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Gender gap? Exercise 11.1 showed a \(2 \times 3\) table relating gender and political party identification, shown again here. The chi-squared statistic for these data equals 10.04 with a P-value of \(0.0066 .\) Conduct all five steps of the chi-squared test. \begin{tabular}{lcccc} \hline & \multicolumn{3}{c} { Political Party Identification } & \\ \cline { 2 - 4 } Gender & Democrat & Independent & Republican & Total \\ \hline Female & 421 & 398 & 244 & 1063 \\ Male & 278 & 367 & 198 & 843 \\ \hline \end{tabular}

Short Answer

Expert verified
Reject the null hypothesis; gender and political party identification are not independent.

Step by step solution

01

Define Hypotheses

In a chi-squared test for independence, we start by defining our null and alternative hypotheses. The null hypothesis is that gender and political party identification are independent. The alternative hypothesis is that they are not independent.
02

Calculate Expected Frequencies

Next, calculate the expected frequencies for each cell in the table using the formula: \(E_{ij} = \frac{(\text{{row total}}_i \times \text{{column total}}_j)}{\text{{Grand Total}}}\). For example, for Females identifying as Democrats, the expected frequency \(E_{11}\) is calculated as: \(E_{11} = \frac{(1063 \times 699)}{1906} \approx 390.01\). Compute similarly for all cells.
03

Compute Chi-Squared Statistic

The chi-squared statistic is calculated using the formula: \(\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}\), where \(O_{ij}\) is the observed frequency and \(E_{ij}\) is the expected frequency. Apply this to each cell and sum the results. For example, for the Female-Democrat cell: \(\chi^2 = \frac{(421 - 390.01)^2}{390.01}\). Do this for all cells to get the total chi-squared value of 10.04.
04

Determine Degrees of Freedom

The degrees of freedom for a chi-squared test of independence in a \(r \times c\) table is given by \((r-1) \times (c-1)\). Here, \(r=2\) (genders) and \(c=3\) (political parties), so the degrees of freedom is \((2-1) \times (3-1) = 2\).
05

Compare the Chi-Squared Value to Critical Value

Using a chi-squared distribution table, find the critical value for 2 degrees of freedom at the desired significance level (usually 0.05). The P-value given is 0.0066, which is less than 0.05, indicating the result is statistically significant.

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Key Concepts

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

Null Hypothesis
In the context of the chi-squared test for independence, the null hypothesis is a key starting point. It represents a statement that there is no association between the variables being studied. Specifically, for our exercise, the null hypothesis asserts that gender and political party identification are independent of one another.

This means that any observed differences in political party preference between males and females in the data are due to random chance. Consequently, this hypothesis suggests that the gender of a person does not affect which political party they identify with.

The null hypothesis helps provide a baseline from which we can measure whether any observed data deviations are significant. Rejecting or failing to reject this hypothesis will guide us in making data-driven conclusions about the relationship between the variables involved.
Degrees of Freedom
Degrees of freedom are an essential part of statistical tests like the chi-squared test. They refer to the number of values in the final calculation of a statistic that is free to vary. For a chi-squared test, understanding degrees of freedom helps determine the shape of the chi-squared distribution and the critical values we compare our test statistic to.

For a table with rows and columns, like our political party and gender example, the degrees of freedom are calculated using the formula \[(r-1) \times (c-1)\] where \(r\) is the number of rows and \(c\) is the number of columns.

In our case, we have 2 genders and 3 political parties, leading to: \[(2-1) \times (3-1) = 2\] degrees of freedom. These degrees of freedom are crucial because they affect the threshold we use to determine statistical significance. Lower degrees of freedom typically correspond to critical values that are closer to zero.
Independence Test
An independence test, like the chi-squared test, is a statistical method used to determine whether two categorical variables are independent. It helps us understand if there is a significant association between the variables being studied.

The test compares the observed frequencies in each category with the frequencies we would expect if the variables were truly independent. This involves calculating the expected frequencies, then using these to assess the observed data.

If the calculated chi-squared statistic is greater than the critical value for the given degrees of freedom and confidence level, we reject the null hypothesis. For our gender and political party example, the small P-value of 0.0066 indicates that there is a substantial statistical relationship between gender and political party preference, suggesting they are not independent.
Expected Frequencies
Expected frequencies play a critical role in performing a chi-squared test. They represent the frequencies we would anticipate in each category if the variables were independent.

