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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 1000 randomly selected adult Americans preferred to watch movies at home rather than at a movie theater. Is there convincing evidence that the majority of adult Americans prefer to watch movies at home? Test the relevant hypotheses using a .05 significance level.

Short Answer

Expert verified
Yes, there is convincing evidence at the 0.05 significance level that the majority of adult Americans prefer to watch movies at home over going to a movie theater.

Step by step solution

01

Formulate Hypotheses

Let \( p \) be the proportion of all adults who prefer to watch movies at home. The null hypothesis \(H_0\) is that the majority of adult Americans do not prefer to watch movies at home, so \( p \leq 0.5 \). The alternative hypothesis \(H_1\) is that the majority do prefer to watch movies at home, so \( p > 0.5 \).
02

Compute Test Statistic

For a sample of size \(n = 1000\), the sample proportion \(\hat{p} = 730/1000 = 0.73\). The test statistic is computed as \( Z = (\hat{p} - p_0) / \sqrt{(p_0*(1-p_0)/n)} \), where \(p_0 = 0.5\) under the null hypothesis. Substituting values, we have: \( Z = (0.73 - 0.5) / \sqrt{(0.5*(1-0.5)/1000)} = 9.2 \).
03

Acceptance / Rejection of Null Hypothesis

As we're dealing with a one-tailed test with significance level of 0.05, we have to calculate the critical Z value. For 0.05 in the one-tailed test, the critical value is approximately 1.645. As our test statistic (\(Z = 9.2\)) is greater than the critical value (\(Z = 1.645\)), we reject the null hypothesis.
04

Interpretation

Thus, there is strong evidence at the 0.05 significance level to suggest that the majority of adult Americans do prefer to watch movies at home over going to a movie theater.

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Key Concepts

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

Significance Level
In hypothesis testing, the significance level is crucial. It's a threshold we set for how much error we're willing to accept when deciding whether to reject the null hypothesis.
In simpler terms, it's the probability of making a mistake by rejecting a true null hypothesis.
  • Typically, a significance level is denoted by the symbol \( \alpha \).
  • Common choices for \( \alpha \) are 0.05, 0.01, and 0.10.
  • A 0.05 significance level means there's a 5% chance you'll mistakenly reject the null hypothesis.
In our example, the significance level is set at 0.05. This means we have a standard 5% risk of ruling out the null hypothesis incorrectly. By setting this level, we decide how strict we will be with our evidence, ensuring that findings are robust and reliable.
Null Hypothesis
The null hypothesis, often symbolized as \( H_0 \), is the default assumption in hypothesis testing.
It asserts that any observed difference is due to chance alone and there's no effect or relationship.
  • The purpose of \( H_0 \) is to provide a baseline or standard to compare against.
  • In our movie preference study, \( H_0 \) states that the majority of adult Americans do not prefer to watch movies at home, mathematically represented as \( p \leq 0.5 \).
  • We always test \( H_0 \) to see if there's enough evidence to refute it.
The null hypothesis acts as a skeptic, assuming the status quo until we provide convincing data suggesting otherwise.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_1 \) or \( H_a \), proposes a new state of affairs, in contrast to the null hypothesis.
It's what researchers aim to support with their data.
  • \( H_1 \) is typically set up to capture the effect or relationship under investigation.
  • In our scenario, \( H_1 \) suggests that the majority of adult Americans do prefer to watch movies at home, expressed as \( p > 0.5 \).
  • The acceptance of \( H_1 \) implies that the evidence significantly contradicts \( H_0 \).
If our testing leads to rejecting the null hypothesis, we're essentially accepting the alternative hypothesis as a more accurate reflection of reality.
Test Statistic
A test statistic is a standardized value that helps decide whether to reject the null hypothesis.
It measures how far the sample data deviates from the null hypothesis assumption.
  • In many cases, the test statistic follows a known distribution, such as the normal distribution.
  • For our study, a Z-test is used, resulting in a Z statistic of the form \( Z = (\hat{p} - p_0) / \sqrt{(p_0(1 - p_0)/n)} \).
  • This formula incorporates the difference between the sample proportion \( \hat{p} = 0.73 \) and the null hypothesis proportion \( p_0 = 0.5 \).
  • The calculated Z statistic is 9.2, which tells us how many standard deviations our sample proportion is from the population proportion under \( H_0 \).
The test statistic provides a way to evaluate whether the observed data is significantly different from what we would expect under the null hypothesis.
Critical Value
In hypothesis testing, the critical value is a threshold that the test statistic must exceed to reject the null hypothesis.
It defines the boundary for the likelihood of data corresponding to the null hypothesis.
  • Critical values are derived from the chosen significance level and the statistical distribution of the test statistic.
  • In a normal distribution, the critical value can be looked up in Z-tables when conducting a Z-test.
  • Here, for a significance level of 0.05 in a one-tailed test, the critical value is approximately 1.645.
  • The test statistic (9.2) significantly exceeds this critical value, prompting the rejection of the null hypothesis.
Therefore, when the test statistic surpasses the critical value, it indicates that the sample data is inconsistent with the null hypothesis, supporting the alternative hypothesis instead.

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Most popular questions from this chapter

Occasionally, warning flares of the type contained in most automobile emergency kits fail to ignite. A consumer advocacy group wants to investigate a claim against a manufacturer of flares brought by a person who claims that the proportion of defective flares is much higher than the value of .1 claimed by the manufacturer. A large number of flares will be tested, and the results will be used to decide between \(H_{0}: p=.1\) and \(H_{a}: p>.1,\) where \(p\) represents the proportion of defective flares made by this manufacturer. If \(H_{0}\) is rejected, charges of false advertising will be filed against the manufacturer. a. Explain why the alternative hypothesis was chosen to be \(H_{a}: p>.1 .\) b. In this context, describe Type I and Type II errors, and discuss the consequences of each.

Assuming a random sample from a large population, for which of the following null hypotheses and sample sizes \(n\) is the large-sample \(z\) test appropriate: a. \(H_{0}: p=.2, n=25\) b. \(H_{0}: p=.6, n=210\) c. \(H_{0}: p=.9, n=100\) d. \(H_{0}: p=.05, n=75\)

A credit bureau analysis of undergraduate students credit records found that the average number of credit cards in an undergraduate's wallet was 4.09 ("Undergraduate Students and Credit Cards in 2004," Nellie Mae, May 2005\()\). It was also reported that in a random sample of 132 undergraduates, the sample mean number of credit cards that the students said they carried was 2.6. The sample standard deviation was not reported, but for purposes of this exercise, suppose that it was 1.2 . Is there convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of \(4.09 ?\)

Explain why the statement \(\bar{x}=50\) is not a legitimate hypothesis.

A county commissioner must vote on a resolution that would commit substantial resources to the construction of a sewer in an outlying residential area. Her fiscal decisions have been criticized in the past, so she decides to take a survey of constituents to find out wherher they favor spending money for a sewer system. She will vote to appropriate funds only if she can be reasonably sure that a majority of the people in her district favor the measure. What hypotheses should she test?

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