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Chapter 10: Problem 54

The article "Theaters Losing Out to Living Rooms" (San Luis Obispo Tribune, June 17,2005 ) states that movie attendance declined in \(2005 .\) The Associated Press found that 730 of 1,000 randomly selected adult Americans prefer to watch movies at home rather than at a movie theater. Is there convincing evidence that a majority of adult Americans prefer to watch movies at home? Test the relevant hypotheses using a 0.05 significance level.

Short Answer

Expert verified
Yes, there is convincing evidence that a majority of adult Americans prefer to watch movies at home. The p-value is effectively zero which is less than the 0.05 significance level, so the null hypothesis that at most 50% of adult Americans prefer to watch movies at home is rejected, supporting the alternate hypothesis.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis is that not the majority, or at the most 50%, of adult Americans prefer to watch movies at home (i.e \(p \leq 0.5\)). Formally, this is expressed as \(H_0: p \leq 0.5\). The alternative hypothesis is that over half of adult Americans do prefer to watch movies at home (i.e \(p > 0.5\)). Formally, this is expressed as \(H_1: p > 0.5\).
02

Compute the Test Statistic

The test statistic for a hypothesis test for a proportion is \(Z = (p虃 - p_0) / \sqrt{(p_0 * (1 - p_0)) / n}\), where \(p虃\) is the sample proportion, \(p_0\) is the assumed population proportion under the null hypothesis, and \(n\) is the sample size. In this case, \(p虃 = 730/1000 = 0.73\), \(p_0 = 0.5\), and \(n = 1000\). Plugging in these numbers we find \(Z = (0.73 - 0.5) / \sqrt{(0.5 * (1 - 0.5)) / 1000} \approx 4.60\).
03

Find the P-value

The P-value is obtained by referring to the normal distribution table for the obtained test statistic value. Since this is a one-tailed test (greater than), the P-value is the probability that a standard normal random variable is greater than the Z score. For a Z score of 4.60, this probability is effectively zero.
04

Make Your Conclusion

Since the P-value is less than our significance level of 0.05, we must reject the null hypothesis at this level. Therefore, there is sufficient evidence at the 95% confidence level to conclude that a majority of adult Americans prefer to watch movies at home as opposed to going to a theater.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a foundational concept. It represents a statement or assumption that is generally accepted as true until shown otherwise. In this exercise, the null hypothesis (\(H_0\)) is formulated as there being no majority preference; specifically, that at most 50% of adult Americans prefer watching movies at home. This is expressed mathematically as \(H_0: p \leq 0.5\).
  • The null hypothesis is essentially the default position.
  • We aim to gather evidence to reject or fail to reject it.
The reason we focus on potentially rejecting the null hypothesis is to determine if there is strong enough evidence to support an alternative claim鈥攊n this case, that more than 50% of Americans have a different movie-watching preference.
Sample Proportion
When conducting a hypothesis test, one critical component is the sample proportion, denoted as \(\hat{p}\). This value represents the proportion of the sample that possesses a particular characteristic鈥攊n our scenario, it's the fraction of surveyed individuals who prefer watching movies at home. From the given data, 730 out of 1000 people prefer home viewing, resulting in a sample proportion of \(\hat{p} = 730/1000 = 0.73\).
  • The sample proportion is an estimate of the true population proportion.
  • It helps in determining whether the observed effect in the sample reflects a real-world phenomenon.
This proportion is juxtaposed against the hypothesized population proportion under the null hypothesis, helping us calculate the test statistic that guides our decision-making in hypothesis testing.
Significance Level
The significance level, also known as alpha \((\alpha)\), is a threshold set by the researcher to determine whether to reject the null hypothesis. It dictates the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true. In this example, the significance level is set at 0.05, or 5%.
  • A common choice, it indicates tolerance for a 5% chance of incorrectly rejecting the null.
  • It influences the outcome of hypothesis tests by setting a benchmark for the P-value.
When the calculated P-value is less than 0.05, as in this scenario, it implies that the observed data is highly improbable under the null hypothesis. Consequently, we reject the null, supporting the alternative hypothesis that a majority of Americans prefer home movie viewing.

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Most popular questions from this chapter

The article "The Benefits of Facebook Friends: Social Capital and College Students' Use of Online Social Network Sites" (Journal of Computer-Mediated Communication [2007]: \(1143-1168\) ) describes a study of \(n=286\) undergraduate students at Michigan State University. Suppose that it is reasonable to regard this sample as a random sample of undergraduates at Michigan State. You want to use the survey data to decide if there is evidence that more than \(75 \%\) of the students at this university have a Facebook page that includes a photo of themselves. Let \(p\) denote the proportion of all Michigan State undergraduates who have such a page. (Hint: See Example 10.10\()\) a. Describe the shape, center, and spread of the sampling distribution of \(\hat{p}\) for random samples of size 286 if the null hypothesis \(H_{0}: p=0.75\) is true. b. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.83\) for a sample of size 286 if the null hypothesis \(H_{0}: p=0.75\) were true? Explain why or why not. c. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.79\) for a sample of size 286 if the null hypothesis \(H_{0}: p=0.75\) were true? Explain why or why not. d. The actual sample proportion observed in the study was \(\hat{p}=0.80 .\) Based on this sample proportion, is there convincing evidence that the null hypothesis \(H_{0}: p=\) 0.75 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.

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Do state laws that allow private citizens to carry concealed weapons result in a reduced crime rate? The author of a study carried out by the Brookings Institution is reported as saying, "The strongest thing I could say is that I don't see any strong evidence that they are reducing crime" (San Luis Obispo Tribune, January 23,2003 ). a. Is this conclusion consistent with testing \(H_{0}:\) concealed weapons laws reduce crime versus \(H_{a}:\) concealed weapons laws do not reduce crime or with testing \(H_{0}:\) concealed weapons laws do not reduce crime versus \(H_{a}:\) concealed weapons laws reduce crime Explain. b. Does the stated conclusion indicate that the null hypothesis was rejected or not rejected? Explain.

The paper "I Smoke but I Am Not a Smoker" (Journal of American College Health [2010]: 117-125) describes a survey of 899 college students who were asked about their smoking behavior. Of the students surveyed, 268 classified themselves as nonsmokers, but said yes when asked later in the survey if they smoked. These students were classified as "phantom smokers," meaning that they did not view themselves as smokers even though they do smoke at times. The authors were interested in using these data to determine if there is convincing evidence that more than \(25 \%\) of college students fall into the phantom smoker category.

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