/*! 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
The required sample size is 669.

Step by step solution

01

Understanding the Problem

We want to estimate the proportion of Independents supporting the health plan to within 1% margin of error at 90% confidence level. The reported proportion is 55% or 0.55.
02

Using the Confidence Interval Formula

The confidence interval for a proportion is calculated using the formula: \[ \hat{p} \pm z \cdot \sqrt{ \frac{ \hat{p}(1-\hat{p})}{n} } \]Where \(\hat{p}\) is the sample proportion, \(z\) is the z-score for the confidence level, and \(n\) is the sample size.
03

Finding the Z-Score for 90% Confidence

For a 90% confidence level, the z-score is approximately 1.645. This is the value we need to use in our sample size calculation.
04

Setting Up the Margin of Error Formula

We want a margin of error (ME) of 1%, or 0.01: \[ ME = z \cdot \sqrt{ \frac{ \hat{p}(1-\hat{p})}{n} } \] Rearranging to find \(n\): \[ n = \left( \frac{z}{ME} \right)^2 \cdot \hat{p}(1-\hat{p}) \]
05

Solving for Sample Size

Plug \(z = 1.645\), \(\hat{p} = 0.55\), and \(ME = 0.01\) into the formula: \[ n = \left( \frac{1.645}{0.01} \right)^2 \cdot 0.55(1-0.55) \]Calculate \(n\).
06

Calculating the Sample Size

First, find \(\left( \frac{1.645}{0.01} \right)^2\) which is 27006.25. Then calculate the rest of the formula: \[ n = 27006.25 \cdot 0.55 \cdot 0.45 \approx 668.35 \]Since the sample size must be whole, round up to 669.

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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 (ME) is a key concept in statistics, especially when dealing with surveys or polls. It defines the range within which we can expect our true population parameter to lie. In simpler terms, it tells us how "off" our sample estimate might be from the actual population value.
Margin of error is influenced by how much uncertainty we're willing to accept in our results. Smaller margins of error mean more precise estimates, but they require larger sample sizes to achieve. Conversely, allowing for a larger margin of error might reduce the burden of sample sizes, but at the cost of precision.
In practice, if we specify a margin of error of 1%, we are saying that we are comfortable with our sample proportion being within plus or minus 1% of the true population proportion. This is crucial in planning surveys because it helps in determining how many people we need to poll to ensure our results are meaningful.
Confidence Interval
Confidence intervals provide a range of values within which we expect the true population parameter, like a proportion or mean, to lie. This range is established based on the sample data and how confident we want to be in our estimate's accuracy.
For instance, a 90% confidence interval means that if we were to take 100 different samples and compute a confidence interval for each one, we would expect 90 of these intervals to contain the true population parameter.
The width of the confidence interval is affected by several factors, including the margin of error, the sample size, and the variability of the data. Why is this important? Because it gives us a cushion of certainty around our estimate, allowing us to make informed decisions fueled by data.
Z-Score
A z-score is a statistical measurement that describes a value's relation to the mean of a group of values, expressed in terms of standard deviations. In the context of confidence intervals, the z-score determines how wide or narrow the interval will be for a given confidence level.
For a typical bell-shaped curve, z-scores help categorize how likely events are under a normal distribution with a specific average. In the example provided, a 90% confidence level corresponds to a z-score of 1.645. This value is chosen from statistical tables that tell us, for a normal distribution, what proportion of data falls within that range.
In practical terms, a higher confidence level (like 95% or 99%) would require a higher z-score, thus widening the confidence interval and necessitating a larger sample size to maintain the same level of precision.
Proportion Estimation
Estimating proportions involves determining what fraction of the population exhibits a particular characteristic, usually based on sample data. In our poll example, the goal is to estimate what proportion of Independents support a National Health Plan.
To accurately estimate a proportion, we use sample data to infer what the entire population might think. This estimation requires us to carefully consider the size of the sample and the margin of error we're willing to accept.
In practice, once we have the sample proportion, we calculate its confidence interval using the formula involving the margin of error, z-score, and sample size. Understanding proportion estimation is vital because it helps paint a picture of a larger population based on a manageable and practical subset.

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Most popular questions from this chapter

Exercise 6.22 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is \(8.0 \%,\) while this proportion is \(8.8 \%\) for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. (a) Conduct a hypothesis test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states. (Reminder: Check conditions) (b) It is possible the conclusion of the test in part (a) is incorrect. If this is the case, what type of error was made?

The General Social Survey asked a random sample of 1,390 Americans the following question: "On the whole, do you think it should or should not be the government's responsibility to promote equality between men and women?" \(82 \%\) of the respondents said it "should be". At a \(95 \%\) confidence level, this sample has \(2 \%\) margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning. (a) We are \(95 \%\) confident that between \(80 \%\) and \(84 \%\) of Americans in this sample think it's the government's responsibility to promote equality between men and women. (b) We are \(95 \%\) confident that between \(80 \%\) and \(84 \%\) of all Americans think it's the government's responsibility to promote equality between men and women. (c) If we considered many random samples of 1,390 Americans, and we calculated \(95 \%\) confidence intervals for each, \(95 \%\) of these intervals would include the true population proportion of Americans who think it's the government's responsibility to promote equality between men and women. (d) In order to decrease the margin of error to \(1 \%\), we would need to quadruple (multiply by 4 ) the sample size. (e) Based on this confidence interval, there is sufficient evidence to conclude that a majority of Americans think it's the government's responsibility to promote equality between men and women.

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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. \({ }^{7}\) (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.

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