Proportions

Categorical Variables - Frequency data

• Proportion is the fraction of individuals having a particular attribute
• # of successes / # of trials

• We can estimate the population proportion by: \[ \hat{p}=\frac{X}{n} \]

• where X denotes number of “successes”

• 531 of 648 people struck by lightning in the US from 1995 to 2008 were men.
• Men make up 49.2% of the U.S. population

\[ \hat{p} = \frac{X}{n} = \frac{531}{648} = 0.8194 \]

\[ \hat{p} = \frac{X}{n} = \frac{531}{648} = 0.8194 \]

We can use this to solve for the standard error of our proportion.

\[ SE_\hat{p} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{(0.8194)(1-0.8194)}{648}} = 0.0151 \]

There are many ways to calculate confidence intervals for proportions. Your book recommends using the Agresti-Coull confidence intervals, which is calculated in two steps.

\[ p= \pm(1.96)\sqrt{\frac{p'(1-p')}{n+4}} \] \[ where p' = \frac{X+2}{n+4}=\frac{533}{652}=0.8174 \]

Thus leading us to…

\[ 0.7932 \pm(1.96)\sqrt{\frac{0.8174(1-0.8174)}{648+4}}= [0.7878, 0.8471] \]

Remember that equations will generally be set to solve for one thing.

The ability to taste phenylthiocarbamide (PTC) is genetically controlled in humans. In Europe and Asia, about 70% of people are “tasters”. Suppose a study is being planned to estimate the relative frequency of tasters in a certain Asian population and the desired standard error of the estimated relative frequency is 0.01. How many people should be included in the study?

Which one of our equations holds n?

A binomial random variable is a variable for which:

• B - There are two possible outcomes for each trial (e.g. binary outcomes of success and failure);

• I - The outcomes of the trial are independent of each other;

• N - The number of trials, n, is fixed in advance; and

• S - The probability of a success on a single trial is the same for all trials (i.e. the same p value) DIFFERENT P!

If X is the count of successes in a binomial setting, then X has the binomial distribution with parameters n and p. The possible values of X are integers from 0 to n.

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

where (Binomial Coefficient) \[ \binom{n}{X} = \frac{n!}{k!(n-k)!} \]

and where \[ n! = n \cdot (n-1) \cdot (n-2) \cdot ... 3 \cdot 2 \cdot 1 \]

Example: The probability of rolling “4” exactly three times in five rolls of a die can be written as: \[ \binom{5}{3} = \frac{n!}{k!(n-k)!} \]

The probability of rolling “4” exactly three times in five rolls of a die can be written as:

• p= Probability of a single 4 = \( \frac{1}{6} \).
• n = 5 dice rolls
• k = 3 successes
• X = also k, so 3

\[ \binom{5}{3} = \frac{n!}{k!(n-k)!}p^X(1-p)^{(n-X)} \]

\[ \binom{5}{3} = \frac{5!}{3!(5-3)!}\frac{1}{6}^{3}(1-\frac{1}{6})^{(5-3)} \]

\[ \binom{5}{3} = \frac{5!}{3!(5-3)!}\cdot\frac{1}{6}^{3}\cdot(1-\frac{1}{6})^{(5-3)} \rightarrow \]

\[ \binom{5}{3} = \frac{120}{6(2)}\cdot\frac{1}{6}^{3}\cdot(\frac{5}{6})^{2} \rightarrow \]

\[ \binom{5}{3} = 10\cdot.00469\cdot(.69\bar{4}) = .032569 \]

Instead, if we wanted to find the probability of seeing a single X, we would use dbinom().

dbinom(x=3,  size=5, prob=(1/6))

[1] 0.03215021

Now we have the result we were expecting.

Let's get back to our lightning strikes. As a reminder, 531 out of 648 people struck by lightning were men, and we calculated that means that 81.94% of people who get struck by lightning are men.

If you attended a conference for people who had been struck by lightning and observed that three of ten talks were given by men, is there evidence that men are more likely to talk about their experience?

H₀: The relative frequency of successes (men giving talks who have been struck by lightning) in the population is 81.94%

H_A: The relative frequency of successes in the population is not 81.94%.

n= 10; X =3 ; p=.8194

531 of 648 people were men.

We know the proportion is .8194

We observed 10 speakers.

Mean number of successes:

\[ \mu = n\cdot p = (10)(0.8194) = 8.194 \]

SD of the number of successes: \[ \sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{(10)\cdot(0.8194)\cdot(1-0.8194)}=1.2165 \]

H₀: The proportion of people who are struck by lightning who are men is 0.5.

H_A: The proportion of people who are struck by lightning who are men is not 0.5.

H₀: The proportion of people who are struck by lightning who are men is 0.5.

H_A: The proportion of people who are struck by lightning who are men is not 0.5.

Mean number of successes: \[ \mu = n\cdot p = (648)\cdot(0.5) = 324 \]

Standard deviation of number of successes \[ \sigma = \sqrt{n\cdot p \cdot (1-p)} = \sqrt{(648)\cdot(0.5)\cdot(0.5)} = 12.72792 \]

Why? When calculating exact binomial probabilities with large n.

If n * p and n * (1-p) are both greater than five:

\[ P = 2(Pr[X \ge Observed]) = 2 (Pr [Z>\frac{Observed - \frac{1}{2}-n\cdot p}{\sqrt{n\cdot p \cdot (1-p)}}]) \]

The ½ is what we call the continuity correction.

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dbinom(10, 25, .061)+dbinom(11, 25, .061)+dbinom(12, 25, .061)+
  dbinom(13, 25, .061)+dbinom(14, 25, .061)+dbinom(15, 25, .061)+
  dbinom(16, 25, .061)+dbinom(17, 25, .061)+dbinom(18, 25, .061)+
  dbinom(19, 25, .061)+dbinom(20, 25, .061)+dbinom(22, 25, .061)+
  dbinom(23, 25, .061)+dbinom(24, 25, .061)+dbinom(25, 25, .061)

[1] 9.93988e-07

This is the sum of 10 to the possible maximum of 25.

This is the same thing as running

 binom.test(10, 25, .061)$p.value

[1] 9.93988e-07

 2*binom.test(10, 25, .061)$p.value

[1] 1.987976e-06