/*! 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 25 A national television network ta... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A national television network takes an exit poll of 1400 voters after each has cast a vote in a state gubernatorial election. Of them, 660 say they voted for the Democratic candidate and 740 say they voted for the Republican candidate. a. Treating the sample as a random sample from the population of all voters, would you predict the winner? Base your decision on a \(95 \%\) confidence interval. b. Base your decision on a \(99 \%\) confidence interval. Explain why you need stronger evidence to make a prediction when you want greater confidence.

Short Answer

Expert verified
At 95% confidence, the Democratic candidate is less likely to win; at 99%, the prediction is uncertain. Greater confidence requires broader evidence.

Step by step solution

01

Define the Problem

We need to determine whether we can predict a winner based on the exit poll data using confidence intervals. Specifically, we need to construct two confidence intervals (CI) for the proportion of voters who voted for the Democratic candidate: one at a 95% confidence level and another at a 99% confidence level.
02

Calculate Sample Proportion

The sample proportion (9) of voters who voted for the Democratic candidate is calculated using the formula: \[ p = \frac{\text{number of voters for Democratic candidate}}{\text{total number of voters}} = \frac{660}{1400} \approx 0.4714 \]
03

Calculate Standard Error

The standard error (SE) of the sample proportion is calculated using the formula: \[ SE = \sqrt{\frac{\u00bfp(1-\u00bfp)}{n}} = \sqrt{\frac{0.4714 \times 0.5286}{1400}} \approx 0.0133 \]
04

Calculate 95% Confidence Interval

To calculate the 95% CI, we use the z-score for 95% confidence, which is approximately 1.96. The CI is given by: \[ \u00bfp \pm (z \times SE) = 0.4714 \pm (1.96 \times 0.0133) = (0.4453, 0.4975) \] This interval does not include 0.5, indicating that we cannot be confident the Democratic candidate will definitely win.
05

Calculate 99% Confidence Interval

To calculate the 99% CI, we use the z-score for 99% confidence, which is approximately 2.576. The CI is given by: \[ \u00bfp \pm (z \times SE) = 0.4714 \pm (2.576 \times 0.0133) = (0.4375, 0.5053) \] This interval includes 0.5, indicating we lack sufficient evidence to predict a definitive winner with 99% confidence.
06

Interpretation

For 95% confidence, since 0.5 is not within the CI, the Democratic candidate is less likely to win. At 99% confidence level, the CI includes 0.5, suggesting more uncertainty. Greater confidence (99%) requires broader intervals, demanding stronger evidence to be conclusive.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sample Proportion
A sample proportion represents a way to understand the fraction of the population that exhibits a particular characteristic, based on a sample. In our context, it's about voters who chose the Democratic candidate. To calculate the sample proportion, you divide the number of favorable outcomes (in this exercise, 660 Democratic votes) by the total number of outcomes (1400 voters total). This gives us \[ p = \frac{660}{1400} \approx 0.4714 \]Thus, about 47.14% of the sampled voters, based on our exit poll, favored the Democratic candidate.
This calculation is essential because it is our best estimate of what we expect to see in the entire population. The sample proportion provides a bridge between our specific data and the larger question about the entire group of voters. It gives a rough estimate but by itself doesn't tell us everything; we need more statistical tools, like the standard error and confidence intervals, to get a fuller picture.
Standard Error
Standard error (SE) is crucial in statistics as it measures the variability or dispersion of the sample proportion obtained from different possible samples. It tells us how much the sample proportion is expected to fluctuate from the actual population proportion. In simple terms, it helps us understand how 'off' our sample statistic might be from reality.
The standard error is calculated with the formula:\[ SE = \sqrt{\frac{p(1-p)}{n}} \]where:\
  • \(p\) is the sample proportion (0.4714 in our case),
  • \(1-p\) is the proportion for all other possibilities (0.5286),
  • \(n\) is the sample size (1400)
Plugging in the values gives us \[ SE = \sqrt{\frac{0.4714 \times 0.5286}{1400}} \approx 0.0133 \]This means that if we took multiple samples, we'd expect the sample proportion to vary by about 1.33% from the true population proportion. SE is a key component in constructing confidence intervals, helping us to judge how reliable our sample estimates are.
Z-Score
The z-score is a statistical measure that represents the number of standard deviations a data point is from the mean. It is critical when calculating confidence intervals as it scales the standard error to reflect the desired level of confidence. Different confidence levels have corresponding z-scores that expand or narrow the confidence interval.
  • For a 95% confidence interval, the z-score is approximately 1.96.
  • For a 99% confidence interval, the z-score is approximately 2.576.
To apply the z-score, we multiply it by the standard error to determine by how much we should adjust our sample proportion to determine the upper and lower bounds of our confidence interval.
For a 95% confidence interval, we use:\[ 0.4714 \pm (1.96 \times 0.0133) = (0.4453, 0.4975) \]For a 99% confidence interval:\[ 0.4714 \pm (2.576 \times 0.0133) = (0.4375, 0.5053) \]These intervals help us predict the likelihood of our sample proportion truly reflecting the population proportion. A wider interval, as seen with the 99% confidence interval, means a greater need for evidence before confidently making any predictions about the winner of the election.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Length of hospital stay for childbirth Data was collected from the records of 2962 patients admitted to a hospital in 2015 to estimate the mean length of stay for childbirth. It was observed that the sample mean was 2.372 days and the margin of error for estimating the population mean is 0.029 at a confidence level of \(95 \% .\) Explain the meaning of the last sentence, showing what it suggests about the \(95 \%\) confidence interval. Find the sample standard deviation.

In 2014, news reports worldwide alleged that the U.S. government had hacked German chancellor Angela Merkel's cell phone. A Pew Research Center survey of German citizens at about that time asked whether they find it acceptable or unacceptable for the U.S. government to monitor communications from their country's leaders. Results from the survey show that of 1000 citizens interviewed, 900 found it unacceptable. (Source: Pew Research Center, July 2014, "Global Opposition to U.S. Surveillance and Drones, But Limited Harm to America's Image") a. Find the point estimate of the population proportion of German citizens who find spying unacceptable. b. The Pew Research Center reports a margin of error at the \(95 \%\) confidence level of \(4.5 \%\) for this survey. Explain what this means.

Catalog mail-order sales A company that sells its products through mail-order catalogs wants information about the success of its most recent catalog. The company decides to estimate the mean dollar amount of items ordered from those who received the catalog. For a random sample of 100 customers from their files, only 5 made an order, so 95 of the response values were \(\$ 0 .\) The overall mean of all 100 orders was \(\$ 10,\) with a standard deviation of \(\$ 10\) a. Is it plausible that the population distribution is normal? Explain and discuss how much this affects the validity of a confidence interval for the mean. b. Find a \(95 \%\) confidence interval for the mean dollar order for the population of all customers who received this catalog. Normally, the mean of their sales per catalog is about \(\$ 15\). Can we conclude that it declined with this catalog? Explain.

USA TODAY conducted a survey of 1000 likely voters in February \(2016 .\) In the survey, 134 respondents said that CNN is the TV news or commentary source they trust the most. The interval estimation at the \(95 \%\) confidence level for the proportion who trust CNN the most is \((0.113,0.155) .\) Explain how to interpret the given confidence interval.

Population variability Explain the reasoning behind the following statement: "In studies about a very diverse population, large samples are often necessary, whereas for more homogeneous populations smaller samples are often adequate." Illustrate for the problem of estimating mean income for all medical doctors in the United States compared to estimating mean income for all entry-level employees at McDonald's restaurants in the United States.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.