Lecture 7: Binomial Probability Distribution and Hypothesis Testing Notes

Categorical Data and Binomial Probabilities

• Categorical data consists of variables categorized into discrete, non-numeric groups. In binary categorical data, observations fall into exactly two mutually exclusive categories:

  • Live or dead

  • Success or failure

  • Affected by disease or not affected

• Binomial probability tools are utilized to analyze binary categorical data and test hypotheses regarding underlying population probabilities.

Mendelian Genetics Example and Sample Space

• Single-gene inheritance traits governed by dominant ($A$) and recessive ($a$) alleles serve as a foundational application of binomial probability.

• Assuming random mating and equal allele frequencies ($P(A) = 0.5$ and $P(a) = 0.5$):

  • Alleles inherited from the mother: $A$ or $a$

  • Alleles inherited from the father: $A$ or $a$

  • Resulting offspring genotypes: $AA$, $aA$, $Aa$, and $aa$

  • Probability of recessive genotype ($aa$): $P(R) = 0.25$

  • Probability of dominant phenotype ($AA$, $Aa$, $aA$): $P(D) = 0.75$

• When sampling 3 individuals randomly from this population, the complete sample space of outcomes for recessive ($R$) and dominant ($D$) phenotypes comprises $2^3 = 8$ possibilities:

  • if the probability of one phenotype doesn’t directly relate to the probability of another, then they are considered independent

  • faster way to do this is through binomial distribution

    • $RRR$: $(0.25) \times (0.25) \times (0.25) = 0.015625$

    • $RRD$: $(0.25) \times (0.25) \times (0.75) = 0.046875$

    • $RDR$: $(0.25) \times (0.75) \times (0.25) = 0.046875$

    • $RDD$: $(0.25) \times (0.75) \times (0.75) = 0.140625$

    • $DRR$: $(0.75) \times (0.25) \times (0.25) = 0.046875$

    • $DRD$: $(0.75) \times (0.25) \times (0.75) = 0.140625$

    • $DDR$: $(0.75) \times (0.75) \times (0.25) = 0.140625$

    • $DDD$: $(0.75) \times (0.75) \times (0.75) = 0.421875$

Mathematical Formulation of the Binomial Distribution

• The binomial distribution describes the probability of obtaining a given number of "successes" ($X$) from a fixed number of independent trials ($n$).

• Individual trials in a binomial process are known as Bernoulli trials, defined as random variables with exactly two possible outcomes.

• Parameters of the binomial distribution:

  • $n$: total number of independent trials

  • $p$: probability of success on any single trial

  • $1 - p$: probability of failure on any single trial

  • $X$: observed number of successful trials (X = 0, 1, 2, \ndots, n)

• General formula for binomial probability:   $P(X) = \binom{n}{X} p^X (1 - p)^{n - X}$

• Combinatorial notation $\binom{n}{X}$ (also written as $nCX$ or $_nC_X$) represents the number of ways to choose $X$ successes from $n$ trials:   $\binom{n}{X} = \frac{n!}{X! (n - X)!}$

• Factorial definition and rules:

  • $n! = n \times (n - 1) \times (n - 2) \times \dots \times 3 \times 2 \times 1$

  • $0! = 1$

  • $1! = 1$

  • Example: $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

Worked Calculation: Sampling Recessive Phenotypes

• To calculate the probability that exactly $2$ out of $3$ randomly sampled individuals express a recessive phenotype ($p = 0.25$):

  • Number of combinations yielding 2 recessives: $\binom{3}{2}$

  • Probability of 2 recessive phenotypes: $(0.25)^2$

  • Probability of 1 dominant phenotype: $(1 - 0.25)^{3 - 2} = (0.75)^1$

• Step-by-step substitution and evaluation:   $P(X = 2) = \binom{3}{2} (0.25)^2 (1 - 0.25)^{3 - 2}$   $P(X = 2) = \frac{3!}{2! \times 1!} (0.25)^2 (0.75)^1$   $P(X = 2) = 3 \times (0.25)^2 \times (0.75)$   $P(X = 2) = 3 \times 0.0625 \times 0.75 = 0.140625 \approx 0.141$

Hypothesis Testing Example: Red Uniforms in Combat Sports

• Biological Context: Male animals frequently display bright red colorations as visual signals of aggression, competitive dominance, and threat intensity.

• Empirical Investigation: A study evaluated whether uniform color influences outcome success in human combat sports (Hill, RA, and RA Barton 2005. "Red enhances human performance in contests." Nature 435:293).

  • Dataset: Examined round outcomes in wrestling, taekwondo, and boxing during the 2004 Olympic Games.

  • Experimental Design: Competitors were randomly assigned red or blue shirt colors.

  • Experimental Findings: In $16$ of $20$ combat rounds, the athlete assigned the red shirt won the match.

• Formulating Hypotheses:

  • Null Hypothesis ($H_0$): Red- and blue-shirted athletes are equally likely to win, or red-shirted athletes are less likely to win (proportion of red winners $\le 0.5$).

  • Alternative Hypothesis ($H_A$): Red-shirted athletes are more likely to win (proportion of red winners $> 0.5$).

  • remember non directional hypothesis as opposed to directional hypotheses, which specify a direction of the expected effect. In a non-directional hypothesis, we would simply state that there is a difference in the probabilities of winning between red- and blue-shirted athletes without indicating which group is expected to be favored.

• Sample Estimations:

  • Observed proportion of red-shirted winners: $\hat{p} = \frac{16}{20} = 0.8$

  • Discrepancy: The observed sample proportion ($0.8$) deviates by $0.3$ from the null hypothesis parameter ($p = 0.5$).

Statistical Null Distribution and P-Value Calculation

• Null Distribution Properties:

  • Under $H_0$, the expected number of red winners follows a binomial distribution with $n = 20$ and $p = 0.5$

  • Expected mean number of red winners: $\mu = n \times p = 20 \times 0.5 = 10$

• P-Value Calculation:

  • The P-value is the probability of obtaining a sample result as extreme as or more extreme than the observed data, assuming $H_0$ is true.

  • For a one-tailed test with $X = 16$ observed successes out of $n = 20$:     $P = P(16) + P(17) + P(18) + P(19) + P(20)$

  • Applying the binomial formula $P(X) = \binom{20}{X} (0.5)^X (0.5)^{20 - X}$ for each outcome:     $P = 0.0059$

• Statistical Decision Rule:

  • Standard significance threshold: $\alpha = 0.05$

  • Result: $P = 0.0059$

  • Since $P < 0.05$, the null hypothesis $H_0$ is rejected.

  • Conclusion: Red-shirted athletes were statistically significantly more likely to win their contests.