Lecture 9
Analyzing Proportions

ABD 3e Chapter 7

Chris Merkord

Learning Objectives

By the end of this lecture, you should be able to:

• Determine whether binary data follow a binomial model
• Calculate and interpret binomial probabilities
• Describe the distribution of counts and proportions
• Estimate a population proportion from a sample proportion
• Explain how sample size affects variability in the sampling distribution of a proportion
• Construct and interpret a confidence interval for a population proportion
• Use the binomial distribution to test hypotheses about a population proportion

Binary Outcomes in Biology

Many biological variables have only two possible outcomes.

These are called a binary variable (or Bernoulli variable).

We record each observation as:

• 1 = success
• 0 = failure

Each observation is a single trial with two possible results.

Examples:

• Individual survives / dies
• Seed germinates / does not germinate
• Nest occupied / not occupied
• Species present / absent
• Offspring male / female

Two Wisconsin Fast Plants growing cups showing a binary outcome: one seed germinates, the other does not.

From a Single Trial to Many Trials

Suppose we observe a binary outcome repeatedly.

Let:

\[ X = \text{number of successes in } n \text{ trials} \]

Now we are no longer modeling a single 0/1 outcome.

We are modeling a count:

• How many successes occur in a fixed number of trials?

Sample of eight binary trials showing five successes and three failures (X = 5, n = 8).

When Does the Binomial Model Apply?

A dataset follows a binomial model if:

1. There is a fixed number of trials ( \(n\) )
2. Each trial has two possible outcomes
3. Trials are independent
4. The probability of success ( \(p\) ) is constant

If these conditions hold, then:

\[ X \sim \text{Binomial}(n, p) \]

Binomial model checklist: Binary? → Fixed n? → Independent? → Constant p?

If the Binomial Model Applies…

If:

\[ X \sim \text{Binomial}(n, p) \]

Then we can compute the probability of any specific number of successes:

\[ \operatorname{Pr}[X = x] \]

We need a formula that tells us:

1. How many ways can ( \(x\) ) successes occur?
2. What is the probability of each arrangement?

The Binomial Probability Formula

For a binomial random variable:

\[ \operatorname{Pr}[X = x] = \binom{n}{x} p^{x} (1 - p)^{n - x} \]

Where:

• \(\binom{n}{x}\) = number of ways to have \(x\) successes out of \(n\) trials, read as “n choose x”
  • E.g. There are two ways to have one head when you flip a coin twice: a head then a tail, or a tail then a head
  • Calculated as \(\frac{n!}{x!\times (n-x)!}\)
  • where \(n!=n \times (n-1) \times (n-2) \times \dots \times 2 \times 1\)
• \(p^{x}(1 - p)^{n - x}\) = probability of any one specific arrangement

Example: Wasp sex ratio

Suppose you randomly sample \(n=5\) wasps from a population where each wasp has the probability \(p=0.2\) of being male. The probability then that exactly \(3\) of the wasps in your sample are male is:

\[ \operatorname{Pr}[3 \text{ males}] = \binom{5}{3} (0.2)^{3} (1-0.2)^{5-3} \]

\[ = \frac{(5\times4\times3\times2\times1)}{(3\times 2\times 1) \times (2\times1)}(0.2)^3(0.8)^2 \]

\[ = \frac{120}{6 \times 2} \times 0.008 \times 0.64 \]

\[ =0.234 \]

…chance of getting 3 males in a sample of 5 wasps

Male and female wasps illustrating a binary outcome (male vs. female) in the wasp offspring example.

The Binomial Distribution

For a fixed ( n ) and ( p ),

\[ X \sim \text{Binomial}(n, p) \]

The binomial distribution is:

• All possible values of ( X )
• And the probability of each value

\[ X = 0, 1, 2, \dots, n \]

From One Probability to Many

For the wasp example, we calculated:

\[ \operatorname{Pr}[X = 3] \]

But we could also calculate:

\[ \operatorname{Pr}[X = 0], \operatorname{Pr}[X = 1], \dots, \operatorname{Pr}[X = n] \]

Together, these probabilities form the binomial distribution.

Figure 1: Animated binomial distribution for n = 5 and p = 0.2. Each frame highlights one possible value of X and updates the title to show Pr[X = x].

Why This Matters

The binomial distribution tells us:

• If we repeated this process many times,

• what counts of successes we would expect to see.

• It describes the distribution of possible counts we would observe if we repeated the process many times.

Example: Mud Plantains Avoid Self-Fertilization

System description

• Mud plantains are hermaphroditic (each flower has male and female structures).
• The style bends either left or right.
• This creates two floral morphs.

Bees transfer pollen:

• From left-styled flowers
• To right-styled flowers

This reduces self-fertilization (“selfing”).

Mud plantain flowers showing left- and right-styled morphs that promote cross-pollination and reduce self-fertilization. Source: Whitlock & Schluter, The Analysis of Biological Data, 3rd ed.

