/*! 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 15 Exercise 6.11 presents the resul... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercise 6.11 presents the results of a poll evaluating support for a generic "National Health Plan" in the US in 2019 , reporting that \(55 \%\) of Independents are supportive. If we wanted to estimate this number to within \(1 \%\) with \(90 \%\) confidence, what would be an appropriate sample size?

Short Answer

Expert verified
A sample size of 6698 is needed.

Step by step solution

01

Understand the Problem

We are tasked to find the appropriate sample size needed to estimate the proportion of Independents supporting a Health Plan within 1% margin of error at 90% confidence level. The proportion given is 55%.
02

Identify the Formula

To calculate the sample size, we use the formula for sample size estimation: \[ n = \left( \frac{Z^2 \times p \times (1-p)}{E^2} \right)\]where \( n \) is the sample size, \( Z \) is the Z-score corresponding to the confidence level, \( p \) is the estimated proportion (0.55 in this case), and \( E \) is the margin of error (0.01 for 1%).
03

Determine the Z-score

For a 90% confidence level, the Z-score commonly used is approximately 1.645. This is calculated from the standard normal distribution table.
04

Substitute Values into the Formula

Now substitute the known values into the formula:\[ n = \left( \frac{1.645^2 \times 0.55 \times (1-0.55)}{0.01^2} \right)\]which simplifies to:\[ n = \left( \frac{2.706025 \times 0.55 \times 0.45}{0.0001} \right)\]
05

Perform the Calculations

Calculate the value inside the parenthesis:\[0.55 \times 0.45 = 0.2475\]Then:\[2.706025 \times 0.2475 = 0.669740625\]Finally, divide by the margin of error squared:\[\frac{0.669740625}{0.0001} \approx 6697.41\]
06

Conclusion on Calculation

Since sample size must be a whole number, we round up our result to the nearest whole number to ensure the estimate meets the precision requirement. Thus, the sample size should be 6698.

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Key Concepts

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

Margin of Error
The margin of error is a crucial component when estimating how far off your sample results could be from the actual population parameter. This term represents the likely maximum difference between the sample statistic and the population parameter. In simpler terms, if you see results from a survey or poll, the margin of error tells you how much you can expect these results to vary if the survey were conducted multiple times under similar conditions. For example, a margin of error of 1% means that the reported percentage could be off by up to 1% in either direction. If you expect 55% support, it might be as low as 54% or as high as 56% because of this error margin. This level of precision is important when high accuracy is needed, such as in policy-making or business decisions. Margin of error is influenced by both sample size and confidence level. Smaller margins mean larger samples are needed, which leads to more precise estimates.
Confidence Level
The confidence level tells us how sure we are that our sample accurately reflects the overall population. It is expressed as a percentage, typically common levels are 90%, 95%, or 99%. Thus, a 90% confidence level means we can be 90% confident that the actual population parameter lies within our calculated interval defined by the margin of error. In practical terms, if we were to take infinite samples and calculate confidence intervals from each, we would expect 90% of these intervals to capture the true population parameter. This doesn't guarantee the true value is captured in every sample but gives a level of assurance in generality. Higher confidence levels require larger sample sizes. This is because we want broader assurance that we are capturing the true parameter, which demands a larger "net" to catch all possible variance within the population.
Z-score
The Z-score, also known as a standard score, is a statistical measurement that describes a value's position in relation to the mean or average of a group of values. When it comes to sampling and confidence intervals, the Z-score indicates how many standard deviations an element is from the mean. To connect Z-scores with the confidence level, remember they come from the standard normal distribution. At a 90% confidence level, the Z-score is approximately 1.645. This Z-score essentially tells us that to have 90% confidence, the estimated value must be within 1.645 standard deviations of the mean. The Z-score helps in adjusting the width of our confidence interval to match how confident we want to be. For a 90% confidence level, the Z-score of 1.645 provides the weights we multiply by the standard error to specify the margin around our estimate.
Proportion Estimation
Proportion estimation is the process of determining the percentage of a particular characteristic within a population based on a sample. It is a common technique in statistics used to draw inferences about a population using a smaller, manageable subset.Key to proportion estimation is the formula: \[ n = \left( \frac{Z^2 \times p \times (1-p)}{E^2} \right) \]In this formula, "\(n\)" represents the sample size needed, "\(p\)" is the estimated proportion of the population, "\(E\)" is the margin of error allowed, and "\(Z\)" represents Z-score connected to the confidence level.For instance, with an assumed proportion of 0.55 (or 55%) supporting a particular policy and wishing for this estimate to be accurate within 1% (margin of error) at a 90% confidence level, using these values in the formula provides the necessary sample size.Proportion estimation helps policymakers and researchers make predictions about population traits without surveying everyone, making it a widely used method in polls and surveys.

