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A poll conducted in 2013 found that \(52 \%\) of U.S. adult Twitter users get at least some news on Twitter. \(^{13}\). The standard error for this estimate was \(2.4 \%\), and a normal distribution may be used to model the sample proportion. Construct a \(99 \%\) confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter, and interpret the confidence interval in context.

Short Answer

Expert verified
The 99% confidence interval is (45.82%, 58.18%).

Step by step solution

01

Identify Given Information

From the problem, we know that the sample proportion \(\hat{p}\) is 52% or 0.52. The standard error (SE) is 2.4% or 0.024. We are asked to construct a 99% confidence interval for this proportion.
02

Find the Z-score for 99% Confidence Level

A 99% confidence level corresponds to a Z-score of approximately 2.576, found using a standard normal distribution (Z-table or calculator).
03

Calculate the Margin of Error

The margin of error (ME) can be calculated using the formula \( ME = Z \times SE \). Substitute the values: \( ME = 2.576 \times 0.024 \). Calculate the result to get the margin of error.
04

Calculate the Margin of Error Result

Calculate \( 2.576 \times 0.024 = 0.061824 \). Thus, the margin of error is approximately 0.0618 or 6.18%.
05

Construct the Confidence Interval

The confidence interval is given by \((\hat{p} - ME, \hat{p} + ME)\). Substitute the values: \( (0.52 - 0.0618, 0.52 + 0.0618) \). Calculate the interval.
06

Calculate Confidence Interval Bounds

Calculate the lower bound: \( 0.52 - 0.0618 = 0.4582 \). Calculate the upper bound: \( 0.52 + 0.0618 = 0.5818 \). Therefore, the confidence interval is \( (0.4582, 0.5818) \).
07

Interpret the Confidence Interval

We are 99% confident that the true proportion of U.S. adult Twitter users who get some news on Twitter is between 45.82% and 58.18%. This means if we were to take many samples and build a confidence interval from each one, 99% of those intervals would contain the true proportion.

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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, denoted as \( \hat{p} \), represents the fraction of the total sample that has a particular characteristic. In this context, the sample proportion is \( 0.52 \) or 52%. This means that 52% of the sampled U.S. adult Twitter users reported obtaining some news from Twitter. The sample proportion is a key component of inferential statistics because it provides an estimate of the population proportion.
  • It is calculated by dividing the number of favorable outcomes by the total number of observations in the sample.
  • This value acts as a point estimate for the true proportion in the population from which the sample was drawn.
  • It is used to compute the margin of error and construct confidence intervals for the population proportion.
Understanding the sample proportion is essential because it forms the baseline from which we assess the variability or error in our estimate through the standard error.
Standard Error
The standard error (SE) quantifies the variability or dispersion of the sample proportion from the true population proportion. It is essentially the standard deviation of the sampling distribution of the sample proportion. In our example, the standard error is given as \( 0.024 \), or 2.4%.
  • Standard error helps us understand how much the sample proportion is likely to fluctuate if we were to draw multiple samples from the same population.
  • A smaller standard error indicates that the sample proportion is a highly precise estimate of the population proportion.
  • The standard error is calculated using the formula \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \) where \( n \) is the sample size.
The standard error plays a crucial role in determining the confidence interval and is directly used to calculate the margin of error, which measures the extent to which our sample proportion may differ from the true population proportion.
Z-score
The Z-score is a statistical measurement that describes a value's position in relation to the mean of a group of values. In the context of confidence intervals, the Z-score is used to determine how many standard errors away from the sample proportion we should go to achieve a specific level of confidence. For a 99% confidence level, a Z-score of approximately 2.576 is used. This score can be found in Z-tables, which relate percentages to standard deviations in a standard normal distribution.
  • The Z-score tells us how "extreme" a value is, in standard deviation terms, from the mean.
  • Higher confidence levels require larger Z-scores, signifying extended intervals.
  • This concept helps to ensure that even with variations in sample values, we can still have confidence in the interval prediction.
Simply put, the Z-score helps determine the width of the confidence interval, with higher scores meaning wider intervals for greater confidence in results potentially encompassing the true population parameter.
Margin of Error
The margin of error (ME) provides a range that describes where the true proportion of the population is likely to lie. It quantifies the uncertainty around the sample proportion.For our example, the margin of error is computed from the formula \( ME = Z \times SE \), giving us \( ME = 2.576 \times 0.024 = 0.061824 \), which approximates to 6.18%. This tells us the amount we need to add and subtract from the sample proportion to construct the confidence interval.
  • A smaller margin of error indicates a more precise estimate of the population proportion.
  • The margin of error helps communicate the reliability and potential variability in an estimate.
  • It allows us to build confidence intervals that provide a range for the estimated population proportion.
In summary, the margin of error offers an understanding of the variability expected in sample estimates and is crucial for measuring the reliability of these estimates within certain confidence levels.

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Most popular questions from this chapter

In Exercise \(5.19,\) we learned that a Rasmussen Reports survey of \(1,000 \mathrm{US}\) adults found that \(42 \%\) believe raising the minimum wage will help the economy. Construct a \(99 \%\) confidence interval for the true proportion of US adults who believe this.

A study suggests that the \(25 \%\) of 25 year olds have gotten married. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. From a random sample of 25 year olds in census data with size 776 , you find that \(24 \%\) of them are married. A friend of yours offers to help you with setting up the hypothesis test and comes up with the following hypotheses. Indicate any errors you see. $$ \begin{array}{l} H_{0}: \hat{p}=0.24 \\ H_{A}: \hat{p} \neq 0.24 \end{array} $$

It is believed that nearsightedness affects about \(8 \%\) of all children. In a random sample of 194 children, 21 are nearsighted. Conduct a hypothesis test for the following question: do these data provide evidence that the \(8 \%\) value is inaccurate?

A food safety inspector is called upon to investigate a restaurant with a few customer reports of poor sanitation practices. The food safety inspector uses a hypothesis testing framework to evaluate whether regulations are not being met. If he decides the restaurant is in gross violation, its license to serve food will be revoked. (a) Write the hypotheses in words. (b) What is a Type 1 Error in this context? (c) What is a Type 2 Error in this context? (d) Which error is more problematic for the restaurant owner? Why? (e) Which error is more problematic for the diners? Why? (f) As a diner, would you prefer that the food safety inspector requires strong evidence or very strong evidence of health concerns before revoking a restaurant's license? Explain your reasoning.

A USA Today / Gallup poll asked a group of unemployed and underemployed Americans if they have had major problems in their relationships with their spouse or another close family member as a result of not having a job (if unemployed) or not having a full-time job (if underemployed). \(27 \%\) of the 1,145 unemployed respondents and \(25 \%\) of the 675 underemployed respondents said they had major problems in relationships as a result of their employment status. (a) What are the hypotheses for evaluating if the proportions of unemployed and underemployed people who had relationship problems were different? (b) The p-value for this hypothesis test is approximately \(0.35 .\) Explain what this means in context of the hypothesis test and the data.

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