Test of association: chi-squared

Question 1

Overview of Chi-Squared Test

Answer 1

• Definition: A statistical test used to determine whether there is a significant association between categorical variables.
• Purpose: To assess how likely it is that an observed distribution of data fits with a specific distribution.

Question 2

Types of Chi-Squared Tests

Answer 2

1. Chi-Squared Test for Independence:
   o Tests whether two categorical variables are independent of each other.
   o Example: Examining if gender is associated with preference for a specific product.
2. Chi-Squared Test for Goodness of Fit:
   o Tests whether the observed frequency distribution of a single categorical variable matches an expected distribution.
   o Example: Determining if a die is fair based on observed rolls.

Question 3

Formula for Chi-Squared Test

Answer 3

χ2=∑(O−E)2E\chi^2 = \sum \frac{(O - E)^2}{E}χ2=∑E(O−E)2
where:
* OOO = Observed frequency
* EEE = Expected frequency

Question 4

Steps to Conduct a Chi-Squared Test

Answer 4

1. State the Hypotheses:
   o Null Hypothesis (H0H_0H0): Assumes no association between variables (for independence).
   o Alternative Hypothesis (HaH_aHa): Assumes there is an association.
2. Create a Contingency Table: For independence, list observed frequencies for each category.
3. Calculate Expected Frequencies:
   E=(Row Total×Column Total)Grand TotalE = \frac{(Row\ Total \times Column\ Total)}{Grand\ Total}E=Grand Total(Row Total×Column Total)
4. Calculate Chi-Squared Statistic: Use the formula provided.
5. Determine Degrees of Freedom (df):
   df=(r−1)(c−1)df = (r - 1)(c - 1)df=(r−1)(c−1)
   where rrr is the number of rows and ccc is the number of columns.
6. Compare to Critical Value: Use a chi-squared distribution table to find the critical value at a specified significance level (usually 0.05).
7. Make a Decision: If χ2\chi^2χ2 calculated > critical value, reject H0H_0H0.

Question 5

Interpretation of Results

Answer 5

• Outcome:
  o If H0H_0H0 is rejected: Conclude there is a significant association between the variables.
  o If H0H_0H0 is not rejected: Conclude there is no significant association.
• Example Reporting: “The chi-squared test revealed a significant association between smoking status and lung disease, χ2(1,N=100)=15.32,p<0.001\chi^2(1, N = 100) = 15.32, p < 0.001χ2(1,N=100)=15.32,p<0.001.”

Question 6

Advantages and Limitations

Answer 6

• Advantages:
  o Simple to compute and interpret.
  o Useful for analyzing categorical data.
• Limitations:
  o Cannot provide information about the strength or direction of the association.
  o Sensitive to sample size; large samples can lead to statistically significant results even for trivial associations.

Question 7

Assumptions of Chi-Squared Test

Answer 7

• Observations should be independent.
• The sample size should be sufficiently large (expected frequencies should generally be 5 or more).
• Data should be in the form of counts (frequencies) rather than percentages or proportions.

Question 8

Example of Chi-Squared Test for Independence

Answer 8

• Scenario: Investigating the relationship between smoking status (smoker, non-smoker) and lung disease (yes, no).
• Data Collection: Collect data from a sample of individuals and create a contingency table.
  Lung Disease Lung Disease-No Total
  Smoker 30 10 40
  Non-Smoker 5 55 60
  Total 35 65 100
• Expected Frequencies Calculation:
  o For smokers with lung disease: E=(40×35)100=14E = \frac{(40 \times 35)}{100} = 14E=100(40×35)=14
  o For non-smokers with lung disease: E=(60×35)100=21E = \frac{(60 \times 35)}{100} = 21E=100(60×35)=21
• Chi-Squared Calculation: Calculate χ2\chi^2χ2 and compare it with the critical value.