Expected ratio of offspring is 1:3

• Under a simple model of inheritance, crossing left- and right-handed strains should yield offspring with a 1:3 ratio of left- to right-handed flowers.

• I.e. 25% of offspring should be left-handed and 75% should be right-handed.

• For this example, let’s call left-handedness “success” and right-handedness “failure”

A given sample size will produce a corresponding sampling distribution

• Let’s try looking at a complete sampling distribution instead of a single outcome

• Image we randomly sample \(n=27\) individuals from a population in which \(p=0.25\)

• What is the probability that the sample contains exactly \(X\) successes?

• For example, the probability of getting exactly six left-handed flowers is:

\[ \operatorname{Pr}[6 \text{ left-handed flowers}] \\= \binom{27}{6} (0.25)^{6} (1-0.25)^{27-6}\\= 296010 \times 0.000244 \times 0.002378\\=0.1719 \]

• Now, repeat to calculate the probability of each outcome

Filling out the sampling distribution

Repeat previous step to calculate the probability of each outcome.

X	Pr[X]
0	4.2×10-4
1	0.0038
2	0.0165
3	0.0459
4	0.0927
5	0.1406
6	0.1719

X	Pr[X]
7	0.1719
8	0.1432
9	0.1008
10	0.0605
11	0.0312
12	0.0138
13	0.0053

X	Pr[X]
14	0.0018
15	5.1×10-4
16	1.3×10-4
17	2.8×10-5
18	5.1×10-6
19	8.1×10-7
20	1.1×10-7

X	Pr[X]
21	1.2×10-8
22	1.1×10-9
23	7.9×10-11
24	4.4×10-12
25	1.8×10-13
26	4.5×10-15
27	5.5×10-17

The collection of probabilities is the sampling distribution

• The plot shows the probability of obtaining \(X\) left-handed flowers out of \(n=27\) randomly sampled, if the proportion of left-handed plants in the population is \(p=0.25\)

Binomial probability distribution for the number of left-handed mud plantain flowers (X) in a sample, showing the probability of observing each possible count.

Sample size determines the spread of the sampling distribution

• Larger sample sizes produce narrower sampling distributions

Sampling distributions illustrating the effect of sample size on the distribution of a proportion. The top panel (n = 10) shows a wider, more variable distribution, while the bottom panel (n = 100) shows a narrower, more concentrated distribution around the true proportion.

From Counts to Proportions

So far, we have modeled the count:

\[ X = \text{number of successes in } n \text{ trials} \]

But we often report results as a proportion:

\[ \frac{X}{n} \]

Same underlying binomial randomness.

Different scale.

Estimating a proportion from a single sample

• We can estimate a population proportion from a single sample.
• If there are \(X\) successes out of \(n\) trials in a random sample, then the estimated proportion of successes is \(\hat{p}\):

\[ \hat{p}=\frac{X}{n} \]

• Law of large numbers - as sample size increases, \(\hat{p}\) approaches \(p\)

Standard error of \(\hat{p}\)

• The standard deviation of the sampling distribution of \(\hat{p}\) (designated by \(\sigma_\hat{p}\)) is

\[ \sigma_\hat{p}=\sqrt{\frac{p(1-p)}{n}} \]

• This is also called the standard error of \(p\)

• As sample size increases, standard error goes down

• In reality, we can almost never calculate this standard error because we don’t know true \(p\)

• We estimate standard error by replacing \(p\) with \(\hat{p}\)

\[ \operatorname{SE}_\hat{p}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]

Approximate confidence interval for a proportion

• 2SE rule of thumb only works when sampling distribution is bell-shaped, which is not true for \(\operatorname{SE}_\hat{p}\)

• Agresti-Coull method provides an approximate the 95% confidence interval for a proportion

\[ p' - 1.96 \sqrt{\frac{p'(1 - p')}{n + 4}} \;<\; p \;<\; p' + 1.96 \sqrt{\frac{p'(1 - p')}{n + 4}} \]where:

\[ p' = \frac{X + 2}{n + 4} \]

• The Wald method is commonly used, but it has been shown not to work well in some situations.

Interpreting a 95% Confidence Interval

If we repeated this sampling process many times,

about 95% of intervals constructed this way

would contain the true population proportion (p).

It does NOT mean:

There is a 95% probability that this specific interval contains (p).

Use the binomial test to test hypotheses about a binomial sampling distribution

• Use data to test whether a population proportion ( \(p\) ) matches a null expectation ( \(p_0\) ) for the proportion

• 𝐻₀: The proportion \(p\) of successes in the population is equal to \(p_0\)

• 𝐻_𝐴: The proportion \(p\) of successes in the population is not equal to \(p_0\)

Example: Distribution of spermatogenesis genes

• Evolutionary theory predicts genes for spermatogenesis (sperm formation) should occur disproportionately more often on the X chromosome

• The X chromosome contains 6.1% of the genes in the genome

• If genes for spermatogenesis occur randomly throughout the genome, we’d expect 6.1% of them to fall on the X chromosome

• Each gene = trial

• Success = gene is on X

Mouse chromosomes. Whitlock & Schluter (2015)

Observed distribution of spermatogenesis genes (data)

• Wang et al. (2001) identified 25 genes involved in spermatogenesis in mice.

