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The report "Parents, Teens and Digital Monitoring" (Pew Research Center, January \(7,2016,\) www.pewinternet .org/2016/01/07/parents-teens-and-digital- monitoring, retrieved May 5,2017 ) reported that \(61 \%\) of parents of teens aged 13 to 17 said that they had checked which web sites their teens had visited. The \(61 \%\) figure was based on a representative sample of 1060 parents of teens in this age group. a. Use the given information to estimate the proportion of parents of teens age 13 to 17 who have checked which web sites their teen has visited. What statistic did you use? b. Use the sample data to estimate the standard error of \(\hat{p}\). c. Calculate and interpret the margin of error associated with the estimate in Part (a).

Short Answer

Expert verified
(a) The estimated proportion of parents of teens age 13 to 17 who have checked their teens' web history is \(0.61\) or \(61\%\). We used the sample proportion \(\hat{p}\) as our statistic. (b) The standard error of the sample proportion \(\hat{p}\) is approximately \(0.0153\). (c) The margin of error is approximately \(0.03\) or \(3\%\). We can be 95% confident that the true proportion of parents checking their teens' web history lies between \(58\%\) and \(64\%\).

Step by step solution

01

(a) Estimate proportion and identify the statistic

To estimate the proportion of parents of teens age 13 to 17 who have checked their teens' web history, we can use the given value from the sample. We are given a sample proportion, denoted as \(\hat{p}\), of 61%. Therefore, \(\hat{p} = 0.61\). b. Estimate the standard error of the sample proportion
02

(b) Estimate the standard error

The standard error of the sample proportion (\(\hat{p}\)) can be calculated using the formula: \(SE(\hat{p}) = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) where \(n\) is the sample size and \(\hat{p}\) is the sample proportion. In our case \(n = 1060\) and \(\hat{p} = 0.61\). Therefore, \(SE(\hat{p}) = \sqrt{\frac{0.61(1-0.61)}{1060}} = \sqrt{\frac{0.61(0.39)}{1060}} \approx 0.0153\) c. Calculate and interpret the margin of error
03

(c) Calculate and interpret the margin of error

The margin of error (ME) can be calculated using the formula: \(ME = 1.96 \times SE(\hat{p})\) where 1.96 is the z-score associated with a 95% confidence level. We found the standard error to be 0.0153, thus \(ME = 1.96 \times 0.0153 \approx 0.03\) So, the margin of error is approximately 0.03 or 3%. This means that we can be 95% confident that the true proportion of parents who check their teens' web history lies within 3% of our sample estimate. In other words, the true proportion lies between 58% and 64%.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values that is likely to include a population parameter with a certain degree of confidence. In simpler terms, it's like using a fishing net to estimate the number of fish in a pond. You throw the net in multiple times, and based on the fish you catch each time, your results give you an idea of how many fish you have in the whole pond. This idea applies to statistics, only instead of fish, you are trying to estimate things like averages, proportions, or totals based on samples.
When we talk about confidence in statistics, we often refer to a 95% confidence interval, meaning that we are 95% sure that the true parameter falls within our estimated interval. In the case of our exercise, the confidence interval helps us to understand the true proportion of parents checking their teens' web history.
  • Sample Proportion (\(\hat{p}\)) = 0.61
  • Margin of Error = 0.03
The confidence interval can be calculated as:\[CI = \hat{p} \pm ME\]For our example, the confidence interval is:\[0.61 \pm 0.03 = (0.58, 0.64)\]This indicates that the actual percentage of parents likely falls between 58% and 64%, with 95% confidence.
Standard Error
The standard error is a statistic that tells us how much the sample proportion, \(\hat{p}\), might differ from the true population proportion. It helps us gauge the reliability of our sample results. Think of it as a measure of how much the sample "wiggles" around the true population figure.
To calculate the standard error of the sample proportion, you can use the formula:\[SE(\hat{p}) = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]where \(n\) is the sample size. In the example of the Pew Research Center, the sample size \(n = 1060\) and the sample proportion \(\hat{p} = 0.61\). Using these values:
  • \(1 - \hat{p} = 0.39\)
  • Calculation: \(\sqrt{\frac{0.61(0.39)}{1060}} \approx 0.0153\)
This result lets us know the expected fluctuation of using a sample estimate to approximate a population proportion. The smaller the standard error, the closer the sample proportion is likely to reflect the real-world percentage.
Sample Proportion
A sample proportion, denoted by \(\hat{p}\), represents the fraction of the sample that displays a particular attribute. It is an estimate of the true proportion of the entire population. Like counting red marbles in a jar of mixed colors to figure out how many there might be in total without looking at every marble.
In the exercise, the sample proportion is \(61\%\), which means in the sample of 1060 parents, about \(61\%\) reported checking their teens' web histories. How do you figure this out?
  • The formula for the sample proportion: \[\hat{p} = \frac{x}{n}\]where \(x\) is the number of successes or the favorable outcome count, and \(n\) is the total sample size.
  • In this specific example, \(x\) is not explicitly given, but since \(\hat{p} = 0.61\), and \(n = 1060\), it implies that \(x = 0.61 \times 1060 = 646.6\), rounded based on context interpretation.
Understanding the sample proportion helps in calculating other statistics like the standard error and helps anticipate the confidence interval for predicting population parameters.

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Most popular questions from this chapter

A random sample will be selected from the population of all adult residents of a particular city. The sample proportion \(\hat{p}\) will be used to estimate \(p,\) the proportion of all adult residents who are employed full time. For which of the following situations will the estimate tend to be closest to the actual value of \(p\) ? i. \(n=500, p=0.6\) ii. \(n=450, p=0.7\) iii. \(n=400, p=0.8\)

In \(2010,\) the National Football League adopted new rules designed to limit head injuries. In a survey conducted in 2015 by the Harris Poll, 1216 of 2096 adults indicated that they were football fans and followed professional football. Of these football fans, 692 said they thought that the new rules were effective in limiting head injuries (December 21 , \(2015,\) www.theharrispoll.com/sports/Football-Injuries.html, retrieved May 6,2017 ). a. Assuming that the sample is representative of adults in the United States, construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. adults who consider themselves to be football fans. b. Construct and interpret a \(95 \%\) confidence interval for the proportion of football fans who think that the new rules have been effective in limiting head injuries. c. Explain why the confidence intervals in Parts (a) and (b) are not the same width even though they both have a confidence level of \(95 \%\)

The report "The Politics of Climate" (Pew Research Center, October \(4,2016,\) www.pewinternet.org \(/ 2016 / 10 / 04\) /the-politics-of-climate, retrieved May 6,2017 ) summarized data from a survey on public opinion of renewable and other energy sources. It was reported that \(52 \%\) of the people in a sample from western states said that they have considered installing solar panels on their homes. This percentage was based on a representative sample of 369 homeowners in the western United States. Use the given information to construct and interpret a \(90 \%\) confidence interval for the proportion of all homeowners in western states who have considered installing solar panels.

Data from a representative sample were used to estimate that \(32 \%\) of all computer users in 2011 had tried to get on a Wi-Fi network that was not their own in order to save money (USA TODAY, May 16,2011 ). You decide to conduct a survey to estimate this proportion for the current year. What is the required sample size if you want to estimate this proportion with a margin of error of \(0.05 ?\) Calculate the required sample size first using 0.32 as a preliminary estimate of \(p\) and then using the conservative value of \(0.5 .\) How do the two sample sizes compare? What sample size would you recommend for this study?

Describe how each of the following factors affects the width of the large- sample confidence interval for \(p\) : a. The confidence level b. The sample size c. The value of \(\hat{p}\)

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