/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 A random sample of \(n_{1}=288\)... [FREE SOLUTION] | 91Ó°ÊÓ

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A random sample of \(n_{1}=288\) voters registered in the state of California showed that 141 voted in the last general election. A random sample of \(n_{2}=216\) registered voters in the state of Colorado showed that 125 voted in the most recent general election. (See reference in Problem 25.) Do these data indicate that the population proportion of voter turnout in Colorado is higher than that in California? Use a \(5 \%\) level of significance.

Short Answer

Expert verified
The voter turnout in Colorado is significantly higher than in California at a 5% significance level.

Step by step solution

01

Define the Hypotheses

First, we define the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). \(H_0: p_1 = p_2\) means the proportion of voters who voted in California is equal to the proportion in Colorado. \(H_a: p_2 > p_1\) states that the proportion of voters in Colorado is greater than the proportion in California.
02

Calculate Sample Proportions

Calculate the sample proportions for both states. For California, \(\hat{p}_1 = \frac{141}{288} \approx 0.4896\). For Colorado, \(\hat{p}_2 = \frac{125}{216} \approx 0.5787\).
03

Compute the Pooled Proportion

The pooled proportion \(\hat{p}\) combines the proportions from both samples: \(\hat{p} = \frac{141 + 125}{288 + 216} = \frac{266}{504} \approx 0.5278\).
04

Calculate the Standard Error

The standard error for the difference in proportions is calculated using the formula: \(SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}\). Substituting the values, we get \(SE = \sqrt{0.5278(1-0.5278)\left(\frac{1}{288} + \frac{1}{216}\right)} \approx 0.0441\).
05

Compute the Test Statistic

The test statistic \(Z\) is calculated using the formula: \(Z = \frac{\hat{p}_2 - \hat{p}_1}{SE}\). Substituting the values, we have \(Z = \frac{0.5787 - 0.4896}{0.0441} \approx 2.02\).
06

Determine the Critical Value

Since we are testing at a 5% level of significance for a one-tailed test, we compare the test statistic against the critical Z-value of 1.645 that corresponds to the 95th percentile of the standard normal distribution.
07

Make a Decision

The calculated test statistic \(Z = 2.02\) is greater than the critical value 1.645. Hence, we reject the null hypothesis \(H_0\) and accept the alternative hypothesis \(H_a\).
08

Conclusion

Based on the sample data, we can conclude with 95% confidence that the proportion of voters who voted in the most recent general election in Colorado is significantly higher than in California.

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Key Concepts

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

Population Proportion
When discussing voter turnout, we want to understand the broader picture of participation in elections within different groups. The population proportion, denoted as \( p \), represents the fraction of the total population demonstrating a particular characteristic—in this case, voting in an election. If we consider the population of registered voters in each state, the sample proportion (\( \hat{p} \)) acts as an estimate of the true population proportion. It is calculated by dividing the number of voters who actually cast their vote by the total number of registered voters. In our exercise, the sample proportion for California is \( \hat{p}_1 = \frac{141}{288} \approx 0.4896 \) and for Colorado, it's \( \hat{p}_2 = \frac{125}{216} \approx 0.5787 \). These figures help us estimate the actual voting proportions in the entire voter populations of each state.
Significance Level
The significance level, often denoted by \( \alpha \), is a threshold set by the researcher to determine when to reject the null hypothesis \( H_0 \). It essentially represents the probability of making a Type I error, which is incorrectly rejecting a true null hypothesis. In our example, a 5% significance level (\( \alpha = 0.05 \)) was used, indicating we are willing to accept a 5% risk of this type of error. This level is crucial when testing for differences or changes in statistics as it guides the threshold for making decisions. If our calculated test statistic exceeds the critical value corresponding to this significance level, we have enough evidence to reject \( H_0 \) and conclude in favor of the alternative hypothesis \( H_a \).
Standard Error
Standard error is a measure of the variability or uncertainty in a statistic. It indicates how much a sample proportion will likely deviate from the true population proportion. In hypothesis testing, especially when comparing two population proportions, the standard error takes into account the combined variance from both samples. The formula used is \( SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \).
  • In this exercise, \( \hat{p} \) is the pooled sample proportion, \( \hat{p} = \frac{266}{504} = 0.5278 \).
  • The calculated standard error \( SE \approx 0.0441 \) informs us about the reliability of our estimate of the difference between two proportions.
This value helps us determine the distribution of our test statistic.
Test Statistic
A test statistic is a standardized value derived from sample data to determine the degree of compatibility with the null hypothesis. In our case of comparing population proportions, we use the \( Z \)-test for hypothesis testing. The test statistic \( Z \) is computed using the formula: \( Z = \frac{\hat{p}_2 - \hat{p}_1}{SE} \).
  • Here, \( \hat{p}_2 \) and \( \hat{p}_1 \) are the sample proportions for Colorado and California respectively, while \( SE \) is the standard error.
  • The calculated \( Z \) value in our example is approximately \( 2.02 \).
This \( Z \) value is compared to the critical value from the standard normal distribution table (here, 1.645 for a 5% significance level and one-tailed test) to make a decision. If \( Z \) exceeds this critical value, as it does in our example, we reject the null hypothesis, suggesting a significant difference favoring the proportion in Colorado over California.

