Chi-Square Tests Notes

Chapter 16: Inference for Counts (Chi-Square Tests)

16.1 Goodness-of-Fit Tests

• Deals with count data at the nominal or ordinal level.
• Examines counts of items in different categories.
• Asks if the actual distribution differs from a predicted model due to random error or a poor model fit.
• Example: Analyzing "up" days in the stock market and comparing it to a population distribution.

16.1 Logic of Goodness-of-Fit Tests

• Individuals are classified and counted into categories.
• Observed frequencies are tabulated for each category.
• Each individual is counted in only one category.
• Expected frequencies are constructed based on the null hypothesis (H0).
• Expected frequency: The frequency predicted by H0 and the sample size, representing an idealized distribution.
• Compares observed frequencies from sample data with expected frequencies predicted by H0.
• Uses the Chi-Square distribution to represent values for all possible random samples when H0 is true.

16.2 Interpreting Chi-Square Values

Chi-Square Distribution

• Positively skewed distribution.
• A family of distributions, each determined by degrees of freedom (df).
• Each df value results in a slightly different shape.

16.2 Hypothesis Tests: Chi-Square Goodness-of-Fit

Step 1: Hypotheses

• Case I: No preference (equal proportions) among categories
  • H0: p1 = p2 = \cdots = pk = a, where 'a' is the probability value to test, and 'k' is the number of categories.
  • HA: At least one pi is not equal to its specified value.
• Case II: Specified proportions or preference from another known population
  • H0: p1 = a1, p2 = a2, \cdots = pk = ak, where a1, a2, \cdots , ak are the probability values or relative preferences of each category.
  • HA: At least one pi is not equal to its specified value.

Step 2: Decision Zone and Type of Test

• \chi^2 is the test statistic (Chi-square).
• Right-tailed test (0 < \chi^2 < \infty).
• Reject H0 if \chi^2 > \chi^2{cv}, where \chi^2{cv} is the critical value from the chi-square distribution with (k - 1) degrees of freedom.

Step 3 and 4: Observed and Expected Frequencies

• Observed Frequencies: Obs(oi). Property: \sum{i=1}^{k} o_i = n
• Expected Frequencies: Exp(ei) = npi. Property: \sum{i=1}^{k} oi = \sum{i=1}^{k} ei = n, where p_i are the probabilities of categories specified in H0.

Step 5: Test Statistic

• \chi^2 = \sum{i=1}^{k} \frac{(oi - ei)^2}{ei}

Step 6: Make a Decision (Critical Value Approach)

• Reject H0 if the test statistic (\chi^2) is in the critical region.
• Fail to reject H0 if the test statistic (\chi^2) is not in the critical region.
• Rejection Region: Reject H0 if \chi^2 > \chi^2_{cv} (Right-tailed Test).

Step 6: Make a Decision (p-value Approach)

• p-value: Probability of seeing the observed data (or more extreme) given H0 is true.
• Calculation of p-value: P(\chi^2 > \chi^2_0) (Right-tailed Test).
• Decision Rule:
  • If p-value < \alpha, reject H0.
  • If p-value \geq \alpha, do not reject H0.

Assumptions and Conditions

• Counted Data Condition: Data must be counts for categories of a categorical variable.
• Independence Assumption: Counts should be independent of each other.
• Randomization Condition: Counted individuals should be a random sample of the population.
• Sample Size Assumption: Enough data for the methods to work.
• Expected Cell Frequency Condition: Expected cell frequency (ei) in each cell/category should be at least 5. If not, collapse adjacent cells meaningfully until ei \geq 5 and adjust df accordingly.

Example 1: Case I (No Preference or Equal Proportion)

• Problem: Are technical support calls equal across all days of the week (uniform distribution)?
• Sample data: Technical support calls for 10 days per day of week.
• Step 1: Hypotheses
  • H0: p1 = p2 = \cdots = pk = \frac{1}{7}
  • HA: At least one p_i is not equal to \frac{1}{7}.
• Step 2: Decision Zone and Critical Value
  • \chi^2{\alpha,k-1} = \chi^2{.05,6} = 12.5916
• Steps 3 and 4: Observed and Expected Frequencies
• Assumptions and Conditions checked
• Steps 5 and 6: Test Statistic and Make a Decision (CV and p-value Approach)
  • Decision: Reject H0. The distribution of technical support calls is not uniform across the days of the week.

Example 2: Case II (Specified Preference or Proportion)

• Problem: Market share analysis of fabric softener companies A and B after advertising campaigns.
• Before campaigns: Company A (45%), Company B (40%), Others (15%).
• Sample: 200 customers; A (102), B (82), Others (16).
• Question: Have customer preferences changed at a 5% significance level?
• Step 1: Hypotheses
  • H0: p1 = 0.45, p2 = 0.40, p3 = 0.15
  • HA: At least one p_i is not equal to its specified value.
• Step 2: Decision Zone and Critical Value
  • \chi^2{\alpha,k-1} = \chi^2{.05,2} = 5.99147
• Steps 3 - 5: Observed, Expected Frequencies, and Test Statistic
• Assumptions and Conditions checked
• Step 6: Make a Decision (CV and p-value Approach)
  • Decision: Reject H0. Market preferences have changed since the advertising campaigns.

Goodness-of-Fit Test for Normal Distribution: Case II (Specified Preference or Proportion)

• Used to check the validity of the assumption of a normal distribution used in many statistical methods.
• Can use frequency distributions, stem-and-leaf displays, histograms, and normal plots.
• Alternatively can conduct a chi-square goodness-of-fit test.

