/*! 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

Suppose that \(8 \%\) of college students are vegetarians. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of the sample proportions of vegetarians in random samples of size 60 is approximately normal since \(n \geq 30\). (b) The distribution of the sample proportions of vegetarian college students in random samples of size 50 is right skewed. (c) A random sample of 125 college students where \(12 \%\) are vegetarians would be considered unusual. (d) A random sample of 250 college students where \(12 \%\) are vegetarians would be considered unusual. (e) The standard error would be reduced by one-half if we increased the sample size from 125 to 250 .

The General Social Survey asked 1,578 US residents: "Do you think the use of marijuana should be made legal, or not?" \(61 \%\) of the respondents said it should be made legal. \(^{20}\) (a) Is \(61 \%\) a sample statistic or a population parameter? Explain. (b) Construct a \(95 \%\) confidence interval for the proportion of US residents who think marijuana should be made legal, and interpret it in the context of the data. (c) A critic points out that this \(95 \%\) confidence interval is only accurate if the statistic follows a normal distribution, or if the normal model is a good approximation. Is this true for these data? Explain. (d) A news piece on this survey's findings states, "Majority of Americans think marijuana should be legalized." Based on your confidence interval, is this news piece's statement justified?

A local news outlet reported that \(56 \%\) of 600 randomly sampled Kansas residents planned to set off fireworks on July \(4^{\text {th }}\). Determine the margin of error for the \(56 \%\) point estimate using a \(95 \%\) confidence level. \({ }^{17}\)

Exercise 6.11 presents the results of a poll evaluating support for a generically branded "National Health Plan" in the United States. \(79 \%\) of 347 Democrats and \(55 \%\) of 617 Independents support a National Health Plan. (a) Calculate a \(95 \%\) confidence interval for the difference between the proportion of Democrats and Independents who support a National Health Plan \(\left(p_{D}-p_{I}\right),\) and interpret it in this context. We have already checked conditions for you. (b) True or false: If we had picked a random Democrat and a random Independent at the time of this poll, it is more likely that the Democrat would support the National Health Plan than the Independent.

A survey of 2,254 American adults indicates that \(17 \%\) of cell phone owners browse the internet exclusively on their phone rather than a computer or other device. \(^{59}\) (a) According to an online article, a report from a mobile research company indicates that 38 percent of Chinese mobile web users only access the internet through their cell phones. \(^{60}\) Conduct a hypothesis test to determine if these data provide strong evidence that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of \(38 \%\). (b) Interpret the p-value in this context. (c) Calculate a \(95 \%\) confidence interval for the proportion of Americans who access the internet on their cell phones, and interpret the interval in this context.

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