Genetics for Medical Students: Chi-Square Test and Statistical Analysis Notes (copy)

Definition and Fundamental Applications of the Chi-Square Test

• Definition: A chi-square test is a statistical test used to compare observed results with expected results.

• Primary Purpose: To determine if a difference between observed data and expected data is due to chance, or if it is due to a relationship between the variables being studied.

• General Application in Genetics:

  • Analyzing Mendelian inheritance patterns.

  • Determining if observed genetic crosses align with expected ratios.

  • Understanding gene expression and inheritance mechanisms.

• Pearson’s Chi-Square Tests: This is one of the most common applications of chi-square distributions. These are statistical tests designed for categorical data to determine whether the data significantly differs from what was expected.

• Frequency Distribution Hypotheses: The test is used to address questions such as:

  • Does the frequency distribution of kernel colour in maize follow a hypothetical genetic ratio?

  • Do individuals from several treatments in the same experiment belong to the same population distribution?

  • Are the frequency distributions of two or more populations independent of each other?

Classification of Experimental Data: Measurement vs. Attribute

• Measurement Data: Data specified along a continuous numerical scale. Examples include:

  • Yield.

  • Plant height.

  • Protein content.

• Attribute Data: Concerned with a finite number of discrete classes. Classes are often defined by the presence or absence of an attribute. Examples include:

  • Two-class systems: Male or female, success or failure, effective or ineffective, dead or alive.

  • Multiple-class systems: Varietal classification, colour classification.

• Classification Criteria:

  • One-way classification: Only one criterion is used (e.g., presence/absence of one character, colour classification of plant tissue, or tenure status of farmers).

  • Two-way (and higher) classification: More than one criterion is used to specify the classes.

• Discrete Classification of Quantitative Measurements: In special cases, quantitative data can be grouped into discrete classes:

  • Classifying plants into three height classes ($\text{tall}$, $\text{intermediate}$, and $\text{short}$) instead of measuring in centimeters ($cm$).

  • Scoring vertical resistance to insect pests on a scale from $0$ through $9$ instead of measuring the actual percentage of plant damage.

Statistical Tests for Homogeneity of Variance (Bartlett’s Test)

• Bartlett’s Test: A statistical procedure for testing the equality (homogeneity) of several variances.

• Usage in Agricultural Research:

  • Verifying homogeneity of variances as a requirement for a valid analysis of variance ($\text{ANOVA}$).

  • Verifying homogeneity of error variances when combining data from a series of experiments.

  • Verifying homogeneity of variances in genetic studies where materials consist of genotypes from different filial generations.

  • Verifying homogeneity of sampling variances among samples taken from two or more populations.

• Application Constraint: The chi-square test for homogeneity of variances is applied whenever more than two variances are tested.

Probability Distributions in Agricultural Research

• Test of Goodness of Fit: Determines whether a set of observed data conforms to a specified probability distribution.

• Normal Distribution:

  • The most important continuous distribution in agricultural research (e.g., data on rice yield or protein).

  • Characteristics: Bell-shaped, asymmetric (according to the transcript), and governed by two parameters ($\text{mean}$ and $\text{variance}$).

• Poisson Distribution:

  • A discrete distribution used to represent the occurrence of rare events over space or time (e.g., spatial distribution of weeds in a rice field, insects caught in traps).

  • Characteristics: Governed by a single parameter which serves as both the $\text{mean}$ and the $\text{variance}$.

Mathematical Formula and the Null Hypothesis

• The Null Hypothesis ($H_0$): States that there is no significant difference between the expected and observed results.

• Chi-Square Formula:     $\chi^2 = \frac{(o - e)^2}{e}$

  • Alternative notation: Chi-square is the sum of the squared difference between observed ($o$) and expected ($e$) data (the deviation, $d$), divided by the expected data in all possible categories.

• General Formula for Test Statistic:     $\chi^2 = \sum \left[ \frac{(O_i - E_i)^2}{E_i} \right]$

  • $O_i$: Observed frequency count for the $i^{\text{th}}$ level.

  • $E_i$: Expected frequency count for the $i^{\text{th}}$ level.

Case Study: Mendelian Inheritance in Pea Plants

• Scenario: A cross between two pea plants yields $880$ plants: $639$ with green seeds and $241$ with yellow seeds.

• Hypothesis: The allele for green is dominant to yellow ($G$ is dominant), and both parents are heterozygous.

• Predicted Ratio: $3:1$ ($3/4$ green; $1/4$ yellow).

