/*! 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

Athabasca Fishing Lodge is located on Lake Athabasca in northern Canada. In one of its recent brochures, the lodge advertises that \(75 \%\) of its guests catch northern pike over 20 pounds. Suppose that last summer 64 out of a random sample of 83 guests did, in fact, catch northern pike weighing over 20 pounds. Does this indicate that the population proportion of guests who catch pike over 20 pounds is different from \(75 \%\) (either higher or lower)? Use \(\alpha=0.05\).

Consider independent random samples from two populations that are normal or approximately normal, or the case in which both sample sizes are at least \(30 .\) Then, if \(\sigma_{1}\) and \(\sigma_{2}\) are unknown but we have reason to believe that \(\sigma_{1}=\sigma_{2}\), we can pool the standard deviations. Using sample sizes \(n_{1}\) and \(n_{2}\), the sample test statistic \(\bar{x}_{1}-\bar{x}_{2}\) has a Student's \(t\) distribution, where \(t=\frac{\bar{x}_{1}-\bar{x}_{2}}{s \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}\) with degrees of freedom d.f. \(=n_{1}+n_{2}-2\) and where the pooled standard deviation \(s\) is $$s=\sqrt{\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}}$$ Note: With statistical software, select the pooled variance or equal variance options. (a) There are many situations in which we want to compare means from populations having standard deviations that are equal. This method applies even if the standard deviations are known to be only approximately equal (see Section \(11.4\) for methods to test that \(\sigma_{1}=\sigma_{2}\) ). Consider Problem 17 regarding average incidence of fox rabies in two regions. For region I, \(n_{1}=16\), \(\bar{x}_{1}=4.75\), and \(s_{1} \approx 2.82\) and for region II, \(n_{2}=15, \bar{x}_{2} \approx 3.93\), and \(s_{2} \approx\) 2.43. The two sample standard deviations are sufficiently close that we can assume \(\sigma_{1}=\sigma_{2}\). Use the method of pooled standard deviation to redo Problem 17 , where we tested if there was a difference in population mean average incidence of rabies at the \(5 \%\) level of significance. (b) Compare the \(t\) value calculated in part (a) using the pooled standard deviation with the \(t\) value calculated in Problem 17 using the unpooled standard deviation. Compare the degrees of freedom for the sample test statistic. Compare the conclusions.

The following data are based on information from the Regis University Psychology Department. In an effort to determine if rats perform certain tasks more quickly if offered larger rewards, the following experiment was performed. On day 1 , a group of three rats was given a reward of one food pellet each time they ran a maze. A second group of three rats was given a reward of five food pellets each time they ran the maze. On day 2, the groups were reversed, so the first group now got five food pellets for running the maze and the second group got only one pellet for running the same maze. The average times in seconds for each rat to run the maze 30 times are shown in the following table. \(\begin{array}{l|cccccc} \hline \text { Rat } & A & B & C & D & E & F \\ \hline \text { Time with one food pellet } & 3.6 & 4.2 & 2.9 & 3.1 & 3.5 & 3.9 \\\ \hline \text { Time with five food pellets } & 3.0 & 3.7 & 3.0 & 3.3 & 2.8 & 3.0 \\ \hline \end{array}\) Do these data indicate that rats receiving larger rewards tend to run the maze in less time? Use a \(5 \%\) level of significance.

Consumer Reports stated that the mean time for a Chrysler Concorde to go from 0 to 60 miles per hour was \(8.7\) seconds. (a) If you want to set up a statistical test to challenge the claim of \(8.7\) seconds, what would you use for the null hypothesis? (b) The town of Leadville, Colorado, has an elevation over 10,000 feet. Suppose you wanted to test the claim that the average time to accelerate from 0 to 60 miles per hour is longer in Leadville (because of less oxygen). What would you use for the alternate hypothesis? (c) Suppose you made an engine modification and you think the average time to accelerate from 0 to 60 miles per hour is reduced. What would you use for the alternate hypothesis? (d) For each of the tests in parts (b) and (c), would the \(P\) -value area be on the left, on the right, or on both sides of the mean? Explain your answer in each case.

Discuss each of the following topics in class or review the topics on your own. Then write a brief but complete essay in which you answer the following questions. (a) What is a null hypothesis \(H_{0} ?\) (b) What is an alternate hypothesis \(H_{1} ?\) (c) What is a type I error? a type II error? (d) What is the level of significance of a test? What is the probability of a type II error?

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