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A political researcher believes that the fraction \(p_{1}\) of Republicans strongly in favor of the death penalty is greater than the fraction \(p_{2}\) of Democrats strongly in favor of the death penalty. He acquired independent random samples of 200 Republicans and 200 Democrats and found 46 Republicans and 34 Democrats strongly favoring the death penalty. Does this evidence provide statistical support for the researcher's belief? Use \(\alpha=.05\)

Short Answer

Expert verified
There is no statistical support, at \( \alpha = 0.05 \), for the claim that more Republicans are in favor of the death penalty than Democrats.

Step by step solution

01

State the Hypotheses

For this hypothesis test, we need to state both the null and alternative hypotheses. \( H_0 : p_1 \leq p_2 \) and \( H_a : p_1 > p_2 \). This is a right-tailed test since we want to test if \( p_1 \) is greater than \( p_2 \).
02

Calculate Sample Proportions

Calculate the sample proportions for both groups: Republicans and Democrats. \( \hat{p}_1 = \frac{46}{200} = 0.23 \) and \( \hat{p}_2 = \frac{34}{200} = 0.17 \).
03

Find the Pooled Sample Proportion

Since we are comparing two proportions, we need a pooled proportion \( \hat{p} \) under the null hypothesis. Calculate it using \( \hat{p} = \frac{46 + 34}{200 + 200} = \frac{80}{400} = 0.20 \).
04

Calculate the Standard Error

The standard error (SE) for the difference in sample proportions is calculated using the formula:\[ SE = \sqrt{\hat{p} (1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} \]Substitute the values: \[ SE = \sqrt{0.20 \times (1 - 0.20) \left( \frac{1}{200} + \frac{1}{200} \right)} = \sqrt{0.20 \times 0.80 \times \frac{1}{100}} = \sqrt{0.0016} = 0.04 \]
05

Compute the Test Statistic

The test statistic for the difference in proportions is calculated as:\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \]Substitute the sample proportions and standard error:\[ z = \frac{0.23 - 0.17}{0.04} = \frac{0.06}{0.04} = 1.5 \]
06

Determine the Critical Value and Decision

For a right-tailed test at \( \alpha = 0.05 \), the critical value from the standard normal distribution table is approximately 1.645. Compare the test statistic (1.5) with the critical value. Since 1.5 < 1.645, we do not reject the null hypothesis.
07

Conclusion

There is not enough statistical evidence at the \( \alpha = 0.05 \) level to support the researcher's claim that the fraction of Republicans strongly in favor of the death penalty is greater than that of Democrats.

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Key Concepts

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

Sample Proportions
In hypothesis testing, sample proportions provide an estimate of the population proportions based on sampled data.
For our example, the researcher observed two different groups: Republicans and Democrats. Calculating sample proportions involves dividing the number of favorable responses by the total number of respondents in each sample.
  • Republicans: 46 out of 200 sampled, giving a sample proportion, \( \hat{p}_1 = \frac{46}{200} = 0.23 \).
  • Democrats: 34 out of 200 sampled, giving a sample proportion, \( \hat{p}_2 = \frac{34}{200} = 0.17 \).
The sample proportions \( \hat{p}_1 \) and \( \hat{p}_2 \) represent the fraction of each group that strongly favors the death penalty, allowing us to make statistical inferences about the entire population of Republicans and Democrats.
Pooled Proportion
A pooled proportion is used in hypothesis testing when comparing two independent sample proportions.
It provides a combined estimate assuming no difference between the groups under the null hypothesis.Combining data from both groups, we calculate the pooled proportion \( \hat{p} \) as follows:\[ \hat{p} = \frac{46 + 34}{200 + 200} = \frac{80}{400} = 0.20 \]
  • Numerator: Total number of successes from both samples (46 Republicans + 34 Democrats).
  • Denominator: Sum of both sample sizes (200 Republicans + 200 Democrats).
The pooled proportion \( 0.20 \) reflects a combined estimate of those strongly favoring the death penalty under the hypothesis that there is no actual distinction between the two populations.
Standard Error
Standard Error (SE) measures the variability of the sampling distribution of a statistic, particularly concerning the difference between two sample proportions.
It helps gauge the precision of our sample estimates.The formula for calculating SE in this context is:\[ SE = \sqrt{\hat{p} (1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} \]With our data:
  • \( \hat{p} = 0.20 \): the pooled proportion.
  • \( n_1 \) and \( n_2 \): size of the Republican and Democrat samples, respectively.
Substituting the values results in:\[ SE = \sqrt{0.20 \times (1 - 0.20) \left( \frac{1}{200} + \frac{1}{200} \right)} = \sqrt{0.20 \times 0.80 \times \frac{1}{100}} = \sqrt{0.0016} = 0.04 \]The SE of 0.04 indicates the spread of the sample mean differences from the true population mean, informing us about the reliability of our hypothesis test.
Test Statistic
The test statistic quantifies how far the sample statistic is from the null hypothesis, measured in units of the standard error.
It's a crucial part in determining whether or not we can reject the null hypothesis.In this scenario, the test statistic \( z \) is critical for comparing the two group proportions. The formula for \( z \) is:\[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \]Where:
  • \( \hat{p}_1 = 0.23 \): the sample proportion for Republicans.
  • \( \hat{p}_2 = 0.17 \): the sample proportion for Democrats.
  • \( SE = 0.04 \): the previously calculated standard error.
With these values, our calculation becomes:\[ z = \frac{0.23 - 0.17}{0.04} = \frac{0.06}{0.04} = 1.5 \]The test statistic of 1.5 is compared against a critical value (usually from a z-table) to decide whether to reject the null hypothesis. In this case, we compare it to the critical value of 1.645 and conclude whether there's supporting evidence for the researcher's claim.

