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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
Chapter 10, Problem 10.3.20

"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.


Carbon Monoxide Emissions An automobile manufacturer claims that the variance of the carbon monoxide emissions for a make and model of one of its vehicles is less than the variance of the carbon monoxide emissions for a top competitor’s equivalent vehicle. A sample of the carbon monoxide emissions of 19 of the manufacturer’s specified vehicles has a variance of 0.008. A sample of the carbon monoxide emissions of 21 of its competitor’s equivalent vehicles has a variance of 0.045. At α=0.10, can you support the manufacturer’s claim? (Adapted from U.S. Environmental Protection Agency)"

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Step 1: Identify the claim and state the null and alternative hypotheses. The claim is that the variance of the carbon monoxide emissions for the manufacturer’s vehicles is less than the variance for the competitor’s vehicles. The null hypothesis (H₀) assumes that the variances are equal, while the alternative hypothesis (Hₐ) supports the claim that the variance for the manufacturer’s vehicles is less than the competitor’s. Mathematically, H₀: σ₁² = σ₂² and Hₐ: σ₁² < σ₂², where σ₁² is the variance of the manufacturer’s vehicles and σ₂² is the variance of the competitor’s vehicles.
Step 2: Determine the critical value and rejection region. Since this is a two-sample F-test for variances, use the F-distribution table with degrees of freedom for the numerator (df₁ = n₁ - 1) and the denominator (df₂ = n₂ - 1). Here, df₁ = 19 - 1 = 18 and df₂ = 21 - 1 = 20. The significance level is α = 0.10, and since the alternative hypothesis is one-tailed (left-tailed), find the critical value Fₐ for the left tail. The rejection region will be F < Fₐ.
Step 3: Calculate the test statistic F. The formula for the F-test statistic is F = (s₁² / s₂²), where s₁² is the sample variance of the manufacturer’s vehicles and s₂² is the sample variance of the competitor’s vehicles. Substitute the given values: s₁² = 0.008 and s₂² = 0.045. Compute F = (0.008 / 0.045).
Step 4: Compare the test statistic to the critical value. If the calculated F value is less than the critical value Fₐ, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.
Step 5: Interpret the decision in the context of the original claim. If the null hypothesis is rejected, it supports the manufacturer’s claim that the variance of carbon monoxide emissions for their vehicles is less than that of the competitor’s vehicles. If the null hypothesis is not rejected, there is insufficient evidence to support the manufacturer’s 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 represents the effect or difference we suspect. In this case, the null hypothesis would state that the variance of carbon monoxide emissions for the manufacturer's vehicle is greater than or equal to that of the competitor's vehicle.
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F-Test for Variances

The F-test is a statistical test used to compare the variances of two populations to determine if they are significantly different. It calculates the F-statistic by taking the ratio of the two sample variances. In this scenario, the F-statistic will help assess whether the variance of emissions from the manufacturer's vehicle is indeed less than that of the competitor's vehicle, as claimed.
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Critical Value and Rejection Region

The critical value is a threshold that determines the boundary for rejecting the null hypothesis in hypothesis testing. It is derived from the chosen significance level (α) and the distribution of the test statistic. The rejection region is the range of values for the test statistic that leads to rejecting H0. For this problem, with α=0.10, the critical value will help identify whether the calculated F-statistic falls within the rejection region, thus supporting or refuting the manufacturer's claim.
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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.


Life of Appliances Company A claims that the variance of the lives of its appliances is less than the variance of the lives of Company B’s appliances. A sample of the lives of 20 of Company A’s appliances has a variance of 1.8. A sample of the lives of 25 of Company B’s appliances has a variance of 3.9. At α=0.025, can you support Company A’s claim?"

Textbook Question

Finding a Critical F-Value for a Two-Tailed Test In Exercises 9–12, find the critical F-value for a two-tailed test using the level of significance α and degrees of freedom d.f.N and d.f.D.


α=0.01, d.f.N=6, d.f.D=7

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.


Use the contingency table and expected frequencies from Exercise 11. At α=0.10, test the hypothesis that the variables are independent.

Textbook Question

Performing a Chi-Square Goodness-of-Fit Test

In Exercises 7–16, (a) identify the claim and state H₀ and Hₐ, (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.


Births by Day of the Week A doctor claims that the number of births by day of the week is uniformly distributed. To test this claim, you randomly select 700 births from a recent year and record the day of the week on which each takes place. The table shows the results. At α=0.10, test the doctor’s claim. (Adapted from National Center for Health Statistics)


Textbook Question

"Finding a Critical F-Value for a Two-Tailed Test In Exercises 9–12, find the critical F-value for a two-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=60, d.f.D=40"

Textbook Question

Performing a One-Way ANOVA Test In Exercises 5–14, (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, the populations are normally distributed, and the population variances are equal.


[APPLET] Well-Being Index The well-being index is a way to measure how people are faring physically, emotionally, socially, and professionally, as well as to rate the overall quality of their lives and their outlooks for the future. The table shows the well-being index scores for a sample of states from four regions of the United States. At α=0.10, can you reject the claim that the mean score is the same for all regions? (Adapted from Gallup and Healthways)