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91Ó°ÊÓ

The article "Statistical Evidence of Discrimination" (J. Amer. Statist. Assoc., 1982: 773-783) discusses the court case Swain v. Alabama (1965), in which it was alleged that there was discrimination against blacks in grand jury selection. Census data suggested that \(25 \%\) of those eligible for grand jury service were black, yet a random sample of 1050 people called to appear for possible duty yielded only 177 blacks. Using a level \(.01\) test, does this data argue strongly for a conclusion of discrimination?

Short Answer

Expert verified
Yes, the data argues strongly for discrimination at the 0.01 significance level.

Step by step solution

01

Define Hypotheses

We start by defining the null and alternative hypotheses. The null hypothesis \(H_0\) states that there is no discrimination; hence, the proportion of black individuals in the jury selection process \(p\) is equal to \(0.25\). The alternative hypothesis \(H_a\) suggests that the proportion is less than \(0.25\), indicating discrimination. Therefore:\[H_0: p = 0.25 \quad \text{vs} \quad H_a: p < 0.25.\]
02

Calculate Sample Proportion

Compute the sample proportion \(\hat{p}\) of blacks in the jury selection. Given that out of 1050 called to appear, 177 were black, \(\hat{p}\) is computed as:\[\hat{p} = \frac{177}{1050} \approx 0.1686.\]
03

Determine Test Statistic

The test statistic for a hypothesis test about a proportion is calculated using the formula:\[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}},\]where \( p_0 = 0.25 \) and \( n = 1050 \). Substituting the given values, calculate the Z-score:\[Z = \frac{0.1686 - 0.25}{\sqrt{\frac{0.25 \times 0.75}{1050}}} \approx -7.87.\]
04

Determine Critical Value

For a one-tailed test at the \(0.01\) significance level, find the critical Z-value. The critical Z-value for a left-tailed test at \(0.01\) significance level is approximately \(-2.33\).
05

Compare Test Statistic and Critical Value

Compare the calculated Z-score with the critical value. Since \(-7.87\) is less than \(-2.33\), this result is in the critical region.
06

Conclusion

Reject the null hypothesis \(H_0\) because the test statistic is in the critical region. The data provides sufficient evidence to support the claim that there is discrimination against blacks in grand jury selection at the \(0.01\) significance level.

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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 is the foundation of hypothesis testing. It's often the statement researchers seek to test or disprove. In the context of the Swain v. Alabama case, the null hypothesis (\(H_0\)) claims that there is no discrimination in grand jury selection, meaning the proportion of eligible black individuals (\(p\)) is indeed \(0.25\), or 25%. This assumption acts as a starting point for statistical inference and remains true until evidence suggests otherwise. The underlying idea is to have a baseline expectation which we can challenge through collected data and statistical calculation if warranted. By assuming no bias, we allow the data to potentially reveal whether a bias may exist.
Alternative Hypothesis
While the null hypothesis suggests no effect, the alternative hypothesis proposes the presence of an effect. In hypothesis testing, researchers focus on this statement to determine if it holds. In our exercise, the alternative hypothesis (\(H_a\)) posits that the proportion of black individuals in grand jury service (\(p\)) is less than \(0.25\), indicating potential discrimination. Essentially, it suggests that something unusual is happening which needs further investigation. Whenever evidence supports the alternative hypothesis, it suggests that the observations could not have occurred under the null hypothesis conditions. Therefore, proving \(H_a\) can lead to challenging established beliefs or practices.
Z-test
The Z-test is a statistical procedure used to determine if there is a significant difference between sample data and the population it's drawn from. It specifically tests the relative size of the difference or "effect." For this exercise, we used the Z-test to compare the observed sample proportion of black grand jury members (\(\hat{p} = 0.1686\)) to the expected proportion (\(p_0 = 0.25\)). The Z-score, calculated as:\[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]allows us to determine whether the observed proportion is significantly lower than expected. A standard normal distribution can be referred to, converting our observed effect into a standardized measure. This enables the direct comparison to critical values for significance testing. Here, the Z-score of approximately \(-7.87\) indicates a substantially lower observed proportion than expected.
Significance Level
The significance level, typically denoted by alpha (\(\alpha\)), is a key part of hypothesis testing. It determines the threshold for rejection of the null hypothesis. In essence, it establishes the risk of making a Type I error – falsely rejecting a true null hypothesis. Common values for significance levels are \(0.05\) or \(0.01\), which signify a 5% or 1% risk respectively. In this exercise, a \(0.01\) level was used. This means researchers accept a 1% chance of wrongly asserting discrimination is present when it isn't. The critical value associated with a \(0.01\) significance level, for a one-tailed test like this, is approximately \(-2.33\). If our calculated test statistic falls beyond this critical value, it supports rejecting the null hypothesis.

