/*! 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 11 Suppose that in a particular sta... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that in a particular state consisting of four distinct regions, a random sample of \(n_{k}\) voters is obtained from the \(k\) th region for \(k=1,2,3,4\). Each voter is then classified according to which candidate \((1,2\), or 3 ) he or she prefers and according to voter registration \((1=\) Dem., \(2=\) Rep., 3 = Indep.). Let \(p_{\text {igk }}\) denote the proportion of voters in region \(k\) who belong in candidate category \(i\) and registration category \(j\). The null hypothesis of homogeneous regions is \(H_{0}: p_{i j 1}=p_{i j 2}=p_{i j 3}=p_{i j 4}\) for all \(i, j\) (i.e., the proportion within each candidate/registration combination is the same for all four regions). Assuming that \(H_{0}\) is true, determine \(\hat{p}_{i j k}\) and \(\hat{e}_{i j k}\) as functions of the observed \(n_{i j k}\) 's, and use the general rule of thumb to obtain the number of degrees of freedom for the chi-squared test.

Short Answer

Expert verified
Calculate common proportions, determine expected frequencies, and use 12 degrees of freedom.

Step by step solution

01

Understand the Problem Statement

First, we need to comprehend the problem. We have a state divided into four regions, where a sample of voters from each region is categorized based on candidate preference and voter registration status. The null hypothesis suggests that the voter distribution is the same across all regions for each candidate and registration category.
02

Define Proportions and Observed Counts

We define \(p_{ij1}, p_{ij2}, p_{ij3}, p_{ij4}\) as the proportions of voters favoring candidate \(i\) and having registration category \(j\) in regions 1, 2, 3, and 4, respectively. The observed count of such voters is \(n_{ijk}\).
03

Calculate the Proportions under Null Hypothesis

Under \(H_{0}\), the proportions \(p_{ij1}=p_{ij2}=p_{ij3}=p_{ij4}\) are equal across all regions. Denote the common proportion as \(p_{ij}\). To estimate this, we sum up all the observations for a given \((i,j)\) and divide by the total observations across all regions. Thus, \( \hat{p}_{ij} = \frac{\sum_{k=1}^{4} n_{ijk}}{\sum_{k=1}^{4} n_{k}} \).
04

Determine Expected Frequencies

Using the estimated proportions, calculate the expected frequency \(\hat{e}_{ijk}\) for each region using \(\hat{e}_{ijk} = \hat{p}_{ij} \cdot n_{k}\), where \(n_k\) is the total sample size of region \(k\).
05

Calculate Degrees of Freedom

The degrees of freedom for the chi-squared test is calculated based on the number of independent comparisons. For the given problem, with 3 candidate preferences and 3 registration categories, and 4 regions, the formula for the degrees of freedom is \((3-1) \times (3-1) \times (4-1) = 12\.\)

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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, often denoted as \(H_0\), is a foundational concept in statistical hypothesis testing. In this exercise, the null hypothesis states that the distribution of voters across different candidate and registration categories is consistent across all four regions. This means that, regardless of the region, the voting patterns based on candidate preference and voter registration are expected to be similar.

The benefit of a null hypothesis is that it provides a statement of no effect or no difference, which we can then test statistically. When conducting a chi-squared test, we compare the observed data to the expected data under the assumption that the null hypothesis is true. If the observed frequencies significantly deviate from the expected frequencies, then we might reject the null hypothesis, suggesting that there is a difference in voter behavior across regions.

Understanding the null hypothesis is crucial because it serves as the baseline assumption. We rely on this assumption to determine the expected frequencies that play a central role in hypothesis testing.
Degrees of Freedom
Degrees of freedom are an integral part of statistical tests, including the chi-squared test. They represent the number of values in a calculation that are free to vary. In other words, degrees of freedom are the parameters that allow for variations in a statistical distribution. They essentially determine the shape and spread of the distribution we are testing.

In the context of the chi-squared test used in our exercise, the degrees of freedom are computed using the formula \((r-1) \times (c-1)\), where \(r\) is the number of candidate preferences, \(c\) is the number of registration categories, and we additionally consider the number of regions. For this scenario, with 3 candidates and 3 registration categories across 4 regions, the degrees of freedom is calculated as \((3-1) \times (3-1) \times (4-1) = 12\).