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Most popular questions from this chapter

Jan Lindhe conducted a study \(^{\star}\) on the effect of an oral antiplaque rinse on plaque buildup on teeth. Fourteen subjects, whose teeth were thoroughly cleaned and polished, were randomly assigned to two groups of seven subjects each. Both groups were assigned to use oral rinses (no brushing) for a 2-week period. Group 1 used a rinse that contained an antiplaque agent. Group 2, the control group, received a similar rinse except that, unknown to the subjects, the rinse contained no antiplaque agent. A plaque index \(y,\) a measure of plaque buildup, was recorded at \(4,7,\) and 14 days. The mean and standard deviation for the 14-day plaque measurements for the two groups are given in the following table: $$\begin{array}{lcc} \hline & \text { Control Group } & \text { Antiplaque Group } \\ \hline \text { Sample size } & 7 & 7 \\ \text { Mean } & 1.26 & .78 \\ \text { Standard deviation } & 32 & .32 \\ \hline \end{array}$$ a. State the null and alternative hypotheses that should be used to test the effectiveness of the antiplaque oral rinse. b. Do the data provide sufficient evidence to indicate that the oral antiplaque rinse is effective? Test using \(\alpha=.05\) c. Bound or find the \(p\) -value for the test.

What conditions must be met for the \(Z\) test to be used to test a hypothesis concerning a population mean \(\mu ?\)

Shear strength measurements derived from unconfined compression tests for two types of soils gave the results shown in the following table (measurements in tons per square foot). Do the soils appear to differ with respect to average shear strength, at the \(1 \%\) significance level? $$\begin{array}{ll} \hline \text { Soil Type I } & \text { Soil Type II } \\ \hline n_{1}=30 & n_{2}=35 \\ \bar{y}_{1}=1.65 & \bar{y}_{2}=1.43 \\ s_{1}=0.26 & s_{2}=0.22 \\ \hline \end{array}$$

Refer to Exercise \(10.10 .\) Click the button "Clear Summary" to delete the results of any previous simulations. Change the sample size for each simulation to \(n=30\) and set up the applet to simulate testing \(H_{0}: p=.4\) versus \(H_{a}: p>.4\) at the .05 level of significance. a. Click the button "Clear Summary" to erase the results or any previous simulations. Set the real value of \(p\) to .4 and implement at least 200 simulations. What is the percentage simulated tests that result in rejecting the null hypothesis? Does the test work as you expected? b. Leave all settings as they were in part (a) but change the real value of \(p\) to \(.5 .\) Simulate at least 200 tests. Repeat when the real value of \(p\) is .6 and .7 . Click the button "Show Summary." What do you observe about the rejection rate as the true value of \(p\) gets further from .4 and closer to \(1 ?\) Does the pattern that you observe match your impression of how a good test should perform?

Suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) constitute a random sample from a normal distribution with known mean \(\mu\) and unknown variance \(\sigma^{2}\). Find the most powerful \(\alpha\) -level test of \(H_{0}: \sigma^{2}=\sigma_{0}^{2}\) versus \(H_{a}:\) \(\sigma^{2}=\sigma_{1}^{2},\) where \(\sigma_{1}^{2}>\sigma_{0}^{2} .\) Show that this test is equivalent to a \(\chi^{2}\) test. Is the test uniformly most powerful for \(H_{a}: \sigma^{2}>\sigma_{0}^{2} ?\)

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