Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

At a time in U.S. history when there appears to be genuine concern about the number of illegal aliens living in the United States, there also appears to be concern over the number of legal immigrants allowed to move to the United States. In a recent poll that included questions about both legal and illegal immigrants to the United States, \(51 \%\) of the \(n=900\) registered voters interviewed indicated that the U.S. should decrease the number of legal immigrants entering the United States. \(^{3}\) a. What is a point estimate for the proportion of U.S. registered voters who feel that the United States should decrease the number of legal immigrants entering the United States? Calculate the margin of error. b. The poll reports a margin of error of \(\pm 3 \%\). How was the reported margin of error calculated so that it can be applied to all of the questions in the survey?

Short Answer

Expert verified
Based on the given data, the point estimate for the proportion of U.S. registered voters who feel that the United States should decrease the number of legal immigrants entering the country is 0.51 or 51%. The margin of error for this estimate is approximately 3%. The reported margin of error was calculated using the formula: Margin of Error = Z* × √(p(1-p)/n), assuming a point estimate of 50% to account for the maximum possible margin of error, and considering the chosen confidence level and sample size. This ensures that the reported margin of error is suitable for all questions in the survey.

Step by step solution

01

Calculate the Point Estimate

The point estimate is simply the proportion of the sample who responded in a particular way. Since we have 900 registered voters, and 51% of them expressed that the U.S. should decrease the number of legal immigrants, we have a point estimate of: Point estimate = \(0.51\)
02

Calculate the Margin of Error

To calculate the margin of error, we will use the following formula: Margin of error = \(Z* \times \sqrt{\frac{p(1-p)}{n}}\) where: - \(Z*\) is the z-score for a certain confidence level (typically, 1.96 for a 95% confidence interval) - \(p\) is the point estimate - \(n\) is the sample size For this exercise: - \(Z* = 1.96\) - \(p = 0.51\) - \(n = 900\) Margin of error = \(1.96 \times \sqrt{\frac{0.51(1-0.51)}{900}} =0.0302\) The margin of error is approximately \(3 \%\).
03

Explaining the Reported Margin of Error Calculation

The reported margin of error is \(\pm 3 \%\). This margin of error was calculated using the same formula mentioned in Step 2: Margin of Error = \(Z* \times \sqrt{\frac{p(1-p)}{n}}\) For a given sample size \(n\), the margin of error will be maximum when the point estimate is closest to \(0.5\). Since the poll designers wanted to have a margin of error that could be applied to all questions in the survey, they calculated the survey's margin of error at a point estimate of \(50 \%\), which corresponds to a \(3 \%\) margin of error for their chosen confidence level and sample size. This ensures that the reported margin of error will be suitable for all questions in the survey, taking into account that it might not precisely match the actual margin of error for each question.

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 Vaia!

Key Concepts

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

Margin of Error
Understanding the margin of error is crucial in interpreting survey results. It represents the range within which we can expect the true population parameter to lie, with a specific level of confidence. The margin of error is typically expressed as a percentage and indicates how much the survey results could vary. To calculate the margin of error, we use the formula:- Margin of error = \( Z^* \times \sqrt{\frac{p(1-p)}{n}} \)Here:
  • \(Z^*\) is the z-score that corresponds to the desired confidence level, commonly 1.96 for 95% confidence.
  • \(p\) is the point estimate or the sample proportion.
  • \(n\) is the sample size.
In our example, where 51% of 900 registered voters want to reduce legal immigrants, the margin is about 3%. This means the actual proportion of all voters supporting the reduction may be as low as 48% or as high as 54%, given the 95% confidence level. Calculating the margin of error helps us gauge the reliability of survey results.
Survey Sample Size
Survey sample size plays a pivotal role in the accuracy of the results. The sample size determines the scope of the survey and greatly influences the margin of error. In general, a larger sample size will reduce the margin of error because it provides more data points, making the estimates closer to the true population values. When planning a survey, it's essential to determine the appropriate sample size to achieve a desired margin of error and confidence level. The formula for determining how many responses are needed involves careful consideration of the variability in the population and the level of detail required. In the exercise, 900 voters were surveyed. This size was likely chosen to ensure the margin of error remained reasonable (about 3%) while maintaining a 95% confidence interval. A well-sized survey sample ensures that a survey can adequately represent the broader population, giving more credibility to its findings. A larger sample will lead to small changes in the estimate, hence increasing accuracy, but can be more costly and time-consuming.
Confidence Interval
A confidence interval provides a range of values that is likely to contain the true population parameter, such as a proportion or mean. It gives not just a single estimate but a range within which the population parameter is expected to fall, considering the margin of error and sample data. The confidence interval is calculated by adding and subtracting the margin of error from the point estimate: - Confidence Interval = Point Estimate ± Margin of Error For the voters' survey, the point estimate was 51%, and the margin of error was 3%. Therefore, the confidence interval would be from 48% to 54% (51% ± 3%).
  • The width of the confidence interval is influenced by:
  • The sample size: Larger samples yield narrower intervals.
  • The variability in the data: More variability leads to wider intervals.
Having a confidence interval helps to understand the range in which the true measure lies, rather than providing a single cryptic number. It reflects how much uncertainty there is in the survey results and helps us understand the precision of the point estimate, offering a more nuanced view of the data.

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

A random sampling of a company's monthly operating expenses for \(n=36\) months produced a sample mean of \(\$ 5474\) and a standard deviation of \(\$ 764\). Find a \(90 \%\) upper confidence bound for the company's mean monthly expenses.

A random sample of \(n=900\) observations from a binomial population produced \(x=655\) successes. Estimate the binomial proportion \(p\) and calculate the margin of error.

Independent random samples of \(n_{1}=800\) and \(n_{2}=640\) observations were selected from binomial populations 1 and \(2,\) and \(x_{1}=337\) and \(x_{2}=374\) successes were observed. a. Find a \(90 \%\) confidence interval for the difference \(\left(p_{1}-p_{2}\right)\) in the two population proportions. Interpret the interval. b. What assumptions must you make for the confidence interval to be valid? Are these assumptions met?

Find a \(99 \%\) lower confidence bound for the binomial proportion \(p\) when a random sample of \(n=400\) trials produced \(x=196\) successes.

Refer to Exercise \(8.43 .\) In addition to tests involving biology concepts, students were also tested on process skills. The results of pretest and posttest scores, published in The American Biology Teacher, are given below. \({ }^{11}\) $$\begin{array}{lccc} & & \text { Sample } & \text { Standard } \\\& \text { Mean } & \text { Size } & \text { Deviation } \\\\\hline \text { Pretest: All BACC Classes } & 10.52 & 395 & 4.79 \\\\\text { Pretest: All Traditional } & 11.97 & 379 & 5.39 \\\\\text { Posttest: All BACC Classes } & 14.06 & 376 & 5.65 \\\\\text { Posstest: All Traditional } & 12.96 & 308 & 5.93\end{array}$$ a. Find a \(95 \%\) confidence interval for the mean score on process skills for the posttest for all BACC classes. b. Find a \(95 \%\) confidence interval for the mean score on process skills for the posttest for all traditional classes. c. Find a \(95 \%\) confidence interval for the difference in mean scores on process skills for the posttest BACC classes and the posttest traditional classes. d. Does the confidence interval in c provide evidence that there is a real difference in the mean process skills scores between posttest BACC and traditional class scores? Explain.

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.

Sign-up for free