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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

The mean length of long-distance telephone calls placed with a particular phone company was known to be 7.3 minutes under an old rate structure. In an attempt to be more competitive with other long-distance carriers, the phone company lowered long-distance rates, thinking that its customers would be encouraged to make longer calls and thus that there would not be a big loss in revenue. Let \(\mu\) denote the mean length of long-distance calls after the rate reduction. What hypotheses should the phone company test to determine whether the mean length of long-distance calls increased with the lower rates?

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week (Ipsos Reid, August 9,2005 ). The mean of the sample observations was 12.7 hours. a. The sample standard deviation was not reported, but suppose that it was 5 hours. Carry out a hypothesis test with a significance level of .05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than 12.5 hours. b. Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of .05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than 12.5 hours. c. Explain why the null hypothesis was rejected in the test of Part (b) but not in the test of Part (a).

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?

The Economist collects data each year on the price of a Big Mac in various countries around the world. The price of a Big Mac for a sample of McDonald's restaurants in Europe in May 2009 resulted in the following Big Mac prices (after conversion to U.S. dollars): \(\begin{array}{llllll}3.80 & 5.89 & 4.92 & 3.88 & 2.65 & 5.57\end{array}\) \(\begin{array}{ll}6.39 & 3.24\end{array}\) The mean price of a Big Mac in the U.S. in May 2009 was \(\$ 3.57\). For purposes of this exercise, assume it is reasonable to regard the sample as representative of European McDonald's restaurants. Does the sample provide convincing evidence that the mean May 2009 price of a Big Mac in Europe is greater than the reported U.S. price? Test the relevant hypotheses using \(\alpha=.05\).

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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