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Most popular questions from this chapter

About \(25 \%\) of young Americans have delayed starting a family due to the continued economic slump. Determine if the following statements are true or false, and explain your reasoning. \(^{14}\) (a) The distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump in random samples of size 12 is right skewed. (b) In order for the distribution of sample proportions of young Americans who have delayed starting a family due to the continued economic slump to be approximately normal, we need random samples where the sample size is at least 40 . (c) A random sample of 50 young Americans where \(20 \%\) have delayed starting a family due to the continued economic slump would be considered unusual. (d) A random sample of 150 young Americans where \(20 \%\) have delayed starting a family due to the continued economic slump would be considered unusual. (e) Tripling the sample size will reduce the standard error of the sample proportion by one-third.

A Kaiser Family Foundation poll for US adults in 2019 found that \(79 \%\) of Democrats, \(55 \%\) of Independents, and \(24 \%\) of Republicans supported a generic "National Health Plan". There were 347 Democrats, 298 Republicans, and 617 Independents surveyed. (a) A political pundit on TV claims that a majority of Independents support a National Health Plan. Do these data provide strong evidence to support this type of statement? (b) Would you expect a confidence interval for the proportion of Independents who oppose the public option plan to include \(0.5 ?\) Explain.

In July 2008 the US National Institutes of Health announced that it was stopping a clinical study early because of unexpected results. The study population consisted of HIV-infected women in sub-Saharan Africa who had been given single dose Nevaripine (a treatment for HIV) while giving birth, to prevent transmission of HIV to the infant. The study was a randomized comparison of continued treatment of a woman (after successful childbirth) with Nevaripine vs Lopinavir, a second drug used to treat HIV. 240 women participated in the study; 120 were randomized to each of the two treatments. Twentyfour weeks after starting the study treatment, each woman was tested to determine if the HIV infection was becoming worse (an outcome called virologic failure). Twenty-six of the 120 women treated with Nevaripine experienced virologic failure, while 10 of the 120 women treated with the other drug experienced virologic failure. \({ }^{36}\) (a) Create a two-way table presenting the results of this study. (b) State appropriate hypotheses to test for difference in virologic failure rates between treatment groups. (c) Complete the hypothesis test and state an appropriate conclusion.

Exercise 6.12 presents the results of a poll where \(48 \%\) of 331 Americans who decide to not go to college do so because they cannot afford it. (a) Calculate a \(90 \%\) confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it, and interpret the interval in context. (b) Suppose we wanted the margin of error for the \(90 \%\) confidence level to be about \(1.5 \% .\) How large of a survey would you recommend?

About \(77 \%\) of young adults think they can achieve the American dream. Determine if the following statements are true or false, and explain your reasoning. \({ }^{13}\) (a) The distribution of sample proportions of young Americans who think they can achieve the American dream in samples of size 20 is left skewed. (b) The distribution of sample proportions of young Americans who think they can achieve the American dream in random samples of size 40 is approximately normal since \(n \geq 30\). (c) A random sample of 60 young Americans where \(85 \%\) think they can achieve the American dream would be considered unusual. (d) A random sample of 120 young Americans where \(85 \%\) think they can achieve the American dream would be considered unusual.

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