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Most popular questions from this chapter

Compare statistical testing with legal methods used in a U.S. court setting. Then discuss the following topics in class or consider the topics on your own. Please write a brief but complete essay in which you answer the following questions. (a) In a court setting, the person charged with a crime is initially considered to be innocent. The claim of innocence is maintained until the jury returns with a decision. Explain how the claim of innocence could be taken to be the null hypothesis. Do we assume that the null hypothesis is true throughout the testing procedure? What would the alternate hypothesis be in a court setting? (b) The court claims that a person is innocent if the evidence against the person is not adequate to find him or her guilty. This does not mean, however, that the court has necessarily proved the person to be innocent. It simply means that the evidence against the person was not adequate for the jury to find him or her guilty. How does this situation compare with a statistical test for which the conclusion is "do not reject" (i.e., accept) the null hypothesis? What would be a type II error in this context? (c) If the evidence against a person is adequate for the jury to find him or her guilty, then the court claims that the person is guilty. Remember, this does not mean that the court has necessarily proved the person to be guilty. It simply means that the evidence against the person was strong enough to find him or her guilty. How does this situation compare with a statistical test for which the conclusion is to "reject" the null hypothesis? What would be a type I error in this context? (d) In a court setting, the final decision as to whether the person charged is innocent or guilty is made at the end of the trial, usually by a jury of impartial people. In hypothesis testing, the final decision to reject or not reject the null hypothesis is made at the end of the test by using information or data from an (impartial) random sample. Discuss these similarities between statistical hypothesis testing and a court setting. (e) We hope that you are able to use this discussion to increase your understanding of statistical testing by comparing it with something that is a well. known part of our American way of life. However, all analogies have weak points. It is important not to take the analogy between statistical hypothesis testing and legal court methods too far. For instance, the judge does not set a level of significance and tell the jury to determine a verdict that is wrong only \(5 \%\) or \(1 \%\) of the time. Discuss some of these weak points in the analogy between the court setting and hypothesis testing.

Harper's Index reported that \(80 \%\) of all supermarket prices end in the digit 9 or \(5 .\) Suppose you check a random sample of 115 items in a supermarket and find that 88 have prices that end in 9 or \(5 .\) Does this indicate that less than \(80 \%\) of the prices in the store end in the digits 9 or 5 ? Use \(\alpha=0.05\).

Based on information from Harper's Index, \(r_{1}=\) 37 people out of a random sample of \(n_{1}=100\) adult Americans who did not attend college believe in extraterrestrials. However, out of a random sample of \(n_{2}=100\) adult Americans who did attend college, \(r_{2}=47\) claim that they believe in extraterrestrials. Does this indicate that the proportion of people who attended college and who believe in extraterrestrials is higher than the proportion who did not attend college? Use \(\alpha=0.01\).

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