/*! 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 Packer Fans In a Harris Poll con... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 15

Packer Fans In a Harris Poll conducted October \(20-25\) 2004,381 of 2114 randomly selected adults who follow professional football said the Green Bay Packers were their favorite team. (a) Verify that the requirements for constructing a confidence interval about \(\hat{p}\) are satisfied. (b) Construct a \(90 \%\) confidence interval for the proportion of adults who follow professional football who say the Green Bay Packers is their favorite team. Interpret this interval. (c) Construct a \(99 \%\) confidence interval for the proportion of adults who follow professional football who say the Green Bay Packers is their favorite team. Interpret this interval. (d) What is the effect of increasing the level of confidence on the width of the interval?

Short Answer

Expert verified
The 90% CI for Packers fans is (0.165, 0.195). The 99% CI is (0.156, 0.204). Higher confidence increases interval width.

Step by step solution

01

Check requirements for confidence interval

The conditions for constructing a confidence interval for a proportion are: (1) the sample should be a simple random sample, (2) the conditions for the binomial distribution are satisfied, and (3) both np and n(1-p) should be greater than 5. Here, n = 2114 and the number of successes (Green Bay Packers fans) is 381. Let's check the third condition: \[ np = 2114 * \hat{p} \]\[ n(1-p) = 2114 * (1-\hat{p}) \]First, we need to find \( \hat{p} \), the sample proportion.
02

Calculate sample proportion \(\hat{p}\)

The sample proportion \(\hat{p}\) is the number of successes divided by the total sample size:\[ \hat{p} = \frac{381}{2114} \approx 0.1802 \]
03

Verify np and n(1-p) conditions

Now, calculate np and n(1-p):\[ np = 2114 * 0.1802 \approx 380.491 \]\[ n(1-p) = 2114 * (1 - 0.1802) \approx 1733.509 \]Both conditions are satisfied as 380.491 and 1733.509 are greater than 5.
04

Construct the 90% confidence interval

To construct a 90% confidence interval, use the formula:\[ \hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]For a 90% confidence level, the z-value (z_{\alpha/2}) is approximately 1.645:\[ 0.1802 \pm 1.645 \sqrt{\frac{0.1802 (1 - 0.1802)}{2114}} \approx 0.1802 \pm 0.015 \]Thus, the 90% confidence interval is approximately \( (0.1652, 0.1952) \). This interval suggests that we are 90% confident that the proportion of all football fans who prefer the Packers is between 16.52% and 19.52%.
05

Construct the 99% confidence interval

For a 99% confidence level, the z-value is approximately 2.576:\[ 0.1802 \pm 2.576 \sqrt{\frac{0.1802 (1 - 0.1802)}{2114}} \approx 0.1802 \pm 0.024 \]Thus, the 99% confidence interval is approximately \( (0.1562, 0.2042) \). This interval suggests that we are 99% confident that the proportion of all football fans who prefer the Packers is between 15.62% and 20.42%.
06

Analyze the effect of the confidence level on the interval width

Increasing the confidence level (from 90% to 99%) increases the width of the interval. This is because a higher confidence level requires a larger z-value, which results in a larger margin of error.

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 91影视!

Key Concepts

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

Sample Proportion
The sample proportion, symbolized as \( \hat{p} \), is a key concept in statistics when analyzing survey data. It represents the ratio of favorable outcomes to the total sample size. In our given problem, 381 out of 2114 adults express that the Green Bay Packers are their favorite team. To find the sample proportion, divide the number of favorable responses by the total sample size: \[ \hat{p} = \frac{381}{2114} \approx 0.1802 \]. This result tells us that approximately 18.02% of the sampled adults favor the Packers. The sample proportion provides an estimate of the true proportion (denoted as \( p \)) in the entire population, which we aim to capture within a confidence interval.
Binomial Distribution
The binomial distribution is crucial when working with proportions and constructing confidence intervals. It applies to scenarios with two possible outcomes and allows us to model the number of successes in a sample. For our problem, a 'success' would be an adult favoring the Packers. The binomial conditions are based on:
  • A fixed number of trials (sample size)
  • Each trial being independent
  • Two possible outcomes (favor the Packers or not)
  • A constant probability of success.
Our sample meets these criteria, as we randomly sample 2114 individuals. The calculation for the 'successes' and 'failures' in our sample should be greater than 5 to apply the normal approximation: \[ np = 2114 \times \hat{p} \approx 380.491 \], and \[ n(1 - \hat{p}) = 2114 \times (1 - 0.1802) \approx 1733.509 \]. Both values satisfy the required condition, allowing us to use the binomial distribution model.
Margin of Error
The margin of error defines the range within which the true population parameter is expected to lie, given a certain confidence level. It accounts for the variability and uncertainty in the sample proportion estimate. The margin of error formula for a proportion is: \[ z_{\frac{\alpha}{2}} \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]. For a 90% confidence interval, our z-value \( z_{\frac{1.645}{2}} \) gives us: \[ 1.645 \sqrt{\frac{0.1802 (1 - 0.1802)}{2114}} \approx 0.015 \]. This provides a 90% confidence interval of approximately \( (0.1652, 0.1952) \). When the confidence level increases to 99%, the z-value also increases to 2.576, yielding a larger margin of error: \[ 2.576 \sqrt{\frac{0.1802 (1 - 0.1802)}{2114}} \approx 0.024 \]. This change results in a wider confidence interval \( (0.1562, 0.2042) \), showing that higher confidence levels lead to broader intervals to accommodate the increased certainty.

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

Construct the appropriate confidence interval. A simple random sample of size \(n=40\) is drawn from a population. The sample mean is found to be \(\bar{x}=120.5\) and the sample standard deviation is found to be \(s=12.9 .\) Construct a \(99 \%\) confidence interval about the population mean.

A Gallup poll conducted January \(17-\) February \(6,2005,\) asked 1028 teenagers aged 13 to \(17,\) "Typically, how many hours per week do you spend watching TV?" Survey results indicate that \(\bar{x}=13.0\) hours and \(s=2.3\) hours. Construct a \(95 \%\) confidence interval for the number of hours of TV teenagers watch each week. Interpret the interval.

A simple random sample of size \(n\) is drawn from a population that is normally distributed. The sample mean, \(\bar{x}\), is found to be \(50,\) and the sample standard deviation, \(s,\) is found to be \(8 .\) (a) Construct a \(98 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(20 .\) (b) Construct a \(98 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(15 .\) How does decreasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(20 .\) Compare the results to those obtained in part (a). How does decreasing the level of confidence affect the margin of error, \(E ?\) (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Why?

A simple random sample of size \(n\) is drawn from a population whose population standard deviation, \(\sigma,\) is known to be \(5.3 .\) The sample mean, \(\bar{x},\) is determined to be 34.2 (a) Compute the \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is 35 (b) Compute the \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(50 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Compute the \(99 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(35 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the size of the margin of error, \(E ?\) (d) Can we compute a confidence interval about \(\mu\) based on the information given if the sample size is \(n=15 ?\) Why? If the sample size is \(n=15,\) what must be true regarding the population from which the sample was drawn?

The mean age of the 42 presidents of the United States on the day of inauguration is 54.8 years with a standard deviation of 6.2 years. A researcher constructed a \(95 \%\) confidence interval for the mean age of presidents on inauguration day. He wrote that he was \(95 \%\) confident the mean age of the president on inauguration day is between 53.0 and 56.7 years of age. What is wrong with the researcher's analysis?
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation
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.