Example 3: Case II (Specified Preference or Proportion)

• Test of Normality: Histogram of 50 gasoline mileages is symmetrical and bell-shaped.
• Sample selected from a normally distributed population.
• Uses chi-square goodness-of-fit test to check normality.
• Step 1: Hypotheses
  • H0: The population of all mileages is normally distributed
  • HA: The population of all mileages is NOT normally distributed.
• Steps 3 - 4: Observed and Expected Frequencies
  • where \bar{x} = 31.56, s = 0.7977. So, for example, P(X < 30.0) = P(\frac{X-\bar{x}}{S} < \frac{30-\bar{x}}{s} ) = P(Z < \frac{30.0-31.56}{0.7977} ) = P(Z < -1.96) \approx .0256
• Assumptions and Conditions checked
• Steps 3 - 4: Observed and Expected Frequencies (Collapsed Categories)
• Step 6: Make a Decision
  • Adjusted degrees of freedom for Chi-square is df = k - 1 - m = 5 - 1 - 2 = 2
  • Decision: Do NOT reject H0 since \chi^2 = 0.30102 < \chi^2 = .05,df =2 = 5.99147.
  • The population of mileages is normally distributed.

16.3 Examining the Residuals

• Used when the null hypothesis is rejected, to discover which values are extraordinary or contribute most to the aggregate chi-square value.
• Examine standardized residuals to compare cells with different counts: \frac{(oi - ei)}{\sqrt{e_i}}
• Standardized residuals from goodness-of-fit tests are z-scores.

Example: Standardized Residuals for Technical Support Data

• Largest value, Sunday, at −3.44, is impressive when viewed as a z-score.

16.4 The Chi-Square Test of Homogeneity or Independence (Contingency Table Analysis)

• Categorical data summarized in a contingency table (cross-tab or pivot table).
• Sample observations cross-classified according to two or more identifiable characteristics.
• Used to determine independence of characteristics of interest and to test for homogeneity.
• Testing for homogeneity is almost the same as testing for independence.

Contingency Table

• Classifies sample observations according to two or more identifiable characteristics.
• Also called a cross-tabulation or pivot table.
• Referred to by row (r) and column (c) dimension, r \times c.

16.4 Hypothesis Tests: Chi-Square Test of Homogeneity or Independence

• Step 1: Hypotheses
• Using the Word “Independence”:
  • H0: Variable 1 is independent of Variable 2.
  • HA: Variable 1 is NOT independent of Variable 2.
• Using the Word “Association”:
  • H0: Variable 1 is NOT associated with Variable 2.
  • HA: Variable 1 is associated with Variable 2.
• Step 2: Decision Zone and Type of Test:
  • \chi^2 is the lower-case Greek letter Chi.
  • Right-tailed Test (0 < \chi^2 < \infty).
  • Reject H0 if \chi^2 > \chi^2{cv} where \chi^2{cv} is the critical value from chi-square distribution with (r − 1)(c − 1) degrees of freedom.
• Step 3 and 4: Observed and Expected Frequencies
  • Observed Frequencies: Obs(o{ij}) (Property: \sum{i=1}^{r} \sum{j=1}^{c} o{ij} = n)
  • Expected Frequencies: Exp(e_{ij}) = \frac{i^{th} \text{row total } \times j^{th} \text{column total}}{\text{grand total or sample size}}
  • Property: \sum{i=1}^{r} \sum{j=1}^{c} o{ij} = \sum{i=1}^{r} \sum{j=1}^{c} e{ij} = n
  • e{ij} = n p{ij} = n \times pi \times pj
• Step 5: Test Statistic
  • \chi^2 = \sum{i=1}^{r} \sum{j=1}^{c} \frac{(o{ij} - e{ij})^2}{e_{ij}}
• Step 6: Make a Decision (Critical Value Approach)
  • If test statistic (\chi^2) is located in the critical region, the null hypothesis is rejected.
  • If the test statistic (\chi^2) is not located in the critical region, the researcher fails to reject the null hypothesis.
  • Rejection Region: Reject H0 if \chi^2 > \chi^2_{cv} (Right-tailed Test).
• Step 6: Make a Decision (p-value Approach)
  • p-value: The p-value is the probability of seeing the observed data (or more extreme) given the null hypothesis is true.
  • Calculation of p-value: P(\chi^2 > \chi^2_0) (Right-tailed Test).
  • Decision Rule:
    • If p-value < \alpha, reject H0.
    • If p-value \geq \alpha, do not reject H0.
• Assumptions and Conditions:
  • Counted Data Condition: The data must be counts for the categories of a categorical variable.
  • Independence Assumption: The counts should be independent of each other. Think about whether this is reasonable.
  • Randomization Condition: The counted individuals should be a random sample of the population. Guard against auto-correlated samples.
  • Sample Size Assumption: We must have enough data for the methods to work.
  • Expected Cell Frequency Condition: Expected cell frequency (e{ij}) in each cell or category should be at least 5. If the expected frequencies (e{ij}) are less than 5, adjacent cells need to be meaningfully collapsed until benchmark (e_{ij} \geq 5) is achieved and df should be adjusted accordingly.

Example 1:

• Problem: Is sex of yearbook editor independent of college’s funding source?
• Step 1: Hypotheses
  • Using the Word “Independence”:
    • H0: Sex of yearbook editor is independent of the college’s funding source.
    • HA: Sex of yearbook editor is NOT independent of college’s funding source.
  • Using the Word “Association”:
    • H0: Sex of yearbook editor is NOT associated with college’s funding source.
    • HA: Sex of yearbook editor is associated with college’s funding source.
• Steps 3 and 4: Observed and Expected Frequencies
• Assumptions and Conditions
• Steps 5 and 6: Test Statistic and Make a Decision (CV Approach)
• Step 6: Make a Decision (p-value Approach)
  • Decision: Since p-value < 0.005 < α = .05, we may reject the null hypothesis based on sample evidence and conclude that the sex and source of funding are not independent.