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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.

Short Answer

Expert verified
The null hypothesis is rejected, so there is sufficient evidence at the 0.01 significance level to conclude that a majority of U.S. adults would not be bothered if the NSA collected records of personal calls.

Step by step solution

01

Set up the hypotheses

The null hypothesis \(H_0\): \(p = 0.5\) suggests that half of the U.S. adults would not be bothered if the NSA collected records of personal phone calls they made. The alternative hypothesis \(H_1\): \(p > 0.5\) suggests that more than 50% of U.S. adults would not be bothered.
02

Calculate the sample proportion

The sample proportion \(\hat{p}\) is given by the number of successes (those wouldn't be bothered) divided by the sample size, i.e., \(\hat{p}=331/502=0.659\).
03

Perform the test statistic calculation

The test statistic for a hypothesis test for a population proportion is \(Z = (\hat{p} - p_0)/\sqrt{p_0(1 - p_0)/n}\), where \(p_0\) is the proportion under the null hypothesis, \(n\) is the sample size. Substituting in the values gives \(Z = (0.659 - 0.5)/\sqrt{0.5(1 - 0.5)/502} = 7.45.\)
04

Determine the p-value and make a decision

The p-value for a Z-value of 7.45 is nearly 0, much less than the significance level of 0.01. Therefore, reject the null hypothesis. There is sufficient evidence at the 0.01 significance level to conclude that a majority of U.S. adults would not be bothered if the NSA collected records of personal calls.

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Key Concepts

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

Statistical Significance
Statistical significance is a term used to determine if the results observed in a study are unlikely to have occurred by chance alone. When conducting hypothesis tests, like in the provided exercise, researchers set a significance level, often denoted by \( \alpha \), which serves as a threshold for deciding when to reject the null hypothesis. In our example, the significance level is set at 0.01, indicating there is a 1% risk of concluding that a difference exists when there is no actual difference. If the p-value calculated from the test statistic is less than \( \alpha \), we say that the results are statistically significant and the null hypothesis is rejected. Crucially, this does not mean that the result is practically significant or important, only that it is statistically unlikely to be due to random variation in the sample.

In the context of the poll, finding statistical significance means we have enough evidence to reasonably conclude that the observed proportion represents the population's behavior rather than being a fluke of this specific sample.
Population Proportion
Population proportion, denoted as \( p \), is a measure that represents the fraction of the population that has a particular characteristic. In the exercise, we're concerned with the proportion of U.S. adults who wouldn't be bothered by NSA surveillance of personal calls. It’s important to distinguish between the actual population proportion and the sample proportion (denoted as \(\textasciicircum p\)). The latter is an estimate derived from a sample and is used to make inferences about the true population proportion.

The accuracy and reliability of these inferences depend on the size and randomness of the sample. If the sample size is too small, or if it's not representative of the population, our conclusions about the population proportion may be flawed. In the given solution, the sample proportion calculated from the poll data served as the foundation for testing hypotheses about the general sentiment of U.S. adults.
p-value
The p-value is a pivotal concept in hypothesis testing. It is a probability that expresses how extreme the observed data are, assuming the null hypothesis is true. In other words, it tells us how likely it is to obtain a result at least as extreme as the one observed if there were actually no effect in the population. A small p-value suggests that the observed data are unlikely under the null hypothesis and hence provide evidence against it.

In our exercise, a p-value nearly 0 indicated that obtaining such a high sample proportion of U.S. adults who aren't bothered by NSA's records collection would be extremely unlikely if the true proportion was really 50% or less. As a result, we reject the null hypothesis in favor of the alternative hypothesis because the p-value is smaller than the pre-set alpha level of 0.01.
Null Hypothesis
The null hypothesis, represented as \( H_0 \), is a default statement used in statistical hypothesis testing that there is no effect or no difference. It serves as a starting point for statistical significance testing and is presumed true until evidence suggests otherwise. In hypothesis testing, we never prove the null hypothesis; we can only provide enough evidence to reject it.

For the poll regarding NSA surveillance, the null hypothesis states that the proportion of U.S. adults who wouldn't be bothered (\( p \)) is equal to 0.5 - implying no majority either way. The exercise's question revolves around whether we can reject this hypothesis and assert with confidence that the true sentiment of the population deviates from this stated proportion.
Alternative Hypothesis
Contrasting the null hypothesis is the alternative hypothesis, labeled as \( H_1 \) or \( H_a \). It proposes what we suspect might be true instead and is the claim we are trying to find evidence for. The alternative hypothesis is only accepted if the data provides sufficient evidence to reject the null hypothesis.

In the case of our exercise, the alternative hypothesis declares that the proportion of people who would not be bothered by NSA call record collection (\( p \)) is greater than 0.5. When the test statistic led to a nearly 0 p-value, it indicated strong evidence in support of this alternative hypothesis, suggesting that it is likely that a majority of the population is indeed unconcerned by the surveillance in question.

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Most popular questions from this chapter

A manufacturer of hand-held calculators receives large shipments of printed circuits from a supplier. It is too costly and time-consuming to inspect all incoming circuits, so when each shipment arrives, a sample is selected for inspection. Information from the sample is then used to test \(H_{0}: p=.01\) versus \(H_{a}: p>.01\), where \(p\) is the actual proportion of defective circuits in the shipment. If the null hypothesis is not rejected, the shipment is accepted, and the circuits are used in the production of calculators. If the null hypothesis is rejected, the entire shipment is returned to the supplier because of inferior quality. (A shipment is defined to be of inferior quality if it contains more than \(1 \%\) defective circuits.) a. In this context, define Type I and Type II errors. b. From the calculator manufacturer's point of view, which type of error is considered more serious? c. From the printed circuit supplier's point of view, which type of error is considered more serious?

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 \(4.09 ?\)

The report "Highest Paying Jobs for \(2009-10\) Bachelor's Degree Graduates" (National Association of Colleges and Employers, February 2010 ) states that the mean yearly salary offer for students graduating with a degree in accounting in 2010 is \(\$ 48,722\). Suppose that a random sample of 50 accounting graduates at a large university who received job offers resulted in a mean offer of \(\$ 49,850\) and a standard deviation of \(\$ 3300 .\) Do the sample data provide strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the 2010 national average of \(\$ 48,722 ?\) Test the relevant hypotheses using \(\alpha=.05 .\)

The poll referenced in the previous exercise (β€œMilitary Draft Study," AP- lpsos, June 2005 ) also included the following question: "If the military draft were reinstated, would you favor or oppose drafting women as well as men?" Forty-three percent of the 1000 people responding said that they would favor drafting women if the draft were reinstated. Using a .05 significance level, carry out a test to determine if there is convincing evidence that fewer than half of adult Americans would favor the drafting of women.

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 \(150^{\circ} \mathrm{F}\), researchers will take 50 water samples at randomly selected times and record the temperature of each sample. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ} \mathrm{F}\) versus \(H_{d}: \mu>\) \(150^{\circ} \mathrm{F}\). In the context of this example, describe Type I and Type II errors. Which type of error would you consider more serious? Explain.

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