Explain why the chi-square independence test is always a right-tailed test.
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- 8. Sampling Distributions & Confidence Intervals: Proportion2h 10m
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- 13. Chi-Square Tests & Goodness of Fit2h 21m
- 14. ANOVA2h 28m
13. Chi-Square Tests & Goodness of Fit
Independence Tests
Problem 10.2.18
Textbook Question
Performing a Chi-Square Independence Test In Exercises 13–28, perform the indicated chi-square independence test by performing the steps below.
a. Identify the claim and state H₀ and Hₐ
b. Determine the degrees of freedom, find the critical value, and identify the rejection region.
c. Find the chi-square test statistic.
d. Decide whether to reject or fail to reject the null hypothesis.
e. Interpret the decision in the context of the original claim.
Achievement and School Location The contingency table shows the results of a random sample of students by the location of school and the number of those students achieving basic skill levels in three subjects. At α=0.01, test the hypothesis that the variables are independent. (Adapted from HUD State of the Cities Report)

Verified step by step guidance1
Step 1: Identify the claim and state the null hypothesis (H₀) and alternative hypothesis (Hₐ). The claim is that the variables 'location of school' and 'achievement in basic skill levels' are independent. H₀: The variables are independent. Hₐ: The variables are not independent.
Step 2: Determine the degrees of freedom (df), find the critical value, and identify the rejection region. Degrees of freedom are calculated using the formula df = (number of rows - 1) × (number of columns - 1). Here, df = (2 - 1) × (3 - 1) = 2. Using α = 0.01, find the critical value from the chi-square distribution table for df = 2. The rejection region is where the test statistic exceeds the critical value.
Step 3: Calculate the expected frequencies for each cell in the contingency table using the formula E = (row total × column total) / grand total. For example, for the cell corresponding to 'Urban' and 'Reading', E = (total for Urban × total for Reading) / grand total. Repeat this for all cells.
Step 4: Compute the chi-square test statistic using the formula χ² = Σ((O - E)² / E), where O is the observed frequency and E is the expected frequency. Sum this value across all cells in the table.
Step 5: Compare the calculated chi-square test statistic to the critical value. If the test statistic exceeds the critical value, reject H₀; otherwise, fail to reject H₀. Interpret the decision in the context of the original claim: If H₀ is rejected, conclude that the variables are not independent; if H₀ is not rejected, conclude that the variables are independent.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Chi-Square Independence Test
The Chi-Square Independence Test is a statistical method used to determine if there is a significant association between two categorical variables. It compares the observed frequencies in each category of a contingency table to the frequencies expected if the variables were independent. A significant result suggests that the variables are related, while a non-significant result indicates independence.
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Independence Test
Null and Alternative Hypotheses
In hypothesis testing, the null hypothesis (H₀) represents the default position that there is no effect or relationship between the variables, while the alternative hypothesis (Hₐ) posits that there is a significant effect or relationship. For the Chi-Square Independence Test, H₀ typically states that the two categorical variables are independent, and Hₐ states that they are not independent.
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Step 1: Write Hypotheses
Degrees of Freedom and Critical Value
Degrees of freedom (df) in a Chi-Square test are calculated based on the number of categories in the variables being analyzed, specifically as (rows - 1) * (columns - 1). The critical value is a threshold that determines the rejection region for the null hypothesis. If the calculated Chi-Square statistic exceeds the critical value at a specified significance level (e.g., α = 0.01), the null hypothesis is rejected.
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Critical Values: t-Distribution
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