To calculate expected frequencies, use the formula: \[E_{ij} = \frac{(\text{{row total}}_i \times \text{{column total}}_j)}{\text{{Grand Total}}} \] This formula ensures that you consider both the row and column totals relative to the overall total.

For example, for Females identifying as Democrats, the expected frequency might looks like: \[E_{11} = \frac{(1063 \times 699)}{1906} \approx 390.01 \]Calculating expected values for all cells in the table provides a basis for comparison with the observed frequencies.

Ultimately, the chi-squared statistic is derived from these expected values, by examining how much the observed data deviate from what was expected under the null hypothesis.

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Most popular questions from this chapter

Multiple response variables Each subject in a sample of 100 men and 100 women is asked to indicate which of the following factors (one or more) are responsible for increases in crime committed by teenagers: \(\mathrm{A}-\) the increasing gap in income between the rich and poor, \(\mathrm{B}-\) the increase in the percentage of single-parent families, \(\mathrm{C}\) - insufficient time that parents spend with their children. To analyze whether responses differ by gender of respondent, we cross-classify the responses by gender, as the table shows. a. Is it valid to apply the chi-squared test of independence to these data? Explain. b. Explain how this table actually provides information needed to cross- classify gender with each of three variables. Construct the contingency table relating gender to opinion about whether factor \(A\) is responsible for increases in teenage crime. \begin{tabular}{lccc} \hline \multicolumn{3}{l} { Three Factors for Explaining Teenage Crime } \\ \hline Gender & A & B & C \\ \hline Men & 60 & 81 & 75 \\ Women & 75 & 87 & 86 \\ \hline \end{tabular}

Karl Pearson devised the chi-squared goodness-of-fit test partly to analyze data from an experiment to analyze whether a particular roulette wheel in Monte Carlo was fair, in the sense that each outcome was equally likely in a spin of the wheel. For a given European roulette wheel with 37 pockets (with numbers \(0,1,2, \ldots, 36\) ), consider the null hypothesis that the wheel is fair. a. For the null hypothesis, what is the probability for each pocket? b. For an experiment with 3,700 spins of the roulette wheel, find the expected number of times each pocket is selected. c. In the experiment, the 0 pocket occurred 110 times. Show the contribution to the \(X^{2}\) statistic of the results for this pocket. d. Comparing the observed and expected counts for all 37 pockets, we get \(X^{2}=34.4\) Specify the df value and indicate whether there is strong evidence that the roulette wheel is not balanced. (Hint: Recall that the \(d f\) value is the mean of the distribution.)

True or false: Group 1 becomes Group 2 Interchanging two rows or interchanging two columns in a contingency table has no effect on the value of the \(X^{2}\) statistic.

Williams College admission Data from 2013 posted on the Williams College website shows that of all 3,195 males applying, \(18.2 \%\) were admitted, and of all 3,658 females applying, \(16.9 \%\) were admitted. Let \(\mathrm{X}\) denote gender of applicant and \(\mathrm{Y}\) denote whether admitted. a. Which conditional distributions do these percentages refer to, those of \(Y\) at given categories of \(X,\) or those of \(X\) at given categories of \(Y ?\) Set up a table showing the two conditional distributions. b. Are \(X\) and \(Y\) independent or dependent? Explain.

Down syndrome diagnostic test The table shown, from Example 8 in Chapter \(5,\) cross-tabulates whether a fetus has Down syndrome by whether the triple blood diagnostic test for Down syndrome is positive (that is, indicates that the fetus has Down syndrome). a. Tabulate the conditional distributions for the blood test result, given the true Down syndrome status. b. For the Down cases, what percentage was diagnosed as positive by the diagnostic test? For the unaffected cases, what percentage got a negative test result? Does the diagnostic test appear to be a good one? c. Construct the conditional distribution on Down syndrome status for those who have a positive test result. (Hint: You condition on the first column total and find proportions in that column.) Of those cases, what percentage truly have Down syndrome? Is the result surprising? Explain why this probability is small. \begin{tabular}{lrrr} \hline & \multicolumn{2}{c} { Blood Test Result } & \\ \cline { 2 - 3 } Down Syndrome Status & Positive & Negative & Total \\ \hline D (Down) & 48 & 6 & 54 \\ D \(^{c}\) (unaffected) & 1307 & 3921 & 5228 \\ Total & 1355 & 3927 & 5282 \\ \hline \end{tabular}

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