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In a simple random sample of 1,500 voters, \(55 \%\) said they planned to vote for a particular proposition, and \(45 \%\) said they planned to vote against it. The estimated margin of victory for the proposition is thus \(10 \% .\) What is the standard error of this estimated margin? What is an approximate \(95 \%\) confidence interval for the margin?

Short Answer

Expert verified
Standard error is 1.29%. 95% CI is approximately (52.47%, 57.53%).

Step by step solution

01

Identify the Problem

We need to find the standard error and a 95% confidence interval for a population proportion where 55% support a proposition and 45% do not.
02

Calculate the Sample Proportion

Out of 1500 voters, 55% are in favor and 45% are against the proposition. Therefore, the sample proportion in favor is \( \hat{p} = 0.55 \) and against is \( \hat{q} = 0.45 \).
03

Determine the Standard Error Formula

The standard error (SE) for a proportion is calculated using the formula: \[ SE = \sqrt{\frac{\hat{p} \cdot \hat{q}}{n}} \] where \( \hat{p} \) and \( \hat{q} \) are the sample proportions, and \( n \) is the sample size.
04

Compute the Standard Error

Substitute the values into the formula: \( \hat{p} = 0.55 \), \( \hat{q} = 0.45 \), and \( n = 1500 \). Thus, the calculation is: \[ SE = \sqrt{\frac{0.55 \cdot 0.45}{1500}} \approx 0.0129 \]
05

Determine the Confidence Interval Formula

The 95% confidence interval for the proportion can be given by \( \hat{p} \pm z \times SE \), where \( z \) is the z-score for a 95% confidence level, approximately 1.96.
06

Calculate the Confidence Interval

Now, compute the confidence interval: \[ 0.55 \pm 1.96 \times 0.0129 \]. This results in: \[ (0.55 - 0.02528, 0.55 + 0.02528) \] which is approximately (0.5247, 0.5753).
07

Interpret the Results

The standard error of the estimated margin is approximately 1.29%. The 95% confidence interval for the true margin of support is approximately between 52.47% and 57.53%.

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Key Concepts

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

Standard Error
When dealing with sample data, one important concept is the standard error (SE). It measures the variability or spread of a sample statistic, like a proportion, relative to the entire population. This variability arises because different samples from the same population can give different results.

To calculate the standard error for a population proportion, we use the following formula:
  • SE = \( \sqrt{\frac{\hat{p} \cdot \hat{q}}{n}} \)
where:
  • \( \hat{p} \) is the sample proportion of success (e.g., support for a proposition).
  • \( \hat{q} = 1 - \hat{p} \), the proportion of failure (e.g., opposition).
  • \( n \) is the sample size, or the number of participants in the survey.
Standard error is crucial because it provides insight into how much variation we can expect in our sample estimates across different samples. The smaller the SE, the more reliable the sample statistic is as an estimate of the population parameter.
Population Proportion
Understanding population proportion is a fundamental part of statistics, especially in survey analysis. Population proportion refers to the fraction or percentage of the population that exhibits a certain characteristic. In the context of an election or opinion poll, it can represent the proportion of people supporting a proposition.

For a given sample, this proportion (known as the sample proportion) is used as an estimate of the true population proportion. Let's say we have a simple random sample of 1500 voters, and 55% indicate they will vote 'yes'. So, the sample proportion (\( \hat{p} \)) is 0.55.

Sample proportions can be used to construct confidence intervals, which give a range that likely contains the true population proportion. Knowing the sample proportion allows statisticians to make inferences about the entire population with a particular level of confidence.
Simple Random Sample
A simple random sample (SRS) is a method of selecting individuals from a population where each member has an equal chance of being chosen. This randomness ensures that the sample represents the population without bias.

Having a simple random sample is critical because it minimizes systematic errors and makes the findings generalizable to the broader population. When we're estimating a population proportion, like voter support for a proposition, using a simple random sample increases the reliability of our conclusions.

In practice, simple random sampling could be done by techniques such as using a random number generator to select individuals or drawing names from a hat. The goal is to achieve an unbiased representation of the population, helping to ensure that the estimates derived from the sample are accurate reflections of the entire population's characteristics.

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Most popular questions from this chapter

A sample of size \(n=100\) is taken from a population that has a proportion \(p=1 / 5\) a. Find \(\delta\) such that \(P(|\hat{p}-p| \geq \delta)=0.025\) b. If, in the sample, \(\hat{p}=0.25,\) will the \(95 \%\) confidence interval for \(p\) contain the true value of \(p ?\)

True or false? a. The center of a \(95 \%\) confidence interval for the population mean is a random variable. b. A \(95 \%\) confidence interval for \(\mu\) contains the sample mean with probability .95 c. A \(95 \%\) confidence interval contains \(95 \%\) of the population. d. Out of one hundred \(95 \%\) confidence intervals for \(\mu, 95\) will contain \(\mu\).

Suppose that the population size \(N\) is not known, but it is known that \(n \leq N\) Show that the following procedure will generate a simple random sample of size \(n .\) Imagine that the population is arranged in a long list that you can read sequentially. a. Let the sample initially consist of the the first \(n\) elements in the list. b. For \(k=1,2, \ldots,\) as long as the end of the list has not been encountered: i. Read the \((n+k)\) -th element in the list. ii. Place it in the sample with probability \(n /(n+k)\) and, if it is placed in the sample, randomly drop one of the existing sample members.

The value of a population mean increases linearly through time: \(\mu(t)=\alpha+\beta t\) while the variance remains constant. Independent simple random samples of size \(n\) are taken at times \(t=1,2,\) and 3 a. Find conditions on \(w_{1}, w_{2},\) and \(w_{3}\) such that $$\hat{\beta}=w_{1} \bar{X}_{1}+w_{2} \bar{X}_{2}+w_{3} \bar{X}_{3}$$ is an unbiased estimate of the rate of change, \(\beta .\) Here \(\bar{X}_{i}\) denotes the sample mean at time \(t_{i}\) b. What values of the \(w_{i}\) minimize the variance subject to the constraint that the estimate is unbiased?

Consider a population of size four, the members of which have values \(x_{1}, x_{2}, x_{3}, x_{4}\) a. If simple random sampling were used, how many samples of size two are there? b. Suppose that rather than simple random sampling, the following sampling scheme is used. The possible samples of size two are $$\left\\{x_{1}, x_{2}\right\\},\left\\{x_{2}, x_{3}\right\\},\left\\{x_{3}, x_{4}\right\\},\left\\{x_{1}, x_{4}\right\\}$$ and the sampling is done in such a way that each of these four possible samples is equally likely. Is the sample mean unbiased?

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