/*! 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 33 The paper "Bedtime Mobile Phone ... [FREE SOLUTION] | 91Ó°ÊÓ

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The paper "Bedtime Mobile Phone Use and Sleep in Adults" (Social Science and Medicine [2016]: \(93-101\) ) describes a study of 844 adults living in Belgium. Suppose that it is reasonable to regard this sample as a random sample of adults living in Belgium. You want to use the survey data to decide if there is evidence that a majority of adults living in Belgium take their cell phones to bed with them. Let \(p\) denote the population proportion of all adults living in Belgium who take their cell phones to bed with them. (Hint: See Example \(10.10 .)\) a. Describe the shape, center, and variability of the sampling distribution of \(\hat{p}\) for random samples of size 844 if the null hypothesis \(H_{0}: p=0.50\) is true. b. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.52\) for a sample of size 844 if the null hypothesis \(H_{0}: p=0.50\) were true? Explain why or why not. c. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.54\) for a sample of size 844 if the null hypothesis \(H_{0}: p=0.50\) were true? Explain why or why not. d. The actual sample proportion observed in the study was \(\hat{p}=0.59 .\) Based on this sample proportion, is there convincing evidence that the null hypothesis \(H_{0}: p=\) 0.50 is not true, or is \(\hat{p}\) consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation. e. Do you think it would be reasonable to generalize the concusion of this test to adults living in the United States? Explain why or why not.

Short Answer

Expert verified
a. For the distribution of \(\hat{p}\) under the null hypothesis \(H_{0}: p=0.50\), we can expect an approximately normal shape due to large sample size (844 adults), centered at 0.50. The variability, or standard deviation, can be calculated as \(\sigma_{\hat{p}} \approx 0.0172\), using the formula \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\). b. For a sample proportion \(\hat{p}=0.52\), the corresponding z-score is approximately 1.16, which yields a probability of roughly 0.123. Hence, we would not be surprised to observe this sample proportion under the null hypothesis. c. For a sample proportion \(\hat{p}=0.54\), the corresponding z-score is approximately 2.33, which yields a probability of roughly 0.010. This low probability suggests that we would indeed be surprised to observe this sample proportion if the null hypothesis were true. d. For the observed sample proportion \(\hat{p}=0.59\), the corresponding z-score is approximately 5.23, which translates to an extremely low probability of roughly 0.00000087. Thus, this sample proportion provides compelling evidence against the null hypothesis. e. It would not be appropriate to directly apply this conclusion to adults living in the United States due to cultural and lifestyle differences between the two countries and differing rates of mobile device usage. A similar study should be conducted in the United States before generalizing these findings.

Step by step solution

01

Shape

Under the null hypothesis (H0: p = 0.50), the sampling distribution of p̂ should follow a normal distribution if the sample size is large enough. Since the sample size is 844, it is reasonable to assume that the sampling distribution will be approximately normal.
02

Center

The mean (or center) of the sampling distribution of p̂, when the null hypothesis is true, is equal to the assumed population proportion p = 0.50.
03

Variability

The standard deviation of the sampling distribution of p̂ can be calculated using the formula: \[\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\] Substituting p = 0.50 and n = 844, we get: \[\sigma_{\hat{p}} = \sqrt{\frac{0.50(1-0.50)}{844}} \approx 0.0172\] #b. Probability of p̂ = 0.52 under Null Hypothesis# When p̂ = 0.52, we can calculate the z-score as follows: \[z = \frac{p̂ - p}{\sigma_{\hat{p}}} = \frac{0.52 - 0.50}{0.0172} \approx 1.16\] Using a standard normal table, the probability of observing a z-score greater than or equal to 1.16 is approximately equal to P(Z > 1.16) ≈ 0.123. This probability is quite high, so we would not be surprised to observe a sample proportion as large as p̂ = 0.52 for a sample of size 844 if the null hypothesis were true. #c. Probability of p̂ = 0.54 under Null Hypothesis# When p̂ = 0.54, we can calculate the z-score as follows: \[z = \frac{p̂ - p}{\sigma_{\hat{p}}} = \frac{0.54 - 0.50}{0.0172} \approx 2.33\] Using a standard normal table, the probability of observing a z-score greater than or equal to 2.33 is approximately equal to P(Z > 2.33) ≈ 0.010. This probability is low, so we would be surprised to observe a sample proportion as large as p̂ = 0.54 for a sample of size 844 if the null hypothesis were true. #d. Probability of p̂ = 0.59 under Null Hypothesis# When p̂ = 0.59, we can calculate the z-score as follows: \[z = \frac{p̂ - p}{\sigma_{\hat{p}}} = \frac{0.59 - 0.50}{0.0172} \approx 5.23\] Using a standard normal table, the probability of observing a z-score greater than or equal to 5.23 is approximately equal to P(Z > 5.23) ≈ 0.00000087. This probability is extremely low, so we would be very surprised to observe a sample proportion as large as p̂ = 0.59 for a sample of size 844 if the null hypothesis were true. It is compelling evidence that the null hypothesis H0: p = 0.50 is not true. #e. Generalizing the Conclusion to Adults Living in the United States# It would be inappropriate to directly generalize the conclusion of this test to adults living in the United States. The study under consideration was conducted in Belgium, which has a different culture and lifestyle. Furthermore, there is a difference in the adoption of mobile devices between Belgium and the United States. Therefore, the results obtained from this study should not be directly applied to the population of adults living in the United States without conducting a similar study in the United States or by performing a meta-analysis that combines the results of similar studies done in different countries.

