/*! 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 107 From Formula \(7.2\), an estimat... [FREE SOLUTION] | 91Ó°ÊÓ

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From Formula \(7.2\), an estimate for margin of error for a \(95 \%\) confidence interval is \(m=2 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) where \(\mathrm{n}\) is the required sample size and \(\hat{p}\) is the sample proportion. Since we do not know a value for \(\hat{p}\), we use a conservative estimate of \(0.50\) for \(\hat{p}\). Replace \(\hat{p}\) with \(0.50\) in the formula and simplify.

Short Answer

Expert verified
The simplified formula for margin of error is \(m= \sqrt{\frac{1}{n}}\).

Step by step solution

01

Substitution

Replace the symbol \(\hat{p}\) in the formula with the conservative estimate of 0.50: \(m=2 \sqrt{\frac{(0.50)(1-0.50)}{n}}\)
02

Simplification

Simplify the fraction inside the square root: \(m=2 \sqrt{\frac{(0.50)(0.50)}{n}} = 2 \sqrt{\frac{0.25}{n}}\
03

Further Simplification

Factor out the square of 0.5 from under the square root sign to simplify further: \(m=2 \cdot 0.5 \sqrt{\frac{1}{n}} = \sqrt{\frac{1}{n}}\)
04

Conclusion

The simplified formula for margin of error is \(m= \sqrt{\frac{1}{n}}\)

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Key Concepts

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

Confidence Interval
Understanding confidence intervals is crucial when working with statistics, especially in estimating population parameters. A confidence interval provides a range of values, typically centered around a sample statistic, that is believed to contain the true population parameter. Imagine you're trying to measure the average height of trees in a large forest. You can't measure them all, but you take a sample and find the average height within this sample. The confidence interval gives you a range that you're 95% confident includes the true average height of all trees in the forest.

When we talk about a 95% confidence interval, we imply that if we were to take 100 different samples and compute a confidence interval for each sample, approximately 95 out of those 100 intervals would contain the true population parameter. It's a way of expressing certainty—or rather uncertainty—about our estimates. In the original exercise, the 95% confidence interval's margin of error is calculated using a formula that takes into account the sample proportion and size. By using these calculations, researchers can state with a certain degree of confidence that the true value lies within the specified interval.
Sample Size
Sample size, denoted as 'n' in statistical formulas, is essentially the number of observations or replicates included in a statistical sample. It's one of the foundational elements in any statistical analysis because it directly influences the precision of an estimate or the power of a hypothesis test.

For instance, let's say you're conducting a survey to understand the percentage of students who prefer online classes. If you survey only a handful of students, the generalizability of your findings will be questionable. Conversely, surveying a large proportion of the student population would give you more confidence that your findings accurately reflect the overall preference.

In the context of our exercise, the sample size affects the margin of error of our confidence interval: a larger sample size generally yields a smaller margin of error, meaning our estimate is more precise. Therefore, determining the right sample size is a balancing act between the desired precision and the resources available for the study.
Sample Proportion
Sample proportion, commonly represented by \(\hat{p}\), is a statistic that estimates the proportion of the population that exhibits a certain characteristic based on a sample drawn from that population. Essentially, it's the equivalent of a percentage for a specific attribute within your sample.

Let's imagine you're studying voting behaviors and want to estimate the proportion of people favoring a particular candidate. By polling a random group of voters, you calculate the sample proportion of voters supporting that candidate. If 60 out of 100 polled voters prefer your candidate, the sample proportion \(\hat{p}\) would be 0.60 or 60%.

In our original exercise, a conservative estimate of 0.50 is used for \(\hat{p}\) when the actual value is unknown. This is done because 0.50 maximizes the product \(\hat{p}(1-\hat{p})\), leading to the largest possible margin of error for a given sample size. This conservative approach ensures that the calculated confidence interval is wide enough to be likely to capture the true population parameter, regardless of the actual sample proportion.

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Most popular questions from this chapter

Assume your class has 30 students and you want a random sample of 10 of them. Describe how to randomly select 10 people from your class using the random number table.

Suppose it is known that \(20 \%\) of students at a certain college participate in a textbook recycling program each semester. a. If a random sample of 50 students is selected, do we expect that exactly \(20 \%\) of the sample participates in the textbook recycling program? Why or why not? b. Suppose we take a sample of 500 students and find the sample proportion participating in the recycling program. Which sample proportion do you think is more likely to be closer to \(20 \%\) : the proportion from a sample size of 50 or the proportion from a sample size of \(500 ?\) Explain your reasoning.

In 2003 and 2017 Gallup asked Democratic voters about their views on the FBI. In \(2003,44 \%\) thought the \(\mathrm{FBI}\) did a good or excellent job. In \(2017,69 \%\) of Democratic voters felt this way. Assume these percentages are based on samples of 1200 Democratic voters. a. Can we conclude, on the basis of these two percentages alone, that the proportion of Democratic voters who think the FBI is doing a good or excellent job has increase from 2003 to \(2017 ?\) Why or why not? b. Check that the conditions for using a two-proportion confidence interval hold. You can assume that the sample is a random sample. c. Construct a \(95 \%\) confidence interval for the difference in the proportions of Democratic voters who believe the FBI is doing a good or excellent job, \(p_{1}-p_{2}\). Let \(p_{1}\) be the proportion of Democratic voters who felt this way in 2003 and \(p_{2}\) be the proportion of Democratic voters who felt this way in 2017 . d. Interpret the interval you constructed in part c. Has the proportion of Democratic voters who feel this way increased? Explain.

A random sample of likely voters showed that \(49 \%\) planned to support Measure \(X\). The margin of error is 3 percentage points with a \(95 \%\) confidence level. a. Using a carefully worded sentence, report the \(95 \%\) confidence interva for the percentage of voters who plan to support Measure \(\mathrm{X}\). b. Is there evidence that Measure \(\mathrm{X}\) will fail? c. Suppose the survey was taken on the streets of Miami and the measure was a Florida statewide measure. Explain how that would affect your conclusion.

The Gallup poll reported that \(45 \%\) of Americans have tried marijuana. This was based on a survey of 1021 Americans and had a margin of error of plus or minus 5 percentage points with a \(95 \%\) level of confidence. a. State the survey results in confidence interval form and interpret the interval. b. If the Gallup Poll was to conduct 100 such surveys of 1021 Americans, how many of them would result in confidence intervals that did not include the true population proportion? c. Suppose a student wrote this interpretation of the interval: "We are \(95 \%\) confident that the percentage of Americans who have tried marijuana is between \(40 \%\) and \(50 \%\)." What, if anything, is incorrect in this interpretation?

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