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Most popular questions from this chapter

Let the test statistic \(T\) have a \(t\) distribution when \(H_{0}\) is true. Give the significance level for each of the following situations: a. \(H_{\mathrm{a}}: \mu>\mu_{0}, \quad\) df \(=15, \quad\) rejection region \(t \geq 3.733\) b. \(H_{\mathrm{a}}\) : \(\mu<\mu_{0}, \quad n=24\), rejection region \(t \leq-2.500\) c. \(H_{\mathrm{a}}: \mu \neq \mu_{0}, n=31\), rejection region \(t \geq 1.697\) or \(t \leq-1.697\)

A sample of 50 lenses used in eyeglasses yields a sample mean thickness of \(3.05 \mathrm{~mm}\) and a sample standard deviation of \(.34 \mathrm{~mm}\). The desired true average thickness of such lenses is \(3.20 \mathrm{~mm}\). Does the data strongly suggest that the true average thickness of such lenses is something other than what is desired? Test using \(\alpha=.05\).

An aspirin manufacturer fills bottles by weight rather than by count. Since each bottle should contain 100 tablets, the average weight per tablet should be 5 grains. Each of 100 tablets taken from a very large lot is weighed, resulting in a sample average weight per tablet of \(4.87\) grains and a sample standard deviation of \(.35\) grain. Does this information provide strong evidence for concluding that the company is not filling its bottles as advertised? Test the appropriate hypotheses using \(\alpha=.01\) by first computing the \(P\)-value and then comparing it to the specified significance level.

Suppose the population distribution is normal with known \(\sigma\). Let \(\gamma\) be such that \(0<\gamma<\alpha\). For testing \(H_{0}: \mu=\mu_{0}\) versus \(H_{\mathrm{a}}: \mu \neq \mu_{0}\), consider the test that rejects \(H_{0}\) if either \(z \geq z_{\gamma}\) or \(z \leq-z_{\alpha-\gamma}\), where the test statistic is \(Z=\left(\bar{X}-\mu_{0}\right) /(\sigma / \sqrt{n})\). a. Show that \(P\) (type I error) \(=\alpha\). b. Derive an expression for \(\beta\left(\mu^{\prime}\right)\). [Hint: Express the test in the form "reject \(H_{0}\) if either \(\bar{x} \geq c_{1}\) or \(\left.\leq \mathrm{c}_{2} . "\right]\) c. Let \(\Delta>0\). For what values of \(\gamma\) (relative to \(\alpha\) ) will \(\beta\left(\mu_{0}+\Delta\right)<\beta\left(\mu_{0}-\Delta\right)\) ?

For each of the following assertions, state whether it is a legitimate statistical hypothesis and why: a. \(H: \sigma>100\) b. \(H: \tilde{x}=45\) c. \(H: s \leq .20\) d. \(H: \sigma_{1} / \sigma_{2}<1\) e. \(H: \bar{X}-\bar{Y}=5\) f. \(H: \lambda \leq .01\), where \(\lambda\) is the parameter of an exponential distribution used to model component lifetime

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