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The study "Digital Footprints" (Pew Internet \& American Life Project, www.pewinternet.org. 2007\()\) reported that \(47 \%\) of Internet users have searched for information about themselves online. The \(47 \%\) figure was based on a random sample of Internet users. For purposes of this exercise, suppose that the sample size was \(n=300\) (the actual sample size was much larger). Construct and interpret a \(90 \%\) confidence interval for the proportion of Internet users who have searched online for information about themselves.

Short Answer

Expert verified
For example, if the calculated confidence interval came out as (0.42, 0.52), it would mean there is a 90% level of confidence that the proportion of Internet users who have searched online for themselves is between 42% and 52%.

Step by step solution

01

Determine the sample proportion and size

The sample proportion (p) of Internet users who have searched online for information about themselves is given as \(p = 0.47\). The sample size (n) is given as \(n = 300\).
02

Identify the confidence level

The problem requires a 90% confidence interval, which means the confidence level (c) is \(c = 0.90\). The critical value for a 90% confidence level is \(Z=1.645\) when using a standard normal (Z) distribution.
03

Calculate the confidence interval

The formula for the confidence interval of a proportion is \[\(p \pm Z \sqrt{ (p(1 - p) \div n)}\], where \(\sqrt{ (p(1 - p) \div n)}\) is the standard deviation of the proportion. Substituting the known values gives \[0.47 \pm 1.645\sqrt{(0.47 \times 0.53) \div 300}\]. Calculating this expression will give the lower and upper limits of the confidence interval.
04

Interpret the confidence interval

Once calculated, the 90% confidence interval will provide an estimate of the range within which the true proportion of Internet users who have searched for information about themselves online falls 90% of the time.

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Key Concepts

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

Sample Proportion
A sample proportion is a key statistical concept used to make inferences about a population based on a sample. It's a simple ratio, obtained by dividing the number of favorable outcomes in a sample by the total number of observations in the sample.For instance, in the given study, 47% of the Internet users sampled had searched for information about themselves. This directly gives us the sample proportion,which is calculated as \( p = \frac{x}{n} \), where \( x \) is the count of favorable outcomes, and \( n \) is the sample size.
  • In this example: \( p = 0.47 \)
  • Sample size \( n = 300 \)
Understanding this concept is essential because it forms the basis for constructing confidence intervals and other statistical analyses.Thus, the sample proportion serves as an estimate of the unknown actual proportion in the population.This metric is fundamental in statistical sampling and critical in determining the likelihood of various outcomes when generalizing from the sample to the larger population.
Critical Value
The critical value is an essential component in constructing confidence intervals.It represents the point on the standard normal distribution that corresponds with the desired confidence level. In this context, we're aiming for a 90% confidence level. To find the critical value for our confidence interval, we refer to the standard normal (\( Z \)) distribution.
  • For a 90% confidence interval, the critical value is typically \( Z = 1.645 \).
This value can be located using a standard normal distribution table or a statistical calculator.Knowing the critical value is crucial because it determines the width of the confidence interval. Larger critical values will result in wider intervals, providing a more conservative estimate of where the true proportion might fall.The critical value helps you gauge how sure you can be about your interval estimation, making it a fundamental part of inferential statistics.
Standard Normal Distribution
The standard normal distribution is a special type of normal distribution with a mean of 0 and a standard deviation of 1.It plays a vital role in statistics, especially in scenarios that involve standardizing other normal distributions for comparison and analysis.In the context of confidence intervals, the standard normal distribution helps translate the confidence level into a critical value.
When sample sizes are large, the distribution of sample proportions approximates the standard normal.This approximation permits the application of \( Z \)-scores to calculate confidence intervals and perform hypothesis testing.For instance:
  • A 90% confidence interval requires a \( Z \)-score from the distribution, which is \( 1.645 \)
Understanding the standard normal distribution is key to accurately interpreting statistical results and ensuring valid inferences about populations from sample data.Its role in standardizing data means that we can compare results across different studies and ensure consistency in data analysis.
Standard Deviation of the Proportion
The standard deviation of the proportion, often called the "standard error," provides a measure of the variability or spread of the sample proportion relative to the true population proportion.It's essential in calculating the confidence interval, helping determine the margin of error for the estimate.
The formula to calculate this is:\[\text{Standard deviation of the proportion} = \sqrt{ \frac{p(1-p)}{n} }\]where \( p \) is the sample proportion and \( n \) is the sample size.
  • In this exercise: \( p = 0.47 \), \( n = 300 \)
The standard deviation of the proportion indicates how much the sample proportion is expected to fluctuate from the actual population proportion.It allows us to compute the range, or confidence interval, in which the true proportion likely exists.This is crucial as it offers insight into the reliability of our sample proportion and helps in making sound statistical conclusions.

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Most popular questions from this chapter

The report "2005 Electronic Monitoring \& Surveillance Survey Many Companies Monitoring. Recording. Videotaping-and Firing-Employees" (American Management Association, 2005\()\) summarized the results of a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet and 131 had fired workers for e-mail misuse. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the Unired Srates. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of the Internet. b. What are two reasons why a \(90 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of e-mail would be narrower than the \(95 \%\) confidence interval computed in Part (a)?

The formula used to compure a confidence interval for the mean of a normal population when \(n\) is small is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=17\) b. \(90 \%\) confidence, \(n=12\) c. \(99 \%\) confidence, \(n=24\) d. \(90 \%\) confidence, \(n=25\) e. \(90 \%\) confidence, \(n=13\) f. \(95 \%\) confidence, \(n=10\)

For each of the following choices, explain which would result in a wider large-sample confidence interval for \(p\) : a. \(90 \%\) confidence level or \(95 \%\) confidence level b. \(n=100\) or \(n=400\)

The article "Nine Out of Ten Drivers Admit in Survey to Having Done Something Dangerous" (Knight Ridder Newspapers, July 8. 2005 ) reported the results of a survey of 1100 drivers. Of those surveyed, 990 admitted to careless or aggressive driving during the previous 6 months. Assuming that it is reasonable to regard this sample of 1100 as representative of the population of drivers, use this information to construct a \(99 \%\) confidence interval to estimate \(p\), the proportion of all drivers who have engaged in careless or aggressive driving in the previous 6 months.

In a survey of 1000 randomly selected adults in the United States, participants were asked what their most favorite and what their least favorite subject was when they were in school (Associated Press, August 17 . 2005)\(.\) In what might seem like a contradiction, math was chosen more often than any other subject in both categories! Math was chosen by 230 of the 1000 as the favorite subject, and it was also chosen by 370 of the 1000 as the least favorite subject. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the favorite subject in school. b. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults for whom math was the least favorite subject.

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