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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 is convincing evidence with a significance level of \(0.05\) to support the claim that less than half of the adult Americans would favor drafting women.

Step by step solution

01

State the Hypotheses

We will start by identifying the null and alternative hypotheses. In our case, the null hypothesis \(H_0\) is that 50% (0.5) or more of the American population would favor drafting of women, that is \(p \geq 0.5\). The alternative hypothesis \(H_1\) is the claim we're testing: fewer than half of adult Americans would favor the drafting of women, translated it is \(p < 0.5\).
02

Choose the Test and Find the Critical Value

As the problem indicates, we use the significance level of \(\alpha = 0.05\). Since the alternative hypothesis is \(p<0.5\), it's a left-tailed test. So, the critical value is determined from the Z-square distribution table, which corresponds to the significance level of \(0.05\), denoted as \(z_{0.05}\). The critical value from the Z-square distribution table for \(\alpha = 0.05\) is approximately \(-1.645\).
03

Calculate Test Statistic

The test statistic is given by \(Z = \frac{{\hat{p}-p_0}}{{\sqrt{p_0(1-p_0)/n}}}\) where \(\hat{p}\) is the sample proportion, \(p_0\) is the population proportion under the null hypothesis, and \(n\) is the sample size. Substitute the given values: \(\hat{p} = 0.43\) (or 43%), \(p_0 = 0.5\) from null hypothesis, and \(n = 1000\). The calculated value of Z is about \(-2.39\).
04

Determine conclusion

Since the calculated Z value of \(-2.39\) is less than the critical Z of \(-1.645\), the test statistic falls in the rejection region. Hence, we reject the null hypothesis. This means that we have enough evidence to support the claim that less than half of the adult Americans would favor drafting women.

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Key Concepts

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

Null Hypothesis
When beginning a hypothesis test, the first step is to define the null hypothesis. The null hypothesis, often denoted by \(H_0\), serves as a statement of no effect or no difference. It's what we assume to be true until proven otherwise. In the context of the problem, the null hypothesis is that 50% or more of Americans would favor drafting women, expressed as \(p \geq 0.5\). This hypothesis suggests there is no significant evidence to show a preference below half. The null hypothesis offers a baseline for comparison, and it's what we test against with our statistical procedures.
It's important to note:
  • The null hypothesis often contains an equality (e.g., \(\geq, \leq, =\)), which assumes no change from the status quo.
  • Rejecting the null hypothesis suggests that there is enough statistical evidence to support the alternative hypothesis.
  • Failing to reject the null hypothesis does not prove it true, it simply means insufficient evidence to support the alternative.
Alternative Hypothesis
The alternative hypothesis, denoted by \(H_1\) or \(H_a\), is what researchers want to prove. For the exercise at hand, it states that fewer than half of adult Americans would favor drafting women. This is mathematically written as \(p < 0.5\). It represents a condition that indicates a significant effect or a difference. If our test results lead to rejecting the null hypothesis, it supports the claim of the alternative hypothesis.
Considerations related to the alternative hypothesis include:
  • It is typically what you believe is true or what you're trying to find evidence for.
  • It contains a "less than", "greater than", or "not equal to" statement, showing a change or difference.
  • Choosing the correct alternative hypothesis is crucial as it directly impacts the direction of the test.
The alternative hypothesis provides the basis for our Z-test, setting the direction (left-tailed, in this case).
Significance Level
The significance level, denoted by \(\alpha\), is the probability of rejecting the null hypothesis when it is true. In hypothesis testing, it measures the risk of making a Type I error 鈥� falsely identifying an effect or difference. A common choice for \(\alpha\) is 0.05, as used in this example, representing a 5% risk.
The significance level plays several critical roles:
  • Determines the cutoff for deciding whether test results are statistically significant.
  • Guides the rejection region on the distribution curve, helping to show where the sample results are unlikely under the null hypothesis.
  • Balancing \(\alpha\) is vital; a smaller \(\alpha\) means more stringent evidence is required to reject the null.
For our test, \(\alpha = 0.05\) thus means results are considered statistically significant if they fall below this threshold, supporting evidence against the null hypothesis.
Z-test
A Z-test is a type of hypothesis test used when the population variance is known or when the sample size is large (typically \(n > 30\)). Its purpose is to determine if there is a significant difference between population means or proportions.
Key aspects of a Z-test include:
  • With large sample sizes, the central limit theorem allows us to assume the sample mean approximates a normal distribution.
  • The formula for the test statistic in a proportion test is \(Z = \frac{{\hat{p}-p_0}}{{\sqrt{p_0(1-p_0)/n}}}\), where \( \hat{p} \) is the sample proportion, \( p_0 \) is the population proportion under the null hypothesis, and \( n \) is the sample size.
  • The calculated Z value helps determine whether to reject the null hypothesis by comparing it with critical values from the standard normal distribution.
In this exercise, the calculated Z-value of -2.39, compared with the critical value threshold of -1.645, indicates a significant difference, leading to rejection of the null hypothesis. This signifies support for the alternative claim that fewer than half of adults favor drafting women.

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Most popular questions from this chapter

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose that the specifications state that the mean strength of welds should exceed 100 \(\mathrm{lb} / \mathrm{in} .^{2} .\) The inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100 .\) Explain why this alternative hypothesis was chosen rather than \(\mu<100\).

Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40 -amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact \(40 .\) If the mean amperage is lower than 40 , customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40 , the manufacturer might be liable for damage to an electrical system as a result of fuse malfunction. To verify the mean amperage of the fuses, a sample of fuses is selected and tested. If a hypothesis test is performed using the resulting data, what null and alternative hypotheses would be of interest to the manufacturer?

A number of initiatives on the topic of legalized gambling have appeared on state ballots. Suppose that a political candidate has decided to support legalization of casino gambling if he is convinced that more than twothirds of U.S. adults approve of casino gambling. USA Today (June 17,1999 ) reported the results of a Gallup poll in which 1523 adults (selected at random from households with telephones) were asked whether they approved of casino gambling. The number in the sample who approved was 1035 . Does the sample provide convincing evidence that more than two-thirds approve?

When a published article reports the results of many hypothesis tests, the \(P\) -values are not usually given. Instead, the following type of coding scheme is frequently used: \(^{*} p<.05,^{* *} p<.01,^{*} *^{*} p<.001, *\) Which of the symbols would be used to code for each of the following \(P\) -values? a. 037 c. \(.072\) b. \(.0026\) d. \(.0003\)

Let \(\mu\) denote the true average diameter for bearings of a certain type. A test of \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq 0.5\) will be based on a sample of \(n\) bearings. The diameter distribution is believed to be normal. Determine the value of \(\beta\) in each of the following cases: a. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.52\) b. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.48\) c. \(n=15, \alpha=.01, \sigma=0.02, \mu=0.52\) d. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.54\) e. \(n=15, \alpha=.05, \sigma=0.04, \mu=0.54\) f. \(n=20, \alpha=.05, \sigma=0.04, \mu=0.54\) \(\mathrm{g}\). Is the way in which \(\beta\) changes as \(n, \alpha, \sigma\), and \(\mu\) vary consistent with your intuition? Explain.
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