• Expected Numerical Values:

  • Green seeds: $880 \times 0.75 = 660$

  • Yellow seeds: $880 \times 0.25 = 220$

• Chi-Square Calculation (Table 1):

  • Green: Observed ($639$), Expected ($660$), Deviation ($-21$), Deviation squared ($441$), $d^2/e = 0.668$

  • Yellow: Observed ($241$), Expected ($220$), Deviation ($21$), Deviation squared ($441$), $d^2/e = 2$

  • Total $\chi^2 = 0.668 + 2 = 2.668$

• Interpretation:

  • Degrees of Freedom ($df$): $2 \text{ categories} - 1 = 1$.

  • P-Value: At $\chi^2 = 2.668$ and $df = 1$, the $p$ value is approximately $0.10$.

  • Result: Since p > 0.05 ($10\%$ chance), the deviation is non-significant and consistent with Mendel's Law.

Step-by-Step Procedure for Hypothesis Testing

1. State the Hypothesis: Define the null hypothesis and predicted results.

2. Gather Data: Conduct experiments or use provided data.

3. Determine Expected Numbers: Calculate expected values ($E_i$) for each class. Use numerical values, not percentages.

   • Constraint: Chi-square should not be calculated if the expected value in any category is less than $5$.

4. Calculate $\chi^2$: Complete calculations to three significant digits; round the final answer to two significant digits.

5. Determine Significance:

   • Calculate degrees of freedom ($df = k - 1$).

   • Locate the value in the chi-square distribution table.

   • Find the associated $p$ value.

6. State Conclusion:

   • If p > 0.05: Accept the hypothesis (deviation is due to chance alone).

   • If p < 0.05: Reject the hypothesis (factors other than chance are operating).

Statistical Tables: Chi-Square Distribution

• Note: The values between $p = 0.95$ and $p = 0.10$ are generally non-significant. Values at $p = 0.05$ and below are significant.

Advanced Case Study: Acme Toy Company Baseball Cards

• The Claim: 30% rookies, 60% veterans, 10% All-Stars.

• Random Sample ($n = 100$): 50 rookies, 45 veterans, 5 All-Stars.

• Hypotheses:

  • $H_0$: $\text{Rookies} = 30\%$, $\text{Veterans} = 60\%$, $\text{All-Stars} = 10\%$

  • $H_a$: At least one of the proportions in the null hypothesis is false.

• Degrees of Freedom ($df$): $k - 1 = 3 - 1 = 2$.

• Expected Frequencies ($E_i = n \times p_i$):

  • $E_1 = 100 \times 0.30 = 30$

  • $E_2 = 100 \times 0.60 = 60$

  • $E_3 = 100 \times 0.10 = 10$

• Statistical Calculation:

  • $\chi^2 = \frac{(50 - 30)^2}{30} + \frac{(45 - 60)^2}{60} + \frac{(5 - 10)^2}{10}$

  • $\chi^2 = \frac{400}{30} + \frac{225}{60} + \frac{25}{10} = 13.33 + 3.75 + 2.50 = 19.58$

• Interpretation:

  • P(\chi^2 > 19.58) = 0.0001.

  • Since 0.0001 < 0.05, the researcher rejects the null hypothesis.

• Condition Check: The method is appropriate because:

  • Simple random sampling was used.

  • Variable is categorical.

  • All expected frequencies are $\ge 5$.

The Contingency Test (Chi-Square Test of Independence)

• Definition: A test used to determine if there is a relationship between two categorical variables.

• Contingency Table: A table displaying the frequencies of two or more categorical variables, helping to visualize relationships.

• Independence: Two variables are independent if the occurrence of one does not affect the probability of the other.

• Applications:

  1. Social Sciences: Relationships between demographics and opinions/behaviors.

  2. Medical Research: Associations between risk factors and disease outcomes; treatment effectiveness across groups.

  3. Marketing: Customer preferences, purchase patterns, and advertising effectiveness.

  4. Biology: Gene expression patterns or genetic variations across cell types.

  5. Cybersecurity: Comparing pre- and post-attack data to assess security effectiveness.

  6. Cryptanalysis: Comparing plaintext and ciphertext distributions to assess cipher strength.

     • Cipher: An algorithm used to encrypt/encode a message into ciphertext to protect confidentiality.

     • Plaintext: The original, unreadable content requiring a decryption algorithm and key to access.

  7. Goodness-of-Fit: Comparing observed frequencies with theoretical distributions (e.g., binomial).

• Example Scenario: A study investigating the relationship between gender and movie preference (action, comedy, drama). The test determines if preferences are genuine relationships or random chance.