In the context of a chi-square test for independence, what does a large value of the statistic indicate about the relationship between the two categorical variables being tested?
Table of contents
- 1. Intro to Stats and Collecting Data1h 14m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically2h 5m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables3h 6m
- 6. Normal Distribution and Continuous Random Variables2h 11m
- 7. Sampling Distributions & Confidence Intervals: Mean3h 23m
- Sampling Distribution of the Sample Mean and Central Limit Theorem19m
- Distribution of Sample Mean - Excel23m
- Introduction to Confidence Intervals15m
- Confidence Intervals for Population Mean1h 18m
- Determining the Minimum Sample Size Required12m
- Finding Probabilities and T Critical Values - Excel28m
- Confidence Intervals for Population Means - Excel25m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 25m
- 9. Hypothesis Testing for One Sample3h 29m
- 10. Hypothesis Testing for Two Samples4h 50m
- Two Proportions1h 13m
- Two Proportions Hypothesis Test - Excel28m
- Two Means - Unknown, Unequal Variance1h 3m
- Two Means - Unknown Variances Hypothesis Test - Excel12m
- Two Means - Unknown, Equal Variance15m
- Two Means - Unknown, Equal Variances Hypothesis Test - Excel9m
- Two Means - Known Variance12m
- Two Means - Sigma Known Hypothesis Test - Excel21m
- Two Means - Matched Pairs (Dependent Samples)42m
- Matched Pairs Hypothesis Test - Excel12m
- 11. Correlation1h 24m
- 12. Regression1h 50m
- 13. Chi-Square Tests & Goodness of Fit2h 21m
- 14. ANOVA1h 57m
13. Chi-Square Tests & Goodness of Fit
Independence Tests
Problem 11.RE.4
Textbook Question
Does the Treatment Affect Success? The following table lists frequencies of successes and failures for different treatments used for a stress fracture in a foot bone (based on data from “Surgery Unfounded for Tarsal Navicular Stress Fracture,” by Bruce Jancin, Internal Medicine News, Vol. 42, No. 14). Use a 0.05 significance level to test the claim that success of the treatment is independent of the type of treatment. What does the result indicate about the increasing trend to use surgery?
Verified step by step guidance1
Step 1: Understand the problem. The goal is to test whether the success of the treatment is independent of the type of treatment. This involves performing a chi-square test for independence, which is used to determine if there is a significant association between two categorical variables.
Step 2: Organize the data into a contingency table. The table should display the frequencies of successes and failures for each type of treatment. Label the rows as 'Type of Treatment' and the columns as 'Success' and 'Failure'.
Step 3: State the null and alternative hypotheses. The null hypothesis (H₀) is that the success of the treatment is independent of the type of treatment. The alternative hypothesis (H₁) is that the success of the treatment is not independent of the type of treatment.
Step 4: Calculate the expected frequencies for each cell in the contingency table using the formula: , where E is the expected frequency, R is the row total, C is the column total, and N is the grand total.
Step 5: Compute the chi-square test statistic using the formula: , where O is the observed frequency and E is the expected frequency. Compare the test statistic to the critical value from the chi-square distribution table at the 0.05 significance level to determine whether to reject the null hypothesis.
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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 a population based on sample data. It involves formulating a null hypothesis (H0) that assumes no effect or relationship, and an alternative hypothesis (H1) that suggests a significant effect or relationship. The goal is to determine whether the observed data provides enough evidence to reject the null hypothesis at a specified significance level, such as 0.05.
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Step 1: Write Hypotheses
Chi-Square Test of Independence
The Chi-Square Test of Independence is a statistical test used to determine if there is a significant association between two categorical variables. In this context, it assesses whether the success of a treatment is independent of the type of treatment used. The test compares the observed frequencies of outcomes in a contingency table to the expected frequencies if the variables were independent, providing a p-value to evaluate significance.
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Significance Level
The significance level, often denoted as alpha (α), is the threshold used to determine whether to reject the null hypothesis in hypothesis testing. A common significance level is 0.05, which indicates a 5% risk of concluding that a difference exists when there is none (Type I error). If the p-value obtained from the test is less than or equal to the significance level, the null hypothesis is rejected, suggesting a statistically significant effect or relationship.
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Step 4: State Conclusion Example 4
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