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The article "Poll Finds Most Oppose Return to Draft, Wouldn't Encourage Children to Enlist" (Associated Press, December 18,2005 ) reports that in a random sample of 1000 American adults, 700 indicated that they oppose the reinstatement of a military draft. Is there convincing evidence that the proportion of American adults who oppose reinstatement of the draft is greater than two-thirds? Use a significance level of .05 .

Short Answer

Expert verified
Without further calculations, we cannot definitively say whether we reject the null hypothesis or not. The step-by-step process described above will enable us to draw a statistical conclusion based on the calculated P-value.

Step by step solution

01

State the Hypotheses

There are two hypotheses we need: the null hypothesis and the alternative hypothesis. The null hypothesis asserts that the proportion is at most two-thirds (\( p \leq 2/3 \)). The alternative hypothesis asserts that the proportion is greater than two-thirds (\( p > 2/3 \)).
02

Define the Test Statistic

We use the formula \(Z = (p̂ - p0) / \sqrt{p0(1-p0)/n}\) to calculate the test statistic, where \(p̂\) represents the sample proportion, \(p0\) represents a hypothesized population proportion, and \(n\) is the sample size. Here, \(p0\) is two-thirds, \(p̂\) is 0.7 (700 out of 1000 adults), and \(n\) is 1000.
03

Calculate Test Statistic

Substitute our values into the test statistic equation: \(Z = (0.7 - 2/3) / \sqrt{(2/3)(1-2/3)/1000}\). Calculation yields a Z score.
04

Determine the P-value

The P-value corresponds to the probability of obtaining a Z-score equal to or more extreme than the calculated Z-score under the null hypothesis. This can be determined using a standard normal distribution table or a statistical software package.
05

Draw Conclusion

If the P-value is less than the significance level (0.05 in this case), then we reject the null hypothesis. If it's greater than or equal to 0.05, then we do not reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
The null hypothesis in statistics is a statement of no effect or no difference used as a benchmark for testing the statistical evidence. It generally assumes that observations are the result of pure chance or that a certain parameter (like a population mean or proportion) is equal to a specified value. For instance, if we want to test whether a coin is fair, the null hypothesis would state that the probability of getting heads (\( p \) ) is 0.5. In the exercise, the null hypothesis (\( H_0 \) ) posits that the proportion of American adults opposing the draft is at most two-thirds (\( p leq \frac{2}{3} \) ). It's crucial to define the null hypothesis precisely as it sets the stage for statistical testing.
Alternative Hypothesis
Contrasting the null hypothesis, the alternative hypothesis (\( H_A \) or \( H_1 \) ) represents a researcher's conjecture that there is a statistically significant effect or difference between groups, or that a certain parameter differs from the null hypothesis's value. The alternative hypothesis cannot be tested directly; it is accepted by rejecting the null hypothesis. In our study, the alternative hypothesis suggests that more than two-thirds of American adults oppose the draft (\( p > \frac{2}{3} \) ). Whether or not the alternative hypothesis is supported depends on the outcome of the significance test.
Significance Level
The significance level, often denoted by alpha (\( alpha \) ), is the threshold for determining the statistical significance of a test. It is the probability of rejecting the null hypothesis when it is actually true—an error known as a Type I error. Common significance levels are 0.05, 0.01, and 0.10. Choosing a lower alpha value makes the test more stringent. In the provided exercise, a significance level of 0.05 is used, which means we are willing to accept a 5% chance of incorrectly rejecting a true null hypothesis.
Test Statistic
The test statistic is a standardized value calculated from sample data during a hypothesis test. It measures how far the data are from the null hypothesis. Different tests have different test statistics (like Z, t, F, Chi-square, etc.), but they all serve the purpose of helping to determine whether to reject the null hypothesis. In this context, the test statistic is a Z-score which quantifies the standard deviations separating the observed sample proportion from the hypothesized population proportion under the null hypothesis. It is calculated using the formula provided in the exercise step 2.
P-value
The P-value is the probability that the test statistic will take on a value that is at least as extreme as the one derived from the sample data, assuming the null hypothesis is true. It's a measure of the strength of the evidence against the null hypothesis. The smaller the P-value, the stronger the evidence that you should reject the null hypothesis. In the given problem, by comparing the calculated P-value to the significance level, we determine whether the evidence is sufficient to support the alternative hypothesis—that a greater proportion of the population is opposed to the draft than the null hypothesis would suggest.
Sample Proportion
The sample proportion (\( p̂ \) ) is a statistic that estimates the proportion of items in a population that possess a certain attribute, based on a sample taken from the population. It is calculated by dividing the number of items with the attribute by the total number of items in the sample. For the problem given, the sample proportion is derived from the survey data, wherein 700 out of 1000 adult respondents opposed the draft, resulting in a sample proportion of 0.7. This figure is essential in computing the test statistic and subsequently in conducting the hypothesis test.