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Key Concepts

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

Sampling Distribution
When we're talking about sampling distribution in statistics, we refer to the probability distribution of a sample statistic, such as the sample mean or sample proportion. In this example, we're focusing on the sampling distribution of \( \hat{p} \), which is the sample proportion.

For large sample sizes, according to the Central Limit Theorem, this distribution tends to be approximately normal. This is why when the sample from the study was taken, which consisted of 844 adults, it was large enough to assume a normal shape for its sampling distribution.

Important characteristics of the sampling distribution of \( \hat{p} \) include its shape, center, and spread:
  • Shape: With a sufficiently large sample, the distribution is approximately normal.
  • Center: The mean of the sampling distribution of \( \hat{p} \) is equal to the true population proportion, \( p \), under the null hypothesis. In this case, it's 0.50.
  • Variability: This is often expressed as the standard deviation of \( \hat{p} \), calculated using the formula: \[ \sigma_{\hat{p}} = \sqrt{ \frac{p(1-p)}{n} } \]where \( n \) is the sample size.
This mathematical foundation helps us assess the likelihood of observing different sample proportions, like 0.52, 0.54, or 0.59 in the original exercise.
Null Hypothesis
In statistical hypothesis testing, the null hypothesis \( H_0 \) is a statement that there is no effect or no difference. It's a default or starting assumption, which we aim to test against an alternative hypothesis.

In the context of our exercise, the null hypothesis asserts that the true population proportion \( p \) of adults who take their phones to bed is 0.50. This is represented as: \[ H_0: p = 0.50 \]

Testing the null hypothesis involves using sample data to calculate probabilities. We use z-scores to determine how far a sample proportion \( \hat{p} \) is from the assumed population proportion.

If the calculated probability (p-value) is very low, it suggests that observing such a sample proportion would be very unlikely if the null hypothesis were true. In the step-by-step solution, we used probabilities to assess whether values like 0.52, 0.54, and 0.59 could reasonably occur under \( H_0 \).
  • For \( \hat{p} = 0.52 \), the probability of it happening is relatively high, so it does not refute \( H_0 \).
  • For \( \hat{p} = 0.54 \), a smaller probability indicates it's more unlikely under \( H_0 \).
  • For \( \hat{p} = 0.59 \), the likelihood is extremely low, providing strong evidence against the null hypothesis.
Population Proportion
Population proportion, denoted usually by \( p \), is a key concept when dealing with statistics and surveys. It refers to the fraction or percentage of a population that possesses a certain characteristic of interest.

In the exercise at hand, we're investigating the population proportion of adults in Belgium who bring their phones to bed.

The purpose of the study is to understand if a majority does this behavior, and the null hypothesis suggests this proportion is 0.50, meaning equal to half the population.

To estimate a population proportion, researchers use the sample proportion \( \hat{p} \), which is the count of successes (people bringing phones to bed) divided by the sample size.
  • Given that \( \hat{p} = 0.59 \) in the study, the researchers inferred that it is significantly different from 0.50.

Variability in sampling distribution allows us to calculate how unusual the observed \( \hat{p} \) is concerning the assumed \( p \). Large differences between observed and assumed proportions point to a possible rejection of the null hypothesis, suggesting that the true \( p \) might differ.

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Most popular questions from this chapter

Refer to the instructions given prior to this exercise. The paper "College Students' Social Networking Experiences on Facebook" (Journal of Applied Developmental Psychology [2009]: \(227-238\) ) summarized a study in which 92 students at a private university were asked how much time they spent on Facebook on a typical weekday. You would like to estimate the average time spent on Facebook by students at this university.

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The report "A Crisis in Civic Education" (American Council of Trustees and Alumni, January 2016, www.goacta.org /images/download/A_Crisis_in_Civic_Education.pdf, retrieved November 30,2016 ) summarizes data from a survey of a representative sample of 1000 adult Americans regarding their understanding of U.S. government. Only 459 of the adults in the sample were able to give a correct response to a question asking them to choose a correct definition of the Bill of Rights from a list of five possible answers. Using a significance level of \(0.01,\) determine if there is convincing evidence that less than half of adult Americans could identify the correct definition of the Bill of Rights.

Refer to the instructions given prior to this exercise. The paper "College Students' Social Networking Experiences on Facebook" (Journal of Applied Developmental Psychology [2009]: \(227-238\) ) summarized a study in which 92 students at a private university were asked whether they used Facebook just to pass the time. Twenty-three responded "yes" to this question. The researchers were interested in estimating the proportion of students at this college who use Facebook just to pass the time.

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