Textbook Question
What conditions are necessary in order to use a one-way ANOVA test?
Ch. 10 - Chi-Square Tests and the F-Distribution
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470
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Table of contents
- Ch. 1 - Introduction to Statistics87
- Ch. 2 - Descriptive Statistics176
- Ch. 3 - Probability184
- Ch. 4 - Discrete Probability Distributions102
- Ch. 5 - Normal Probability Distributions183
- Ch. 6 - Confidence Intervals154
- Ch. 7 - Hypothesis Testing with One Sample165
- Ch. 8 - Hypothesis Testing with Two Samples134
- Ch. 9 - Correlation and Regression87
- Ch. 10 - Chi-Square Tests and the F-Distribution87
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Not the one you use?Change textbook
Chapter 10, Problem 10.1.8
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.
Coffee A researcher claims that the numbers of cups of coffee U.S. adults drink per day are distributed as shown in the figure. You randomly select 1600 U.S. adults and ask them how many cups of coffee they drink per day. The table shows the results. At α=0.05, test the researcher’s claim. (Adapted from Gallup)
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Step 1: Identify the claim and state the null hypothesis (H₀) and alternative hypothesis (Hₐ). The claim is that the distribution of the number of cups of coffee U.S. adults drink per day matches the given percentages: 0 cups (36%), 1 cup (26%), 2 cups (19%), 3 cups (8%), and 4 or more cups (11%). H₀: The observed frequencies follow the claimed distribution. Hₐ: The observed frequencies do not follow the claimed distribution.
Step 2: Calculate the expected frequencies for each category using the total sample size (1600) and the claimed percentages. For example, for 0 cups, the expected frequency is calculated as 1600 × 0.36 = 576. Repeat this for all categories: 1 cup, 2 cups, 3 cups, and 4 or more cups.
Step 3: Compute the chi-square test statistic using the formula: χ² = Σ((Oᵢ - Eᵢ)² / Eᵢ), where Oᵢ is the observed frequency and Eᵢ is the expected frequency for each category. Perform this calculation for each category and sum the results.
Step 4: Determine the critical value for the chi-square test at α = 0.05 with degrees of freedom (df). The degrees of freedom are calculated as df = (number of categories - 1). In this case, df = 5 - 1 = 4. Use a chi-square distribution table to find the critical value for df = 4 and α = 0.05.
Step 5: Compare the calculated chi-square test statistic to the critical value. If the test statistic exceeds the critical value, reject the null hypothesis (H₀). Otherwise, fail to reject H₀. Interpret the decision in the context of the original claim: If H₀ is rejected, the observed frequencies do not match the claimed distribution; if H₀ is not rejected, the observed frequencies are consistent with the claimed distribution.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Chi-Square Goodness-of-Fit Test
The Chi-Square Goodness-of-Fit Test is a statistical method used to determine if a sample distribution matches an expected distribution. It compares the observed frequencies of outcomes in categorical data to the frequencies expected under a specific hypothesis. This test helps assess whether the differences between observed and expected values are due to chance or indicate a significant deviation.
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10:17
Goodness of Fit Test
Null and Alternative Hypotheses (H₀ and Hₐ)
In hypothesis testing, the null hypothesis (H₀) represents a statement of no effect or no difference, while the alternative hypothesis (Hₐ) suggests that there is an effect or a difference. For the Chi-Square Goodness-of-Fit Test, H₀ typically states that the observed distribution of data matches the expected distribution, while Hₐ posits that there is a significant difference between the two.
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06:21Step 1: Write Hypotheses
Critical Value and Rejection Region
The critical value in hypothesis testing is a threshold that determines the boundary for rejecting the null hypothesis. It is derived from the significance level (α), which indicates the probability of making a Type I error. The rejection region is the range of values for the test statistic that leads to the rejection of H₀; if the calculated test statistic falls within this region, the null hypothesis is rejected in favor of the alternative hypothesis.
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05:50Critical Values: t-Distribution
Related Practice
Textbook Question
"Finding Left-Tailed Critical F-Values In this section, you only needed to calculate the right-tailed critical F-value for a two-tailed test. For other applications of the F-distribution, you will need to calculate the left-tailed critical F-value. To calculate the left-tailed critical F-value, perform the steps below.
1. Interchange the values for d.f.N and d.f.D.
2. Find the corresponding F-value in Table 7.
3. Calculate the reciprocal of the F-value to obtain the left-tailed critical F-value.
In Exercises 27 and 28, find the right- and left-tailed critical F-values for a two-tailed test using the level of significance α and degrees of freedom d.f.N and d.f.D.
α=0.10, d.f.N=20, d.f.D=15"
Textbook Question
Finding Expected Frequencies
In Exercises 3–6, find the expected frequency for the values of n and pᵢ.
n=500, pᵢ=0.9
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 = 0.10.
Sample statistics: s₁² = 773, n₁ = 5 and s₂² = 765, n₂ = 6"
Textbook Question
Conditional Relative Frequencies In Exercises 37–42, use the contingency table from Exercises 33–36, and the information below.
Relative frequencies can also be calculated based on the row totals (by dividing each row entry by the row’s total) or the column totals (by dividing each column entry by the column’s total). These frequencies are conditional relative frequencies and can be used to determine whether an association exists between two categories in a contingency table.
What percent of U.S. adults ages 25 and over who have a degree are not in the civilian labor force?
Textbook Question
Describe the difference between the variance between samples MSB and the variance within samples MSW.
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