• 10 genes (40%) were on the X chromosome.

Do the results support the hypothesis that spermatogenesis genes occur preferentially on the X chromosome?

Cartoon of the mouse genome. Blue ovals represent spermatogenesis genes. Lines represent chromosomes and are proportional in length to real life. Whitlock & Schluter (2015)

Hypotheses and test statistic

STEP 1: State hypotheses

• 𝐻₀: The probability that a spermatogenesis gene falls on the X chromosome is 0.061 ( \(p=0.061\) )

• 𝐻_𝐴: The probability that a spermatogenesis gene falls on the X chromosome is something other than 0.061 ( \(p\ne0.061\) )

STEP 2: Calculate test statistic

• For binomial test, it is the observed number of successes

• For this example: 10

• How many would be expected under 𝐻₀?

  • Answer: \(0.061\times25=1.525\)

Determining the P-value

STEP 3: Determine P-value

• How likely are we to get 10 by chance alone if the 𝐻₀ is true?

• To decide, we need the null distribution, the sampling distribution for the test statistic assuming that 𝐻₀ is true

\[ \operatorname{Pr}[X \text{ successes}] = \\ \binom{25}{X} (0.061)^{X} (1-0.061)^{25-X} \]

Now calculate chance of getting observed outcome or more extreme:

\[ P=2 \times \operatorname{Pr}[\text{successes} \geq 10] \]

\[ = 2\times (9.9\times10^{-7}) \]

\[ = 1.98 \times 10^{-6} \]

Interpreting the results

Decision criteria:

• Using \(\alpha=0.05\) then \(p<\alpha\) , we reject 𝐻₀
• The proportion of spermatogenesis genes on the X chromosome is not equal to 0.061

Biological Interpretation:

• The observed proportion differs from the theoretical expectation.
• This suggests that spermatogenesis genes are not randomly distributed with respect to the X chromosome.
• Some biological process may be influencing their genomic location.

Rejecting 𝐻0 does not tell us why the difference exists, only that the observed pattern is unlikely under the null model.

Next step:

Question

If \(p\ne0.061\) then what is \(p\) ?

Answer

Estimate \(p\) with \(\hat{p}\) to find out

Estimate the proportion of spermatogenesis genes on the X chromosome

• Our best estimate of the proportion of genes on the X chromosome is:

\[ \hat{p}=\frac{X}{n}=\frac{10}{25}=0.40 \]

• Which is much greater than 0.061

• We would report our findings like this:

  A disproportionately large proportion of spermatogenesis genes occur on the X chromosome (0.40, SE=0.10; binomial test, 𝑛=25, 𝑃<0.001).

Estimating proportions: highly irradiated radiologists

• Male radiologists have long suspected that they tend to have fewer sons than daughters.

• What is the proportion of males among the offspring of radiologists?

• In a sample of 87 offspring of “highly irradiated” male radiologists, 30 were male (Hama et al. 2001). Assume that this was a random sample.

• Best estimate of proportion of male offspring:

\[ \hat{p}=\frac{X}{n}=\frac{30}{87}=0.345 \]

• Standard error of \(\hat{p}\) :

\[ \operatorname{SE}_\hat{p}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\=\sqrt{\frac{0.345(1-0.345)}{87}}=0.051 \]

Applying the Agresti-Coull method to calculate an approximate 95% confidence interval

Uses an adjusted estimate \(p'\) for constructing the interval

\[ p' = \frac{X + 2}{n + 4}\\=\frac{30+2}{87+4}\\=0.351 \]

\[ p' - 1.96 \sqrt{\frac{p'(1 - p')}{n + 4}}\;<\;p\;<\;p' + 1.96 \sqrt{\frac{p'(1 - p')}{n + 4}} \]

\[ 0.351 - 1.96 \sqrt{\frac{0.228}{91}}\;<\;p\;<\;0.351 + 1.96 \sqrt{\frac{0.228}{91}} \]

\[ 0.351 - 1.96 \times 0.098\;<\;p\;<\;0.351 + 1.96 \times0.098 \]

\[ 0.253\;<\;p\;<\;0.449 \]

From Confidence Interval to Hypothesis Test

We estimated the proportion of male offspring:

• \(\hat{p} = 0.345\)
• 95% CI ≈ (0.25, 0.45)

Now test:

\(H_0: p = 0.50\)
\(H_A: p \ne 0.50\)

Question:

Does 0.50 fall inside the 95% confidence interval?

Conclusion:

Because 0.50 is not in the interval, we would reject \(H_0\) at \(\alpha = 0.05\).

The data suggest the proportion of male offspring differs from 50%.