Skip to main content
Ch. 10 - Chi-Square Tests and the F-Distribution
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Non è quello che usi tu?Cambia libro di testo
Capitolo 10, Problema 10.2.22

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.


Choosing a College The contingency table shows the results of a survey asking 1858 parents and students of different incomes what their deciding factor was in choosing a college. At α=0.01, can you conclude that the deciding factor in choosing a college is related to the income of the family? (Adapted from Sallie Mae)


Guida verificata passo dopo passo
1
Step 1: Identify the claim and state the hypotheses. The claim is that the deciding factor in choosing a college is related to the family income. Set the null hypothesis H₀: The deciding factor is independent of family income. Set the alternative hypothesis Hₐ: The deciding factor is related to family income.
Step 2: Determine the degrees of freedom (df). Use the formula df = (number of rows - 1) × (number of columns - 1). Here, there are 3 income groups (rows) and 4 deciding factors (columns), so df = (3 - 1) × (4 - 1) = 2 × 3 = 6. Then, find the critical value from the chi-square distribution table at α = 0.01 and df = 6. Identify the rejection region as chi-square values greater than this critical value.
Step 3: Calculate the expected counts for each cell in the contingency table using the formula: Expected = Row total × Column totalGrand total. Do this for all 12 cells.
Step 4: Compute the chi-square test statistic using the formula: χ2 = ∑all cells (Observed − Expected)2Expected. Sum the squared differences between observed and expected counts divided by expected counts for all cells.
Step 5: Compare the calculated chi-square statistic to the critical value. If the statistic is greater than the critical value, reject the null hypothesis; otherwise, fail to reject it. Finally, interpret the decision in context: state whether there is sufficient evidence at the 0.01 significance level to conclude that the deciding factor in choosing a college is related to family income.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
6m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Chi-Square Test of Independence

This test determines whether two categorical variables are related or independent. It compares observed frequencies in a contingency table to expected frequencies calculated under the assumption of independence. A significant result suggests a relationship between the variables.
Video consigliato:
Percorso guidato
06:28
Independence Test

Degrees of Freedom and Critical Value

Degrees of freedom for a chi-square test of independence are calculated as (number of rows - 1) × (number of columns - 1). The critical value is found using the chi-square distribution table at a chosen significance level (α). It defines the rejection region for the test.
Video consigliato:
Percorso guidato
05:50
Critical Values: t-Distribution

Hypothesis Testing and Interpretation

Hypothesis testing involves stating a null hypothesis (H₀: variables are independent) and an alternative hypothesis (Hₐ: variables are related). After calculating the test statistic, compare it to the critical value to decide whether to reject H₀. The conclusion should be interpreted in the context of the problem.
Video consigliato:
Percorso guidato
05:52
Performing Hypothesis Tests: Proportions
Pratica correlata
Domanda del libro di testo

"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=0.025, d.f.N=7, d.f.D=3"

70
views
Domanda del libro di testo

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)


41
views
Domanda del libro di testo

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.


111
views
Domanda del libro di testo

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.


217
views
Domanda del libro di testo

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.


Homicides by Month A researcher claims that the number of homicide crimes in California by month is uniformly distributed. To test this claim, you randomly select 2000 homicides from a recent year and record the month when each happened. The table shows the results. At α=0.10, test the researcher’s claim. (Adapted from California Department of Justice)


50
views
Domanda del libro di testo

In each exercise,

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 1 and 2, use the table, which lists the distribution of educational achievement for people in the United States ages 25 and older. It also lists the results of a random survey for two additional age groups. (Adapted from U.S. Census Bureau)



Use the data for 30- to 34-year-olds and 65- to 69-year-olds to test whether age and educational attainment are related. Use α=0.01.

70
views