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Chapter 10: Problem 38

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

The city council in a large city has become concerned about the trend toward exclusion of renters with children in apartments within the city. The housing coordinator has decided to select a random sample of 125 apartments and determine for each whether children are permitted. Let \(p\) be the proportion of all apartments that prohibit children. If the city council is convinced that \(p\) is greater than 0.75 , it will consider appropriate legislation. a. If 102 of the 125 sampled apartments exclude renters with children, would a level .05 test lead you to the conclusion that more than \(75 \%\) of all apartments exclude children? b. What is the power of the test when \(p=.8\) and \(\alpha=.05 ?\)

A television manufacturer claims that (at least) \(90 \%\) of its TV sets will need no service during the first 3 years of operation. A consumer agency wishes to check this claim, so it obtains a random sample of \(n=100\) purchasers and asks each whether the set purchased needed repair during the first 3 years after purchase. Let \(\hat{p}\) be the sample proportion of responses indicating no repair (so that no repair is identified with a success). Let \(p\) denote the actual proportion of successes for all sets made by this manufacturer. The agency does not want to claim false advertising unless sample evidence strongly suggests that \(p<.9 .\) The appropriate hypotheses are then \(H_{0}: p=.9\) versus \(H_{a}: p<.9\). a. In the context of this problem, describe Type I and Type II errors, and discuss the possible consequences of each. b. Would you recommend a test procedure that uses \(\alpha=.10\) or one that uses \(\alpha=.01 ?\) Explain.

According to a Washington Post-ABC News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bothered if the National Security Agency collected records of personal telephone calls they had made. Is there sufficient evidence to conclude that a majority of U.S. adults feel this way? Test the appropriate hypotheses using a .01 significance level.

The paper "MRI Evaluation of the Contralateral Breast in Women with Recently Diagnosed Breast Cancer" (New England Journal of Medicine \([2007]: 1295-1303)\) describes a study of the use of MRI (Magnetic Resonance Imaging) exams in the diagnosis of breast cancer. The purpose of the study was to determine if MRI exams do a better job than mammograms of determining if women who have recently been diagnosed with cancer in one breast have cancer in the other breast. The study participants were 969 women who had been diagnosed with cancer in one breast and for whom a mammogram did not detect cancer in the other breast. These women had an MRI exam of the other breast, and 121 of those exams indicated possible cancer. After undergoing biopsies, it was determined that 30 of the 121 did in fact have cancer in the other breast, whereas 91 did not. The women were all followed for one year, and three of the women for whom the MRI exam did not indicate cancer in the other breast were subsequently diagnosed with cancer that the MRI did not detect. The accompanying table summarizes this information. Suppose that for women recently diagnosed with cancer in only one breast, the MRI is used to decide between the two "hypotheses" \(H_{0}\) : woman has cancer in the other breast \(H_{a}:\) woman does not have cancer in the other breast (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) a. One possible error would be deciding that a woman who does have cancer in the other breast is cancerfree. Is this a Type I or a Type II error? Use the information in the table to approximate the probability of this type of error. b. There is a second type of error that is possible in this setting. Describe this error and use the information in the given table to approximate the probability of this type of error.

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above \(150^{\circ} \mathrm{F}\), a scientist will take 50 water samples at randomly selected times and will record the water temperature of each sample. She will then use a \(z\) statistic $$ z=\frac{\bar{x}-150}{\frac{\sigma}{\sqrt{n}}} $$ to decide between the hypotheses \(H_{0}: \mu=150\) and \(H_{a}: \mu>150,\) where \(\mu\) is the mean temperature of discharged water. Assume that \(\sigma\) is known to be 10 . a. Explain why use of the \(z\) statistic is appropriate in this setting. b. Describe Type I and Type II errors in this context. \(c\). The rejection of \(H_{0}\) when \(z \geq 1.8\) corresponds to what value of \(\alpha\) ? (That is, what is the area under the \(z\) curve to the right of \(1.8 ?\) ) d. Suppose that the actual value for \(\mu\) is 153 and that \(H_{0}\) is to be rejected if \(z \geq 1.8 .\) Draw a sketch (similar to that of Figure 10.5 ) of the sampling distribution of \(\bar{x},\) and shade the region that would represent \(\beta\), the probability of making a Type II error. e. For the hypotheses and test procedure described, compute the value of \(\beta\) when \(\mu=153\). f. For the hypotheses and test procedure described, what is the value of \(\beta\) if \(\mu=160\) ? g. What would be the conclusion of the test if \(H_{0}\) is rejected when \(z \geq 1.8\) and \(\bar{x}=152.4\) ? What type of error might have been made in reaching this conclusion?
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