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Chapter 10: Problem 29

The poll referenced in the previous exercise ("Military Draft Study," AP- Ipsos, 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
There's convincing evidence that less than half of adult Americans would favor the drafting of women.

Step by step solution

01

Stating the hypotheses

The null hypothesis \(H_0\) is that the true proportion we鈥檙e testing is half (0.5), while the alternative hypothesis \(H_a\) is that the true proportion is less than half, meaning that less than half of adult Americans would favor the drafting of women.
02

Collect and summarize the data

Here, the sample proportion \(\hat{p}\) is \(0.43\). The sample size \(n\) is \(1000\). The sample proportionis a useful statistic for estimating the population proportion.
03

Calculate the test statistic

For this, a test statistic \(z\) is calculated using the formula: \( z = \frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\). With \( p_0 = 0.5\) (the null hypothesis), \(n= 1000\) and \(\hat{p} = 0.43\), the result will be \( z = \frac{0.43 - 0.5}{\sqrt{\frac{0.5 * 0.5}{1000}}} = -4.47\).
04

Compute the P-value

The level of significance for this test is \(0.05\). The P-value is the probability that the z-score is less than \( -4.47\) under the null hypothesis, which can be looked up in a z-distribution table or calculated using statistical software. The P-value is virtually 0, which is less than the level of significance.
05

Conclusion

Given that the P-value is less than the significance level, the null hypothesis is rejected, meaning that there's convincing evidence that less than half of adult Americans would favor the drafting of women.

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Key Concepts

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

Significance Level
The significance level, commonly symbolized as \(\alpha\), plays a crucial role in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. Think of it as a threshold that helps in deciding whether the evidence we have is strong enough to reject the null hypothesis.
In our exercise, the significance level is set at \(0.05\). This is a typical convention. It translates to a 5% risk of concluding that a difference exists when there is no actual difference.
Choosing a significance level involves:
  • Balancing risks: A lower \(\alpha\) reduces Type I error (incorrectly rejecting a true null hypothesis) but increases the risk of a Type II error (failing to reject a false null hypothesis).
  • Contextual relevance: In areas like medicine, smaller levels (e.g., \(0.01\)) might be preferred for higher accuracy.
In summary, the significance level is about setting parameters for how much random chance plays in our decision-making.
P-value
The P-value is another cornerstone of hypothesis testing. It tells us the probability of observing a test statistic as extreme, or more extreme, than the one calculated, assuming the null hypothesis is true.
This value helps in decision-making: if the P-value is less than the significance level, it means the observed data is highly improbable under the null hypothesis, leading us to reject it.
In this exercise, the P-value is almost 0, much lower than the significance level \(\alpha = 0.05\). This strong evidence against the null hypothesis suggests that it's unlikely the proportion favoring drafting women is 50%. Thus, we reject the null hypothesis and accept the alternative hypothesis.
  • Interpretation: A small P-value signals significant results, often leading us away from the null hypothesis.
  • Practical usage: P-values cannot tell us the size or importance of the effect, only the probability related to our hypo-testing.
Essentially, P-values offer a lens on the likelihood that your results do not occur by random sampling error.
Proportion Test
A proportion test is a statistical method used to determine if two proportions differ significantly from each other. It's especially useful when you're comparing a sample proportion with a known population proportion.
In our exercise, we conducted a one-sample proportion test. The hypothesis was whether the true proportion of adults favoring drafting women is less than 0.5.
This involves:
  • Null Hypothesis \(H_0\): Proposes no effect or status quo, here, suggesting the proportion is 0.5.
  • Sample Proportion \(\hat{p}\): The proportion observed in the sample, which was 0.43.
  • Test Statistic \(z\): Calculated to assess how far our sample finds away from the null proportion, leading to a \(z=-4.47\) in our case.
  • P-value: This is determined through the test statistic to conclude the hypothesis test.
In conclusion, the proportion test provides a framework to test claims or assumptions about population proportions, using sample data to infer and generalize findings to a larger population.

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Most popular questions from this chapter

Researchers have postulated that because of differences in diet, Japanese children have a lower mean blood cholesterol level than U.S. children do. Suppose that the mean level for U.S. children is known to be 170 . Let \(\mu\) represent the true mean blood cholesterol level for Japanese children. What hypotheses should the researchers test?

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?

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 \(H_{0}=\pi=.05\) versus \(H_{a}: \pi>.05\), where \(\pi\) is the true 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 \(5 \%\) 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?

Much concern has been expressed in recent years regarding the practice of using nitrates as meat preservatives. In one study involving possible effects of these chemicals, bacteria cultures were grown in a medium containing nitrates. The rate of uptake of radio-labeled amino acid was then determined for each culture, yielding the following observations: \(\begin{array}{llllllll}7251 & 6871 & 9632 & 6866 & 9094 & 5849 & 8957 & 7978\end{array}\) \(\begin{array}{lllllll}7064 & 7494 & 7883 & 8178 & 7523 & 8724 & 7468\end{array}\) Suppose that it is known that the true average uptake for cultures without nitrates is 8000 . Do the data suggest that the addition of nitrates results in a decrease in the true average uptake? Test the appropriate hypotheses using a significance level of \(.10 .\)

In a representative sample of 1000 adult Americans, only 430 could name at least one justice who is currently serving on the U.S. Supreme Court (Ipsos, January 10,2006 ). Using a significance level of \(.01\), carry out ? hypothesis test to determine if there is convincing evidence to support the claim that fewer than half of adult Americans can name at least one justice currently serving on the Supreme Court.
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