Simply put, the degrees of freedom indicate the number of comparisons we've made across regions that are independent. The calculation helps define the critical value needed to determine the significance of our test statistic.
Expected Frequencies
Expected frequencies are calculated to provide a comparison benchmark in hypothesis testing. They represent the frequencies of each category we would expect to observe if the null hypothesis is true. For our chi-squared test, expected frequencies help us determine whether the observed data significantly differ from what was expected under the assumption of homogeneity across regions.

To compute the expected frequencies (\(\hat{e}_{i j k}\)), we use the formula \(\hat{e}_{i j k} = \hat{p}_{ij} \times n_k\). Here, \(\hat{p}_{ij}\) is the estimated common proportion of voters for each candidate and registration category, calculated by averaging over all regions. \(n_k\) is the total number of voters sampled in region \(k\). This formulation indicates how many voters in each category we would expect per region if the regions were homogeneous in terms of voting patterns.

Comparing these expected frequencies with the observed frequencies (actual data collected) enables us to evaluate the null hypothesis. Large deviations from the expected frequencies can suggest that regional voting patterns differ significantly, leading us possibly to reject the null hypothesis.

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Most popular questions from this chapter

Qualifications of male and female head and assistant college athletic coaches were compared in the article "Sex Bias and the Validity of Believed Differences Between Male and Female Interscholastic Athletic Coaches" (Research Quarterly for Exercise and Sport, 1990: 259-267). Each person in random samples of 2225 male coaches and 1141 female coaches was classified according to number of years of coaching experience to obtain the accompanying twoway table. Is there enough evidence to conclude that the proportions falling into the experience categories are different for men and women? Use \(\alpha=.01\).

An information-retrieval system has ten storage locations. Information has been stored with the expectation that the long-run proportion of requests for location \(t\) is given by \(p_{i}=(5.5-|i-5.5|) / 30\). A sample of 200 retrieval requests gave the following frequencies for locations \(1-10\), respectively: \(4,15,23,25,38,31,32,14,10\), and 8 . Use a chi-squared test at significance level . 10 to decide whether the data is consistent with the a priori proportions (use the P-value approach).

Let \(p_{1}\) denote the proportion of successes in a particular population. The test statistic value in Chapter 8 for testing \(H_{0}: p_{1}=p_{10} \quad\) was \(z=\left(\hat{p}_{1}-p_{10}\right) / \sqrt{p_{10} p_{20} / n}\), where \(p_{20}=1-p_{10}\). Show that for the case \(k=2\), the chisquared test statistic value of Section 14.1 satisfies \(\chi^{2}=z^{2}\). [Hint: First show that \(\left.\left(n_{1}-n p_{10}\right)^{2}=\left(n_{2}-n p_{20}\right)^{2}\right]\)

Say as much as you can about the \(P\)-value for an upper-tailed chi-squared test in each of the following situations: a. \(x^{2}=7.5\), df \(=2\) b. \(x^{2}=13.0\), df \(=6\) c. \(X^{2}=18.0, \mathrm{df}=9\) d. \(\chi^{2}=21.3\), df \(=5\) e. \(x^{2}=5.0, k=4\)

The authors of the article "Predicting Professional Sports Game Outcomes from Intermediate Game Scores" (Chance, 1992: 18-22) used a chi-squared test to determine whether there was any merit to the idea that basketball games are not settled until the last quarter, whereas baseball games are over by the seventh inning. They also considered football and hockey. Data was collected for 189 basketball games, 92 baseball games, 80 hockey games, and 93 football games. The games analyzed were sampled randomly from all games played during the 1990 season for baseball and football and for the \(1990-1991\) season for basketball and hockey. For each game, the late-game leader was determined, and then it was noted whether the late-game leader actually ended up winning the game. The resulting data is summarized in the accompanying table. The authors state that "Late-game leader is defined as the team that is ahead after three quarters in basketball and football, two periods in hockey, and seven innings in baseball. The chi-square value on three degrees of freedom is \(10.52\) \((P<.015) .\) a. State the relevant hypotheses and reach a conclusion using \(\alpha=.05\). b. Do you think that your conclusion in part (a) can be attributed to a single sport being an anomaly?

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