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Ch. 10 - Chi-Square Tests and the F-Distribution
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
All textbooksLarson 8th EditionCh. 10 - Chi-Square Tests and the F-DistributionProblem 10.2.6
Chapter 10, Problem 10.2.6

True or False? In Exercises 5 and 6, determine whether the statement is true or false. If it is false, rewrite it as a true statement.


When the test statistic for the chi-square independence test is large, you will, in most cases, reject the null hypothesis.

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1
Understand the context: The chi-square independence test is used to determine whether two categorical variables are independent. The null hypothesis (H₀) states that the variables are independent, while the alternative hypothesis (H₁) states that they are not independent.
Recall the relationship between the test statistic and the null hypothesis: A large test statistic indicates a greater deviation from the expected frequencies under the null hypothesis, suggesting that the observed data does not align well with the assumption of independence.
Review the decision rule: In hypothesis testing, if the test statistic is large enough to exceed the critical value (determined by the chi-square distribution and the significance level, α), you reject the null hypothesis.
Evaluate the statement: The statement is true because a large test statistic typically leads to rejecting the null hypothesis, as it implies strong evidence against the assumption of independence.
If the statement were false, the correction would be: 'When the test statistic for the chi-square independence test is large, you will, in most cases, fail to reject the null hypothesis.' However, this is incorrect based on the logic of the test.

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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 to the frequencies expected under the null hypothesis of independence. A large test statistic indicates that the observed data significantly deviates from what would be expected if the variables were independent.
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Independence Test

Null Hypothesis

The null hypothesis is a statement that assumes no effect or no association between variables in a statistical test. In the context of the chi-square independence test, the null hypothesis posits that the two categorical variables are independent of each other. Rejecting the null hypothesis suggests that there is a statistically significant relationship between the variables.
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Step 1: Write Hypotheses

Test Statistic

A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It measures how far the sample statistic is from the null hypothesis, allowing researchers to determine whether to reject or fail to reject the null hypothesis. In the chi-square test, a larger test statistic typically indicates stronger evidence against the null hypothesis.
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Step 2: Calculate Test Statistic
Related Practice
Textbook Question

Finding Expected Frequencies

In Exercises 7–12, (a) calculate the marginal frequencies and (b) find the expected frequency for each cell in the contingency table. Assume that the variables are independent.


Textbook Question

Performing a Two-Sample F-Test In Exercises 19–26, (a) identify the claim and state H0 and Ha, (b) find the critical value and identify the rejection region, (c) find the test statistic F, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed.


Annual Salaries An employment information service claims that the standard deviation of the annual salaries for public relations managers is less in Louisiana than in Florida. You select a sample of public relations managers from each state. The results of each survey are shown in the figure. At α=0.05, can you support the service’s claim? (Adapted from America’s Career InfoNet)


Textbook Question

Explain how to find the critical value for an F-test.

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Textbook Question

Testing for Normality Using a chi-square goodness-of-fit test, you can decide, with some degree of certainty, whether a variable is normally distributed. In all chi-square tests for normality, the null and alternative hypotheses are as listed below.


H₀: The variable has a normal distribution.


Hₐ: The variable does not have a normal distribution.


To determine the expected frequencies when performing a chi-square test for normality, first estimate the mean and standard deviation of the frequency distribution. Then, use the mean and standard deviation to compute the z-score for each class boundary. Then, use the z-scores to calculate the area under the standard normal curve for each class. Multiplying the resulting class areas by the sample size yields the expected frequency for each class.In Exercises 17 and 18, (a) find the expected frequencies, (b) 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, and (e) interpret the decision in the context of the original claim.


In Exercises 17 and 18, (a) find the expected frequencies, (b) 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, and (e) interpret the decision in the context of the original claim.


Test Scores At α=0.01, test the claim that the 200 test scores shown in the frequency distribution are normally distributed.


Textbook Question

"Finding a Critical F-Value for a Right-Tailed Test In Exercises 5–8, find the critical F-value for a right-tailed test using the level of significance α and degrees of freedom d.f.N and d.f.D.


α=0.05, d.f.N=9, d.f.D=16"

Textbook Question

"In Exercises 13–18, test the claim about the difference between two population variances σ₁² and σ₂² at the level of significance α. Assume the samples are random and independent, and the populations are normally distributed.


Claim: σ₁² = σ₂²; α = 0.05.

Sample statistics: s₁² = 310, n₁ = 7 and s₂² = 297, n₂ = 8"