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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 10.3.26
Chapter 10, Problem 10.3.26

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)


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Identify the claim and state the null and alternative hypotheses: The claim is that the standard deviation of annual salaries for public relations managers in Louisiana is less than in Florida. The null hypothesis (H₀) states that the variances are equal (σ₁² = σ₂²), while the alternative hypothesis (Hₐ) states that the variance in Louisiana is less than in Florida (σ₁² < σ₂²).
Determine the critical value and rejection region: Since this is a one-tailed F-test with α = 0.05, use the F-distribution table to find the critical value. The degrees of freedom for Louisiana are df₁ = n₁ - 1 = 24 - 1 = 23, and for Florida, df₂ = n₂ - 1 = 28 - 1 = 27. Look up the critical value for these degrees of freedom at the 0.05 significance level.
Calculate the test statistic F: The formula for the F-test statistic is F = (s₁² / s₂²), where s₁ and s₂ are the sample standard deviations. Square the given standard deviations (s₁ = \(27,200 and s₂ = \)39,100) and divide the variance of Louisiana by the variance of Florida to compute F.
Decide whether to reject or fail to reject the null hypothesis: Compare the calculated F value to the critical value. If F is less than the critical value, reject the null hypothesis; otherwise, fail to reject the null hypothesis.
Interpret the decision in the context of the original claim: If the null hypothesis is rejected, conclude that there is sufficient evidence to support the claim that the standard deviation of annual salaries for public relations managers in Louisiana is less than in Florida. If the null hypothesis is not rejected, conclude that there is not enough evidence to support the claim.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions about population parameters based on sample data. It involves formulating two competing hypotheses: the null hypothesis (H0), which represents no effect or no difference, and the alternative hypothesis (Ha), which indicates the presence of an effect or difference. In this case, the claim is that the standard deviation of salaries in Louisiana is less than in Florida, leading to H0: σ1 ≥ σ2 and Ha: σ1 < σ2.
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Step 1: Write Hypotheses

F-Test

The F-test is a statistical test used to compare the variances of two populations to determine if they are significantly different. It is particularly useful when assessing the equality of variances, which is a key assumption in many statistical analyses. The test statistic, F, is calculated as the ratio of the two sample variances, and the critical value is determined based on the desired significance level (α) and the degrees of freedom for each sample.
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Tukey Test

Rejection Region

The rejection region in hypothesis testing is the range of values for the test statistic that leads to the rejection of the null hypothesis. It is determined by the significance level (α), which represents the probability of making a Type I error. For a one-tailed F-test at α = 0.05, the rejection region is located in the upper tail of the F-distribution, indicating that if the calculated F statistic falls within this region, the null hypothesis should be rejected in favor of the alternative hypothesis.
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Step 4: State Conclusion
Related Practice
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.


Heart Transplant Waiting Times The table at the left shows a sample of the waiting times (in days) for a heart transplant for two age groups. At α=0.05, can you conclude that the variances of the waiting times differ between the two age groups? (Adapted from Organ Procurement and Transplantation Network)


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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.


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Textbook Question

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

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Textbook Question

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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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"

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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=0.05, test the claim that the 400 test scores shown in the frequency distribution are normally distributed.


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