Chi Square

Chi Square Analysis: Overview

Chi-square analysis is a statistical tool used to quantify whether observed differences between groups or categories are statistically significant.
Purpose: determine if differences are likely due to the treatment (independent variable) rather than random chance.
Core decision rule: compare the calculated chi-square value to a critical value from the chi-square distribution (or equivalently compare to a p-value).
Practical interpretation: if the data are statistically significant, there is a high chance that manipulating the treatment type is causing the change in the dependent variable, leading to REJECTING the Null Hypothesis. If not statistically significant, there is a very low chance of a relationship, leading to FAIL/ACCEPT the Null Hypothesis.
Related Science Practice (SP 5.C):
- Task: calculate the chi-square value and use it to determine the p-value for a given data set.
- Task: draw conclusions about the experiment based on the comparison of the chi-square value to the p-value.

Alternative Hypothesis (H1): States there is a relationship between the independent and dependent variable.
- Example: Treating plant roots with a growth hormone will cause them to grow faster.
Null Hypothesis (H0): States that any observed relationship is due to random chance.
- Example: Treating plant roots with a growth hormone has no effect on how fast the roots grow.

You must have a way to calculate expected values for a “normal” (null) situation.
Steps:
- Compute the expected (E) values based on a model or theoretical probabilities.
- Collect observed (O) data.
- Use chi-square to determine if differences between observed and expected are statistically significant.
Core idea: chi-square assesses whether deviations between O and E are due to chance alone or indicate a real effect.

Chi-square statistic:
\chi^2 = \sumi \frac{(Oi - Ei)^2}{Ei}
Degrees of freedom:
df = k - 1
where k = number of possible outcomes (categories).
Significance decision (general rule):
- If \chi^2 exceeds the critical value from the chi-square distribution with df degrees of freedom (or if the p-value < \alpha), reject H0.
- If \chi^2 does not exceed the critical value (or if the p-value \ge \alpha), fail to reject (accept) H0.
Common alpha level: \alpha = 0.05, corresponding roughly to a critical value that depends on df.

Expected probability: heads = 0.5, tails = 0.5.
Observed (O): Heads = 27, Tails = 23; Total = 50.
Expected (E): Heads = 25, Tails = 25.
Alternative hypothesis (H1): The coin is weighted and the observed difference is significant.
Null hypothesis (H0): There is no significant difference between observed and expected values.
Current question: Can we claim a statistically significant difference based on the current data?
- Answer: No. We need to perform the chi-square test to assess significance.

Step 1: Create a data table
- Columns: Toss, Observed (O), Expected (E), O−E, (O−E)^2, (O−E)^2/E
- Categories: Heads, Tails, Totals
- Total chi-square = ?
Step 2: Fill in Observed values
- Heads: O = 27, Tails: O = 23
Step 3: Calculate Expected values
- Heads: E_{Heads} = 0.5 \times 50 = 25
- Tails: E_{Tails} = 0.5 \times 50 = 25
Step 4: Compute components
- For Heads: O-E = 27-25 = 2; (O-E)^2 = 4; (O-E)^2/E = 4/25 = 0.16
- For Tails: O-E = 23-25 = -2; (O-E)^2 = 4; (O-E)^2/E = 4/25 = 0.16
Step 5: Degrees of freedom
- Two outcomes (Heads, Tails) → df = 2 - 1 = 1
Step 6: Compare to critical value
- Chi-square value: \chi^2 = 0.16 + 0.16 = 0.32
- With df = 1, the critical value at \alpha = 0.05 is \chi^2_{crit} = 3.84
- Since 0.32 < 3.84, the p-value > 0.05; fail to reject H0; differences are not statistically significant.

Scenario: In a cross between two heterozygous plants (Aa x Aa), yellow seeds (A) are dominant over green (a). Observed:
- Yellow: 4400
- Green: 1624
- Total offspring: 6024
Prediction under Mendel's law (phenotypic ratio): Yellow : Green = 3 : 1
Step 1: Determine expected values based on 3/4 vs 1/4 proportions
- Expected yellow = 0.75 × 6024 = 4518
- Expected green = 0.25 × 6024 = 1506
Step 2: Compute deviations and chi-square components
- Observed vs Expected:
- Yellow: O = 4400, E = 4518 → O−E = -118
- Green: O = 1624, E = 1506 → O−E = 118
- Squares and ratio:
- For Yellow: (O−E)^2 = 118^2 = 13924; (O−E)^2/E = 13924/4518 ≈ 3.08
- For Green: (O−E)^2 = 118^2 = 13924; (O−E)^2/E = 13924/1506 ≈ 9.25
Step 3: Sum to get chi-square value
- \chi^2 = 3.08 + 9.25 ≈ 12.3
Step 4: Degrees of freedom
- Number of phenotypes considered = 2 → df = 2 - 1 = 1
Step 5: Critical value and conclusion
- For df = 1 at \alpha = 0.05, \chi^2_{crit} = 3.84
- Since 12.3 > 3.84, p < 0.05; reject H0.
- Conclusion: Differences between observed and expected phenotypic ratios are statistically significant in this data set.

Use cases: chi-square tests are appropriate when you have categorical data with counts in each category and you can specify expected counts under a null model.
Decision rule:
- If \chi^2 > \chi^2_{crit}(df) or equivalently p-value < \alpha: reject H0 (differences are statistically significant).
- If \chi^2 \le \chi^2_{crit}(df) or p-value \ge \alpha: fail to reject H0 (differences may be due to chance).
Relationship to p-value: p-value represents the probability of observing a chi-square as extreme as (or more extreme than) the observed value under H0.
Degrees of freedom: depends on the number of possible outcomes; for a simple two-category test, df = 1; for more categories, df = k − 1.
Practical considerations:
- Ensure expected counts are not too small (commonly E ≥ 5 is a typical guideline for chi-square validity).
- The chi-square test assesses whether observed deviations are inconsistent with the null model, not the magnitude of an effect or its practical significance.
Connections:
- Builds on Mendel’s laws in genetics to predict expected genotype/phenotype frequencies.
- Ties to fundamental probability and sampling concepts from foundational courses.

In genetics, Mendel’s laws help calculate expected offspring counts for a given cross (e.g., Aa x Aa yields 1:2:1 genotype ratio and 3:1 phenotype ratio for dominant vs recessive traits).
Null hypothesis in this context: differences between observed and expected numbers of offspring for each genotype/phenotype are due to random chance.

Observed:
- Yellow = 4400; Green = 1624; Total = 6024
Expected based on 3:1 phenotype ratio:
- Yellow = 4518; Green = 1506
Deviations:
- O−E: Yellow = -118; Green = 118
Squared deviations and standardized contributions:
- (O−E)^2/E: Yellow ≈ 3.08; Green ≈ 9.25
Chi-square total:
- \chi^2 \approx 12.3 with df = 1
Conclusion for this data: Reject Null Hypothesis; the observed distribution deviates significantly from the expected 3:1 ratio.

Core purpose of chi-square analysis: test whether observed categorical data fit a theoretical expectation.
Main outputs: chi-square statistic \chi^2, degrees of freedom df, p-value (or comparison to a critical value).
Practical workflow:
- Define hypotheses (H0, H1).
- Tabulate observed counts (O).
- Compute expected counts (E) under H0.
- Calculate \chi^2 = \sum \frac{(O-E)^2}{E}.
- Determine df and compare to critical value or compute p-value.
- Draw conclusion: reject or fail to reject H0 based on the comparison.

If you have specific data sets, bring them to practice solving a full chi-square problem step by step, including table setup, calculations, df determination, and interpretation.