Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The article "Facebook Etiquette at Work" (USA Today, March 24, 2010) reported that \(56 \%\) of 1,200 social network users surveyed indicated that they thought it was not \(\mathrm{OK}\) for someone to "friend" his or her boss. Suppose that this sample can be regarded as a random sample of social network users. Is it reasonable to conclude that more than half of social network users feel this way? Use what you know about the sampling distribution of \(\hat{p}\) to support your answer.

Short Answer

Expert verified
Based on the results of the constructed confidence interval, determine if it's reasonable to conclude that more than half of social network users think it's not acceptable to friend their boss. Specify the result based on whether the interval contains \(0.5\) or not. If it does, more than half feel this way. If it doesn't, it would be incorrect to conclude that more than half feel this way.

Step by step solution

01

Calculate the Sample Proportion

The sample proportion, \(\hat{p}\), is already provided in the exercise as \(0.56\) or \(56\%\). This value of \(\hat{p}\) means that in the sample of 1,200 social network users, \(56\%\) indicated they thought it was not acceptable to friend their boss.
02

Calculate the Standard Error

The standard error (SE) of the sampling distribution of a proportion is calculated using the formula: \(SE = \sqrt{ \(\hat{p}(1 - \hat{p})/n }\), where \(n\) is the sample size. Substituting the given values: \(SE = \sqrt{ (0.56 \times (1 - 0.56))/1200}\). Calculate this value.
03

Construct a Confidence Interval

A \(95\%\) confidence interval for the population proportion is calculated using the formula: \(\hat{p} \pm 1.96 \times SE\). Substitute the values of \(\hat{p}\) and SE obtained in the previous steps to calculate the confidence interval.
04

Interpret the Results

If the confidence interval constructed in step 3 contains the value \(0.5\) or \(50\%\), it's reasonable to conclude that more than half of social network users feel it's not okay to friend their boss. If it doesn't contain \(0.5\) or \(50\%\), then it's not reasonable to conclude that more than half of social network users feel this way.

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 Vaia!

Key Concepts

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

Sample Proportion
Understanding the sample proportion is crucial when analyzing survey data. The sample proportion, denoted as \(\hat{p}\), represents the percentage of respondents in a sample who have a certain characteristic—in this case, the percentage of social network users who think it is unacceptable to friend their boss. For the survey mentioned, where 1,200 users were asked about their views on social network etiquette related to friending a boss, the sample proportion \(\hat{p}\) was found to be 56% or 0.56.

This tells us that out of the 1,200 individuals surveyed, 672 (\(0.56 \times 1200\)) expressed that friending a boss is not okay. It's important to remember that this is only a sample of the total population of social network users. To make any conclusions about the entire population, statisticians use the sample proportion as a starting point for further calculations such as the standard error and confidence intervals.
Standard Error
The standard error (SE) is a measure of how much we expect our sample's proportion to vary from the true population proportion. It gives us an idea about the precision of our sample proportion estimate. The formula for computing the standard error of the sample proportion is \(SE = \sqrt{ \hat{p}(1 - \hat{p})/n }\), where \(n\) is the sample size.

Using the provided data, the standard error is calculated by plugging in the sample proportion \(0.56\) and the sample size of 1,200. The resulting standard error helps determine the reliability of the sample proportion and is used to construct confidence intervals. A lower standard error indicates a more precise estimate. It is valuable to note that larger sample sizes tend to produce a smaller standard error, indicating a more accurate estimation of the population proportion.
Confidence Interval
A confidence interval is a range of values that is likely to contain the population proportion with a certain level of confidence. The usual level of confidence used is 95%, which means that we can be 95% confident that the true population proportion lies within this interval.

To construct a 95% confidence interval, you take the sample proportion \(\hat{p}\) and add and subtract 1.96 times the standard error. The number 1.96 is tied to the 95% confidence level. From our exercise, using the sample proportion and standard error calculated in the previous steps, we find the range within which we can be confident the true proportion of the population lies. If the value of 50% (the hypothesized threshold for the majority) is not within this interval, then we have evidence to suggest that the true population proportion is indeed different from 50%.
Population Proportion
The population proportion, denoted by \(p\), is the true proportion of all social network users who believe it's unacceptable to friend a boss. Unlike the sample proportion which is based on a sample group, the population proportion encompasses everyone in the population being studied. The goal of survey research, like the example in the exercise, is to estimate this population parameter based on the observed sample proportion.

We can never know the true population proportion without asking every single social network user, but statistics provides tools—like the confidence interval—to infer the population proportion within a certain level of confidence. By calculating the confidence interval around the sample proportion and interpreting whether it encompasses a pivotal value (in this case, 50%), we can make inferences about the population proportion and how social network users as a whole feel about befriending their boss online.

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

The article "Facebook Etiquette at Work" (USA Today, March 24,2010 ) reported that \(56 \%\) of people participating in a survey of social network users said it was not \(\mathrm{OK}\) for someone to "friend" his or her boss. Let \(p\) denote the proportion of all social network users who feel this way and suppose that \(p=0.56\). a. Would \(\hat{p}\) based on a random sample of 50 social network users have a sampling distribution that is approximately normal? b. What are the mean and standard deviation of the sampling distribution of \(\hat{p}\) if the sample size is \(100 ?\)

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21,2006 ) reported that \(37 \%\) of college freshmen carry a credit card balance from month to month. Suppose that the reported percentage was based on a random sample of 1,000 college freshmen. Suppose you are interested in learning about the value of \(p,\) the proportion of all college freshmen who carry a credit card balance from month to month. The following table is similar to the table that appears in Examples 8.4 and \(8.5,\) and is meant to summarize what you know about the sampling distribution of \(\hat{p}\) in the situation just described. The "What You Know" information has been provided. Complete the table by filling in the "How You Know It" column.

A random sample of 1,000 students at a large college included 428 who had one or more credit cards. For this sample, \(\hat{p}=\frac{428}{1,000}=0.428 .\) If another random sample of 1,000 students from this university were selected, would you expect that \(\hat{p}\) for that sample would also be 0.428 ? Explain why or why not.

If a hurricane were headed your way, would you evacuate? The headline of a press release issued January \(21,2009,\) by the survey research company International Communications Research (icrsurvey.com) states, "Thirty-one Percent of People on High-Risk Coast Will Refuse Evacuation Order, Survey of Hurricane Preparedness Finds." This headline was based on a survey of 5,046 adults who live within 20 miles of the coast in high hurricane risk counties of eight southern states. The sample was selected to be representative of the population of coastal residents in these states, so assume that it is reasonable to regard the sample as if it were a random sample. a. Suppose you are interested in learning about the value of \(p\), the proportion of adults who would refuse to evacuate. This proportion can be estimated using the sample proportion, \(\hat{p} .\) What is the value of \(\hat{p}\) for this sample? b. Based on what you know about the sampling distribution of \(\hat{p}\), is it reasonable to think that the estimate is within 0.03 of the actual value of the population proportion? Explain why or why not.

Suppose that a particular candidate for public office is favored by \(48 \%\) of all registered voters in the district. A polling organization will take a random sample of 500 of these voters and will use \(\hat{p}\), the sample proportion, to estimate \(p\). a. Show that \(\sigma_{p}\), the standard deviation of \(\hat{p}\), is equal to \(0.0223 .\) b. If for a different sample size, \(\sigma_{p}=0.0500\), would you expect more or less sample-to-sample variability in the sample proportions than when \(n=500 ?\) c. Is the sample size that resulted in \(\sigma_{\rho}=0.0500\) larger than 500 or smaller than \(500 ?\) Explain your reasoning.

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.

Sign-up for free