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The paper "Sleeping with Technology: Cognitive, Affective and Technology Usage Predictors of Sleep Problems Among College Students" (Sleep Health [2016]: 49-56) summarized data from a survey of a sample of college students. Of the 734 students surveyed, 125 reported that they sleep with their cell phones near the bed and check their phones for something other than the time at least twice during the night. For purposes of this exercise, assume that this sample is representative of college students in the United States. a. Use the given information to estimate the proportion of college students who check their cell phones for something other than the time at least twice during the night. b. Verify that the conditions needed in order for the margin of error formula to be appropriate are met. c. Calculate the margin of error. d. Interpret the margin of error in the context of this problem.

Short Answer

Expert verified
We estimated the proportion of college students who check their cell phones at night to be p̂ ≈ 0.170, and the sample satisfied necessary conditions for calculating the margin of error. The margin of error was found to be 0.0333, and its interpretation in this context is that the true proportion of students who check their phones for something other than the time at least twice during the night is likely between 13.67% and 20.33% with a 95% confidence level.

Step by step solution

01

Estimate the proportion of college students who check their phones during the night

To estimate the proportion p̂ of students who check their cell phones for something other than the time at least twice during the night, divide the number of students who reported this behavior (125) by the total number of students surveyed (734): p̂ = 125 / 734 ≈ 0.170
02

Verify the conditions for the margin of error formula

To use the margin of error formula, we need to ensure that these conditions are met: 1. Random sampling 2. Large enough sample size (np̂ ≥ 10 and n(1-p̂) ≥ 10) We are told that the sample is representative of college students in the United States, so we can assume random sampling. To check if the sample size is large enough: np̂ = 734 * 0.170 ≈ 124.78 > 10 n(1-p̂) = 734 * (1 - 0.170) ≈ 609.22 > 10 Both conditions are satisfied, so we can use the margin of error formula.
03

Calculate the margin of error

The margin of error (ME) is calculated as: ME = Z * √(p̂(1-p̂)/n) For a 95% confidence interval, we use Z = 1.96: ME = 1.96 * √(0.170 * (1 - 0.170) / 734) ME ≈ 0.0333
04

Interpret the margin of error

The margin of error of 0.0333 means that the true proportion of college students who check their cell phones for something other than time at least twice during the night is likely to be within 3.33 percentage points of our estimate, p̂ = 0.170, or between 13.67% and 20.33%, with a 95% confidence level.

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Key Concepts

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

Proportion
In the context of this exercise, a proportion refers to a mathematical representation of the part of the sample exhibiting a certain behavior or attribute. It is expressed as a fraction of the total sample or population size. Here, we aim to estimate the proportion of college students who check their phones for purposes other than checking the time at least twice a night. This is calculated using the formula \( \hat{p} = \frac{x}{n} \), where \( x \) represents the number of students showing the behavior (125), and \( n \) is the total number of surveyed students (734). Hence, the estimated proportion \( \hat{p} \) is approximately 0.17, indicating that 17% of students in the sample have this behavior.

Proportions are useful in statistics as they provide a ratio that makes it easier to understand how one part compares to the entire set. It simplifies comparisons and predictions about a population based on a sample.
Confidence Interval
A confidence interval is a range of values derived from sample data that is likely to contain the true value of an unknown population parameter. It provides a measure of certainty about our sample estimates. In this problem, we're constructing a 95% confidence interval for the proportion \( \hat{p} \) of students who check their phones frequently at night. To calculate this interval, we use the formula: \[ CI = \hat{p} \pm ME \]where \( ME \) is the margin of error.

The margin of error (ME) reflects the degree of uncertainty around our estimate. In this case, it's calculated as \( ME = Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( Z = 1.96 \) for a 95% confidence interval, \( \hat{p} \approx 0.17 \), and \( n = 734 \). The resulting margin of error is approximately 0.0333, which means our 95% confidence interval for the proportion is (0.1367, 0.2033).

This suggests that we are 95% confident that the actual proportion of all college students engaging in this behavior is between 13.67% and 20.33%. Confidence intervals are crucial in research as they provide a range around the estimate, which accounts for natural variations in the sample.
Random Sampling
Random sampling is a fundamental concept in statistics that refers to selecting a subset of individuals from the population where each individual has an equal chance of being chosen. This approach is vital for making sure that the collected sample represents the entire population without bias, which in turn ensures that the results and inferences made about the population are reliable.

In this exercise, we assume the sample of 734 students is representative of the broader population of college students in the United States. Random sampling ensures that our sample estimates, like the proportion of students checking their phones at night, are unbiased and generalizable.
  • Random sampling reduces sampling bias, which can skew data and lead to incorrect conclusions.
  • It helps in creating a diverse sample that mirrors the broader population, making the results more valid across various subgroups.
By assuming random sampling, we confidently apply statistical formulas, such as those for margin of error and confidence intervals, with the assurance that our sample’s characteristics align well with those of the target population. This practice is central to drawing meaningful insights in statistical studies.

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Most popular questions from this chapter

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.

For each of the following choices, explain which one 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 report referenced in the previous exercise also indicated that \(33 \%\) of those in a representative sample of 533 homeowners in southern states said that they had considered installing solar panels. a. Use the given information to construct and interpret a \(90 \%\) confidence interval for the proportion of all homeowners in the southern states who have considered installing solar panels. b. Give two reasons why the confidence interval in Part (a) is narrower than the confidence interval calculated in the previous exercise.

In mid-2016 the United Kingdom (UK) withdrew from the European Union (an event known as "Brexit"), causing economic concerns throughout the world. One indicator that economists use to monitor the health of the economy is the proportion of residential properties offered for sale at auction that are successfully sold. An article titled "Going, going, gone through the roof-sky's the limit at auction" (October \(22,2016,\) www.estateagenttoday .co.uk/features/2016/10/going-going-gone-through-theroof-the-skys-the-limit- at-auction, retrieved May 4,2017 ) reported the success rate of a sample of 26 residential properties offered for sale at auctions in the UK in the summer of \(2016 .\) For this sample of properties, 14 of the 26 residential properties were successfully sold. Suppose it is reasonable to consider these 26 properties as representative of residential properties offered at auction in the post-Brexit UK. a. Would it be appropriate to use the large-sample confidence interval for a population proportion to estimate the proportion of residential properties successfully sold at auction in the post-Brexit UK? Explain. b. Would it be appropriate to use a bootstrap confidence interval for a population proportion to estimate the proportion of residential properties successfully sold at auction in the post-Brexit UK? Explain. c. Use the accompanying output from the "Bootstrap Confidence Interval for One Proportion" Shiny app to report a \(95 \%\) bootstrap confidence interval for the population proportion of residential properties successfully

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