/*! 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 41 America's young people are heavy... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

America's young people are heavy Internet users; \(87 \%\) of Americans ages 12 to 17 are Internet users (The Cincinnati Enquirer, February 7,2006 ). MySpace was voted the most popular website by \(9 \%\) in a sample survey of Internet users in this age group. Suppose 1400 youths participated in the survey. What is the margin of error, and what is the interval estimate of the population proportion for which MySpace is the most popular website? Use a \(95 \%\) confidence level.

Short Answer

Expert verified
The margin of error is approximately 1.41%, and the interval estimate of the population proportion for which MySpace is the most popular website among Internet users aged 12 to 17 in America is (7.59%, 10.41%) with a 95% confidence level.

Step by step solution

01

Identify Given Values

Sample size (n) = 1400 Sample proportion (p) = 0.09 (as 9% of the survey participants voted MySpace as the most popular website) Confidence level = 95%
02

Find the Z-score

For a 95% confidence level, the Z-score (Z) is 1.96. This value can be found using a Z-table or a calculator that provides critical values of Z for different confidence levels.
03

Calculate Margin of Error

Now, we can plug the values into the Margin of Error formula: ME = Z * sqrt(p * (1 - p) / n) ME = 1.96 * sqrt(0.09 * (1 - 0.09) / 1400) ME ≈ 0.0141 The margin of error is approximately 0.0141 or 1.41%.
04

Calculate the Interval Estimate

Now we will calculate the interval estimate using the sample proportion and the margin of error: CI = p ± ME CI = 0.09 ± 0.0141 CI ≈ (0.0759, 0.1041) Thus, the interval estimate of the population proportion for which MySpace is the most popular website among Internet users aged 12 to 17 in America is (0.0759, 0.1041), or (7.59%, 10.41%) with a 95% confidence level.

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.

Margin of Error
The margin of error is a critical concept in statistics that helps you understand the range in which the true population parameter is likely to fall. It reflects the uncertainty or potential error in the sample estimate.
When calculating the margin of error, you multiply the Z-score by the standard error. The standard error measures how much the sample proportion may vary from the population proportion due to random sampling variability.
Consider the formula:
  • Margin of Error (ME) = Z * \( \sqrt{ \frac{p(1 - p)}{n} } \)
In this formula, ME represents the margin of error, \(Z\) is the Z-score, \(p\) is the sample proportion, and \(n\) is the sample size.
Using the example from the exercise, the margin of error was calculated to be 0.0141 or 1.41%. This indicates that the actual proportion of youths who would vote MySpace as the most popular website falls within 1.41% above or below the sample proportion.
This concept helps assess the precision of your survey results, providing a range (confidence interval) that estimates the true population proportion.
Sample Proportion
Sample proportion is a measure that represents the fraction of individuals in your sample who exhibit a certain characteristic. In other words, it's an estimate of the population proportion you derive from your sample data.
In statistics, the sample proportion is often denoted by \( p \).
Using the exercise as an example, since 9% of youths in the survey voted MySpace as the most popular website, the sample proportion \( p \) is 0.09.
This is calculated by dividing the number of favorable responses by the total survey responses.
  • For instance, if 9% of 1400 youths chose MySpace, the calculation would be: \( \frac{9}{100} = 0.09 \).

Knowing the sample proportion helps in calculating the margin of error and constructing the confidence interval. It effectively summarizes the survey's outcome and gives you a point estimate for assessing the whole population.
Z-score
The Z-score is a statistical measure that helps determine the position of a data point in relation to the mean of a group of data. It tells you how many standard deviations the data point is from the mean.
When used in the context of confidence intervals, the Z-score helps in determining the spread or width of the confidence interval around a sample proportion.
For instance, with a 95% confidence level, the standard Z-score used is 1.96. This number comes from the standard normal distribution and can be found using a Z-table.
The Z-score allows us to calculate the margin of error, influencing the width of the confidence interval. A higher Z-score corresponds to a higher confidence level, which means a wider confidence interval. This reflects greater certainty about the range in which the population proportion lies, albeit resulting in a less precise estimate due to the increased width.
Population Proportion
Population proportion is an estimate of a certain characteristic in the entire population, based on the sample data you have. It is denoted by the symbol \( P \) in statistics. While the sample proportion is a measure of that characteristic within a sample, the population proportion estimates it for everyone in the population.
In the exercise, the true population proportion of young Internet users who consider MySpace the most popular site is unknown. However, using the sample proportion and the margin of error, we can estimate the population proportion within a range, known as the confidence interval.
This interval provides a range of values where the actual population proportion is likely to fall.
For example:
  • If the interval estimate is (0.0759, 0.1041), it suggests that between 7.59% and 10.41% of all young Internet users consider MySpace the most popular website.
This estimation allows researchers and statisticians to make informed conclusions about the whole population based on the sample data.

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

Consumption of alcoholic beverages by young women of drinking age has been increasing in the United Kingdom, the United States, and Europe (The Wall Street Journal, February 15,2006 ). Data (annual consumption in liters) consistent with the findings reported in The Wall Street Journal article are shown for a sample of 20 European young women. $$\begin{array}{crrrr} 266 & 82 & 199 & 174 & 97 \\ 170 & 222 & 115 & 130 & 169 \\ 164 & 102 & 113 & 171 & 0 \\ 93 & 0 & 93 & 110 & 1300 \end{array}$$ a. What is the \(95 \%\) confidence interval estimate for the mean ticket sales revenue per theater? Interpret this result. b. Using the movie ticket price of \(\$ 7.16\) per ticket, what is the estimate of the mean number of customers per theater? c. The movie was shown in 3118 theaters. Estimate the total number of customers who saw Hannah Montana: The Movie and the total box office ticket sales for the three- day weekend.

A simple random sample of 60 items resulted in a sample mean of \(80 .\) The population standard deviation is \(\sigma=15\) a. Compute the \(95 \%\) confidence interval for the population mean. b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a \(95 \%\) confidence interval for the population mean. c. What is the effect of a larger sample size on the interval estimate?

For a \(t\) distribution with 16 degrees of freedom, find the area, or probability, in each region. a. To the right of 2.120 b. To the left of 1.337 c. To the left of -1.746 d. To the right of 2.583 e. Between -2.120 and 2.120 f. Between -1.746 and 1.746

A simple random sample of 40 items resulted in a sample mean of \(25 .\) The population standard deviation is \(\sigma=5\) a. What is the standard error of the mean, \(\sigma_{\bar{x}} ?\) b. At \(95 \%\) confidence, what is the margin of error?

Older people often have a hard time finding work. AARP reported on the number of weeks it takes a worker aged 55 plus to find a job. The data on number of weeks spent searching for a job contained in the file JobSearch are consistent with the AARP findings \((A A R P\) Bulletin, April 2008). a. Provide a point estimate of the population mean number of weeks it takes a worker aged 55 plus to find a job. b. \(\quad\) At \(95 \%\) confidence, what is the margin of error? c. What is the \(95 \%\) confidence interval estimate of the mean? d. Discuss the degree of skewness found in the sample data. What suggestion would you make for a repeat of this study?

See all solutions

Recommended explanations on Economics 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.