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The poll referenced in the previous exercise (“Military 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.

Short Answer

Expert verified
The actual answer depends on the calculated Z-score and the corresponding P-value. If P-value is less than or equal to significance level 0.05, then we reject the null hypothesis and suggest that there is convincing evidence. If not, there is no sufficient evidence.

Step by step solution

01

State the Hypotheses

The null hypothesis \(H_0\) assumes there is no effect or difference. Here, it is stated as \(H_0: p = 0.5\), where p represents proportion. The alternative hypothesis \(H_a\) is \(H_a: p < 0.5\). This is a one-sided test where it is checked if the real proportion is less than 0.5.
02

Calculate Test Statistic

To calculate the test statistic, we can use the formula for z score for test of proportion which is \(Z = \frac{p - P}{\sqrt{\frac{P(1 - P)}{n}}}\). Here, p is the sample proportion, P is the known proportion (here 0.5), and n is the number of observations in the sample. Substitute the values to get the z-score: \(Z = \frac{0.43 - 0.5}{\sqrt{\frac{0.5(1 - 0.5)}{1000}}}\).
03

Find P-value

The P-value is the probability that the z-score is less than the calculated z-value under \(H_0\). To find it, a topic-specific statistical table (called a z-table) can be referred or a statistical software/tool can be used. P-value corresponds to the z-score calculated in step 2.
04

Compare P-value and significance level

Here the significance level \(\alpha\) is given as 0.05. If the P-value is less than or equal to \(\alpha\), then reject \(H_0\), otherwise fail to reject \(H0\).
05

Draw Conclusions

Based on the comparison made in the previous step, if \(H0\) is rejected, it suggests that there is convincing evidence that fewer than half of adult Americans would favor the drafting of women. If fail to reject \(H0\), it means that there is not enough evidence to support \(H_a\).

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Key Concepts

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

Null and Alternative Hypothesis
When performing hypothesis testing in statistics, understanding the null and alternative hypothesis is paramount. The null hypothesis (\(H_0\)) represents a statement of no effect or no difference that is to be tested—which in our exercise is the belief that half of the adult Americans (\(p = 0.5\)) would favor drafting women. Conversely, the alternative hypothesis (\(H_a\text{ or }H_1\text{ in some texts}\)) is the statement we suspect might be true instead of the null hypothesis. In our exercise, the alternative hypothesis is that fewer than half (\(p < 0.5\text{ or }H_a: p < 0.5\)) of adult Americans would favor the drafting of women. Ideally, the purpose of these hypotheses is to provide a clear statement of what is being tested, laying the groundwork for further statistical analysis.

In hypothesis testing, you either reject the null hypothesis in favor of the alternative hypothesis if there is sufficient evidence, or you fail to reject it when the evidence isn't strong enough. It's important to note that 'failing to reject' a hypothesis is not the same as 'accepting' it; rather, it means we don't have enough statistical evidence to favor the alternative.
Significance Level
The significance level in hypothesis testing is denoted as \(\alpha\) and represents the threshold for determining whether a statistical result is sufficiently unusual to reject the null hypothesis. In our exercise, the significance level is set at 0.05, indicating a 5% risk of concluding that a difference exists when there is no actual difference. This level of significance means we would consider results that have less than a 5% chance of occurring due to random variation in the data as statistically significant enough to reject the null hypothesis.

Choosing a significance level is an important decision in hypothesis testing as it affects how stringent and conservative the test will be. Commonly used significance levels are 0.05, 0.01, and 0.10. This choice should reflect the consequences of making a mistake in rejecting the null hypothesis, known as a 'Type I error'.
Test Statistic
The test statistic is a standardized value calculated from sample data during a hypothesis test. It is used to decide whether to reject the null hypothesis. It essentially measures how far the sample statistic is from the hypothesized value under the assumption that the null hypothesis is true. In the provided exercise, the z-score serves as the test statistic, calculated from the formula \(Z = \frac{p - P}{\sqrt{\frac{P(1 - P)}{n}}}\).

The test statistic can then be compared against a critical value from a statistical distribution (like the Normal distribution), which corresponds to the significance level chosen. A larger absolute value of the test statistic indicates a greater distance from the Null hypothesis, implying more evidence against it.
P-value
The p-value is a crucial concept in hypothesis testing, representing the probability of observing a test statistic as extreme as, or more extreme than, the one observed if the null hypothesis is true. In essence, it tells us how 'surprising' our data is under the assumption that the null hypothesis is correct. With a low p-value, typically below the significance level of 0.05, we would conclude that such data are unlikely to occur by random chance alone, therefore suggesting enough evidence to reject the null hypothesis.

In the exercise’s context, the p-value is found by looking up the calculated z-score in a z-table or by using a statistical software, and it corresponds to the likelihood of seeing a proportion less than 0.43 if the true proportion favoring the drafting of women is actually 0.5. If this calculated p-value is less than 0.05, this would be considered strong evidence against the null hypothesis.
Statistical Inference
Statistical inference encompasses the process of making conclusions about a population based on sample data. It is rooted in probability and deals with the estimation of parameters and testing of hypotheses. The goal of statistical inference is to infer properties about a population that might not be directly observable due to practical constraints on data collection.

In our exercise, the conclusion is drawn about the sentiment of the entire population of adult Americans on the drafting of women, based on the sample of 1000 survey responses. The inference process involves using the p-value and significance level to assess the strength of the evidence against the null hypothesis. Ideally, statistical inference guides decision-making in the presence of uncertainty and variability, and it should always be done with an understanding of the possible sampling errors and assumptions behind the data collection process.

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Most popular questions from this chapter

According to a survey of 1000 adult Americans conducted by Opinion Research Corporation, 210 of those surveyed said playing the lottery would be the most practical way for them to accumulate \(\$ 200,000\) in net wealth in their lifetime ("One in Five Believe Path to Riches Is the Lottery," San Luis Obispo Tribune, January 11, 2006 ). Although the article does not describe how the sample was selected, for purposes of this exercise, assume that the sample can be regarded as a random sample of adult Americans. Is there convincing evidence that more than \(20 \%\) of adult Americans believe that playing the lottery is the best strategy for accumulating \(\$ 200,000\) in net wealth?

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.

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 \(4.09 ?\)

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).

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

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