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Most popular questions from this chapter

In \(2006,\) Boston Scientific sought approval for a new heart stent (a medical device used to open clogged arteries) called the Liberte. This stent was being proposed as an alternative to a stent called the Express that was already on the market. The following excerpt is from an article that appeared in The Wall Street Journal (August 14,2008 ): Boston Scientific wasn't required to prove that the Liberte was 'superior' than a previous treatment, the agency decided - only that it wasn't "inferior' to Express. Boston Scientific proposed - and the FDA okayed - a benchmark in which Liberte could be up to three percentage points worse than Express meaning that if \(6 \%\) of Express patients' arteries reclog, Boston Scientific would have to prove that Liberte's rate of reclogging was less than \(9 \%\). Anything more would be considered 'inferior.'... In the end, after nine months, the Atlas study found that 85 of the patients suffered reclogging. In comparison, historical data on 991 patients implanted with the Express stent show a \(7 \%\) rate. Boston \(S\) cientific then had to answer this question: Could the study have gotten such results if the Liberte were truly inferior to Express?" Assume a \(7 \%\) reclogging rate for the Express stent. Explain why it would be appropriate for Boston Scientific to carry out a hypothesis test using the following hypotheses: \(H_{0}: p=.10\) \(H_{a}: p<.10\) where \(p\) is the proportion of patients receiving Liberte stents that suffer reclogging. Be sure to address both the choice of the hypothesized value and the form of the alternative hypothesis in your explanation.

In a survey of 1005 adult Americans, \(46 \%\) indicated that they were somewhat interested or very interested in having web access in their cars (USA Today, May I. 2009 ). Suppose that the marketing manager of a car manufacturer claims that the \(46 \%\) is based only on a sample and that \(46 \%\) is close to half, so there is no reason to believe that the proportion of all adult Americans who want car web access is less than \(.50 .\) Is the marketing manager correct in his claim? Provide statistical evidence to support your answer. For purposes of this exercise, assume that the sample can be considered as representative of adult Americans.

Researchers at the University of Washington and Harvard University analyzed records of breast cancer screening and diagnostic evaluations ("Mammogram Cancer Scares More Frequent than Thought," USA Today, April 16, 1998). Discussing the benefits and downsides of the screening process, the article states that, although the rate of false-positives is higher than previously thought, if radiologists were less aggressive in following up on suspicious tests, the rate of false-positives would fall but the rate of missed cancers would rise. Suppose that such a screening test is used to decide between a null hypothesis of \(H_{0}:\) no cancer is present and an alternative hypothesis of \(H_{a}:\) cancer is present. (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) a. Would a false-positive (thinking that cancer is present when in fact it is not) be a Type I error or a Type II error? b. Describe a Type I error in the context of this problem, and discuss the consequences of making a Type I error. c. Describe a Type II error in the context of this problem, and discuss the consequences of making a Type II error. d. What aspect of the relationship between the probability of Type I and Type II errors is being described by the statement in the article that if radiologists were less aggressive in following up on suspicious tests, the rate of false-positives would fall but the rate of missed cancers would rise?

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. \(H_{0}: p=.05, n=75\)

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