/*! 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

While the majority of people who are color blind are male, the National Eye Institute reports that \(0.5 \%\) of women of with Northern European ancestry have the common form of red-green color blindness. Suppose a random sample of 100 women with Northern European ancestry is selected. Can we find the probability that less than \(0.3 \%\) of the sample is color blind? If so, find the probability. If not, explain why this probability cannot be calculated.

The Perry Preschool Project was created in the early 1960 s by David Weikart in Ypsilanti, Michigan. In this project, 123 African American children were randomly assigned to one of two groups: One group enrolled in the Perry Preschool, and one group did not enroll. Follow-up studies were done for decades. One research question was whether attendance at preschool had an effect on high school graduation. The table shows whether the students graduated from regular high school or not and includes girls only (Schweinhart et al. 2005). $$\begin{array}{lcc}\hline & \text { Preschool } & \text { No Preschool } \\\\\hline \text { HS Grad } & 21 & 8 \\ \text { No HS Grad } & 4 & 17\end{array}$$ a. Find the percentages that graduated for both groups, and compare them descriptively. Does this suggest that preschool was associated with a higher graduation rate? b. Which of the conditions fail so that we cannot use a confidence interval for the difference between proportions?

Suppose you want to estimate the mean grade point average (GPA) of all students at your school. You set up a table in the library asking for volunteers to tell you their GPAs. Do you think you would get a representative sample? Why or why not?

A 2017 Gallup poll reported that 658 out of 1028 U.S. adults believe that marijuana should be legalized. When Gallup first polled U.S. adults about this subject in 1969 , only \(12 \%\) supported legalization. Assume the conditions for using the CLT are met. a. Find and interpret a \(99 \%\) confidence interval for the proportion of U.S. adults in 2017 that believe marijuana should be legalized. b. Find and interpret a \(95 \%\) confidence interval for this population parameter. c. Find the margin of error for each of the confidence intervals found in parts a and b. d. Without computing it, how would the margin of error of a \(90 \%\) confidence interval compare with the margin of error for the \(95 \%\) and \(99 \%\) intervals? Construct the \(90 \%\) confidence interval to see if your prediction was correct.

According to a 2017 survey conducted by Netflix, \(46 \%\) of couples have admitted to "cheating" on their significant other by streaming a TV show ahead of their partner. Suppose a random sample of 80 Netflix subscribers is selected. a. What percentage of the sample would we expect have "cheated" on their partner? b. Verify that the conditions for the Central Limit Theorem are met. c. What is the standard error for this sample proportion? d. Complete the sentence: We expect ____ \(\%\) of streaming couples to admit to Netflix "cheating," give or take _____ \(\% .\)

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