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A local news outlet reported that \(56 \%\) of 600 randomly sampled Kansas residents planned to set off fireworks on July \(4^{t h} .\) Determine the margin of error for the \(56 \%\) point estimate using a \(95 \%\) confidence level. \(^{10}\)

Short Answer

Expert verified
The margin of error is approximately 3.96%.

Step by step solution

01

Identify the Sample Proportion

From the problem, we know that \(56\%\) of the 600 sampled residents plan to set off fireworks. This is our sample proportion \( \hat{p} \). Convert this percentage to a decimal by dividing by 100: \( \hat{p} = 0.56 \).
02

Calculate the Sample Size

The sample size \( n \) is given as 600. This value will be used in our calculations to determine the margin of error.
03

Find the Z-score for a 95% Confidence Level

A 95% confidence level corresponds to a Z-score of approximately 1.96. This value is commonly used and can be found in standard Z-score tables.
04

Calculate the Standard Error

The standard error (SE) is calculated using the formula for the standard error of a proportion: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \). Substitute \( \hat{p} = 0.56 \) and \( n = 600 \) into the formula: \[ SE = \sqrt{\frac{0.56 \times 0.44}{600}} \approx 0.0202. \]
05

Calculate the Margin of Error

The margin of error (MOE) is found by multiplying the Z-score by the standard error: \( MOE = Z \times SE \). With \( Z = 1.96 \) and \( SE \approx 0.0202 \), calculate: \[ MOE = 1.96 \times 0.0202 \approx 0.0396 \] or \( 3.96\% \).

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Key Concepts

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

Sample Proportion
The sample proportion is an essential concept in statistics that represents the fraction or percentage of a particular group in a sample who share a specific characteristic. In our exercise, this is the percentage of Kansas residents sampled who plan to set off fireworks, which is given as 56%.
To use this figure in calculations, we must convert it to a decimal. Simply divide 56 by 100, giving us a sample proportion denoted by \( \hat{p} = 0.56 \).
Understanding the sample proportion helps us in estimating the population parameter, which is the actual proportion of the entire population with this characteristic. It serves as a point estimate for further statistical analysis.
Z-score
The Z-score is a statistical measure that helps us understand where a specific statistic lies in the context of a normal distribution. It is used for calculating confidence intervals and is essential for determining the margin of error.
A Z-score tells us how many standard deviations away our statistic is from the mean. For a 95% confidence level, which is commonly used in statistics, the Z-score is approximately 1.96.
  • This value, 1.96, can be found in standard Z-score tables that are available in most statistics textbooks.
  • It indicates that we are capturing the middle 95% of data in a normal distribution, which helps define the range of our confidence interval.
This Z-score is multiplied by the standard error to find the margin of error, which we'll learn more about in subsequent sections.
Standard Error
The standard error is a crucial component in the calculation of the margin of error. It provides an estimate of the variability or spread within a sampling distribution.
In simpler terms, the standard error tells us how much our sample proportion is expected to fluctuate due to random sampling variability. It is calculated using the formula for the standard error of a proportion:
\[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
In our example, substituting the sample proportion \( \hat{p} = 0.56 \) and sample size \( n = 600 \) into the formula gives us:
\[ SE = \sqrt{\frac{0.56 \times 0.44}{600}} \approx 0.0202 \]
  • The standard error is used to gauge the reliability of the sample proportion.
  • A smaller standard error suggests the sample proportion is a more accurate reflection of the population proportion.
Confidence Interval
A confidence interval provides a range of values, which is believed to contain the population parameter with a certain level of confidence. In our exercise, we want to estimate the true proportion of the population that plans to set off fireworks based on our sample data.
To construct a 95% confidence interval, you start with the sample proportion as the point estimate and add and subtract the margin of error. In mathematical terms, it is expressed as:
\[ CI = \hat{p} \pm MOE \]
In our case, with a sample proportion of 0.56 and a margin of error of approximately 0.0396, the confidence interval is:
\[ 0.56 \pm 0.0396 \]
  • This means that we can say with 95% confidence that the true proportion of Kansas residents planning to set off fireworks lies between 52.04% and 59.96%.
  • Confidence intervals are a key part of inferential statistics, helping us make informed guesses about population parameters.
They reflect the uncertainty or precision of an estimate and are influenced by the chosen confidence level and the variability of data.

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Most popular questions from this chapter

Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) The chi-square distribution, just like the normal distribution, has two parameters, mean and standard deviation. (b) The chi-square distribution is always right skewed, regardless of the value of the degrees of freedom parameter. (c) The chi-square statistic is always positive. (d) As the degrees of freedom increases, the shape of the chi-square distribution becomes more skewed.

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. \(^{50}\) (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. \(^{51}\) 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.

A national survey conducted among a simple random sample of 1,507 adults shows that \(56 \%\) of Americans think the Civil War is still relevant to American politics and political life. \({ }^{47}\) (a) Conduct a hypothesis test to determine if these data provide strong evidence that the majority of the Americans think the Civil War is still relevant. (b) Interpret the p-value in this context. (c) Calculate a \(90 \%\) confidence interval for the proportion of Americans who think the Civil War is still relevant. Interpret the interval in this context, and comment on whether or not the confidence interval agrees with the conclusion of the hypothesis test.

A survey on 1,509 high school seniors who took the SAT and who completed an optional web survey shows that \(55 \%\) of high school seniors are fairly certain that they will participate in a study abroad program in college. \({ }^{12}\) (a) Is this sample a representative sample from the population of all high school seniors in the US? Explain your reasoning. (b) Let's suppose the conditions for inference are met. Even if your answer to part (a) indicated that this approach would not be reliable, this analysis may still be interesting to carry out (though not report). Construct a \(90 \%\) confidence interval for the proportion of high school seniors (of those who took the SAT) who are fairly certain they will participate in a study abroad program in college, and interpret this interval in context. (c) What does "90\% confidence" mean? (d) Based on this interval, would it be appropriate to claim that the majority of high school seniors are fairly certain that they will participate in a study abroad program in college?

A professor using an open source introductory statistics book predicts that \(60 \%\) of the students will purchase a hard copy of the book, \(25 \%\) will print it out from the web, and \(15 \%\) will read it online. At the end of the semester he asks his students to complete a survey where they indicate what format of the book they used. Of the 126 students, 71 said they bought a hard copy of the book, 30 said they printed it out from the web, and 25 said they read it online. (a) State the hypotheses for testing if the professor's predictions were inaccurate. (b) How many students did the professor expect to buy the book, print the book, and read the book exclusively online? (c) This is an appropriate setting for a chi-square test. List the conditions required for a test and verify they are satisfied. (d) Calculate the chi-squared statistic, the degrees of freedom associated with it, and the p-value. (e) Based on the p-value calculated in part (d), what is the conclusion of the hypothesis test? Interpret your conclusion in this context.

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