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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 10.1.16

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)


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Step 1: Identify the claim and state the null and alternative hypotheses. The claim is that the number of births by day of the week is uniformly distributed. The null hypothesis (H₀) states that the distribution of births is uniform across all days of the week. The alternative hypothesis (Hₐ) states that the distribution of births is not uniform across all days of the week.
Step 2: Calculate the expected frequency for each day. Since the total number of births is 700 and there are 7 days in a week, the expected frequency for each day is calculated as: 7007. This gives the expected frequency for each day.
Step 3: Compute the chi-square test statistic using the formula: (O-E)2E, where O represents the observed frequency and E represents the expected frequency for each day. Substitute the observed frequencies from the table and the calculated expected frequency into the formula.
Step 4: Determine the critical value and rejection region. Use the chi-square distribution table with degrees of freedom (df) equal to the number of categories minus 1. In this case, df = 7 - 1 = 6. At α = 0.10, find the critical value corresponding to df = 6. The rejection region is where the test statistic exceeds the critical value.
Step 5: Compare the test statistic to the critical value and make a decision. If the test statistic falls in the rejection region, reject the null hypothesis (H₀). Otherwise, fail to reject the null hypothesis. Interpret the decision in the context of the original claim about the uniform distribution of births.

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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 events in different categories to the frequencies expected under a specific hypothesis. This test is particularly useful for categorical data, allowing researchers to assess whether the observed data significantly deviates from what would be expected if the null hypothesis were true.
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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) reflects the claim being tested. For the Chi-Square Goodness-of-Fit Test, H₀ typically states that the observed frequencies are consistent with the expected frequencies, while Hₐ suggests that there is a significant difference. Clearly defining these hypotheses is crucial for determining the outcome of the test.
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Step 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 rejecting H₀. If the calculated test statistic falls within this region, it suggests that the observed data is unlikely under the null hypothesis, prompting a rejection of H₀.
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Critical Values: t-Distribution
Related Practice
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=27, d.f.D=19"

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.


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)"

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

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

"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

Contingency Tables and Relative Frequencies In Exercises 33–36, use the information below.

The frequencies in a contingency table can be written as relative frequencies by dividing each frequency by the sample size. The contingency table below shows the number of U.S. adults (in millions) ages 25 and over by employment status and educational attainment. (Adapted from U.S. Census Bureau)


What percent of U.S. adults ages 25 and over (a) are employed and are only high school graduates, (b) are not in the civilian labor force, and (c) are not high school graduates?