when we refer to a 95% confidence interval around the proportion of American adults unaware of the dehydrating effects of caffeine, we're expressing that we're 95% confident that the actual percentage in the entire population falls within this range.

To create a confidence interval, you use the point estimate, which is the most likely value based on our sample data. In this case, that's 20% of the sample not knowing that caffeine dehydrates. Then, we consider the margin of error, which allows for uncertainty by stating how far the true value might be from our point estimate.

In practical terms, if you were to take many samples and calculate a 95% CI for each, we'd expect that approximately 95% of those intervals would contain the true population proportion.
Point Estimate
A point estimate is a single value given to estimate a population parameter. This estimate comes from your sample data and serves as the best guess for the true population value based on the information at hand.

In the context of the textbook exercise, we calculated the proportion of American adults in our sample who didn't know that caffeine dehydrates as 20%. This proportion of 0.20 is the point estimate—we use it to infer about the entire American adult population. It's important to remember that the point estimate is a statistic, which means it's subject to variability since it's derived from a sample rather than the entire population. That's why we use a point estimate as the center of a confidence interval, acknowledging that it's an educated guess rather than an exact measurement.
Margin of Error
Margin of error plays a crucial role in understanding the precision of our estimates. It's the amount by which we concede that our point estimate could be off, reflecting possible errors in our sample's representation of the entire population.

The margin of error is mainly influenced by two factors: the level of confidence we desire (often 90%, 95%, or 99%) and the standard deviation or variability within our sample. Higher confidence levels or greater variability both lead to a larger margin of error, suggesting less precision.

In our original exercise, the margin of error is given as plus or minus 1.8%. This incorporates the level of uncertainty around our point estimate—the 20% of participants unaware of caffeine's effects. The calculation of the margin of error also considers the size of the sample and the variability of the responses (the fact that not everybody gave the same answer).

Connecting this again to our exercise, by stating this margin of error alongside our point estimate, we convey the message that, while we are not 100% certain, we are statistically confident (with the noted confidence level) that the true percentage falls within this error margin of our sample-based estimate.

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Most popular questions from this chapter

It is important that the force required to extract a cork from a wine bottle not have a large standard deviation. Years of production and testing indicate that the no.9 corks in Applied Example 6.13 (p. 285 ) have an extraction force that is normally distributed with a standard deviation of 36 Newtons. Recent changes in the manufacturing process are thought to have reduced the standard deviation. a. What would be the problem with the standard deviation being relatively large? What would be the advantage of a smaller standard deviation? A sample of 20 randomly selected bottles is used for testing. Extraction Force in Newtons $$\begin{array}{llllllllll}\hline 296 & 338 & 341 & 261 & 250 & 347 & 336 & 297 & 279 & 297 \\\259 & 334 & 281 & 284 & 279 & 266 & 300 & 305 & 310 & 253 \\\\\hline\end{array}$$ b. Is the preceding sample sufficient to show that the standard deviation of extraction force is less than 36.0 Newtons, at the 0.02 level of significance? During a different testing, a sample of eight bottles is randomly selected and tested. Extraction Force in Newtons $$\begin{array}{rrrrrr}331.9 & 312.0 & 289.4 & 303.6 & 346.9 & 308.1 & 346.9 & 276.0\end{array}$$ c. Is the preceding sample sufficient to show that the standard deviation of extraction force is less than 36.0 Newtons, at the 0.02 level of significance? d. What effect did the two different sample sizes have on the calculated test statistic in parts b and c? What effect did they have on the \(p\) -value or critical value? Explain. e. What effect did the two different sample standard deviations have on the answers in parts b and c? What effect did they have on the \(p\) -value or critical value? Explain.

A commercial farmer harvests his entire field of a vegetable crop at one time. Therefore, he would like to plant a variety of green beans that mature all at one time (small standard deviation between maturity times of individual plants). A seed company has developed a new hybrid strain of green beans that it believes to be better for the commercial farmer. The maturity time of the standard variety has an average of 50 days and a standard deviation of 2.1 days. A random sample of 30 plants of the new hybrid showed a standard deviation of 1.65 days. Does this sample show a significant lowering of the standard deviation at the 0.05 level of significance? Assume that maturity time is normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

a. The central \(90 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values? b. The central \(95 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values? c. The central \(99 \%\) of the chi-square distribution with 11 degrees of freedom lies between what values?

$$ \text { Show that } \frac{\sqrt{n p q}}{n} \text { simplifies to } \sqrt{\frac{p q}{n}} \text { . } $$

A bank randomly selected 250 checking account customers and found that 110 of them also had savings accounts at the same bank. Construct a \(95 \%\) confidence interval for the true proportion of checking account customers who also have savings accounts.

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