The Binomial Distribution

Karl L. WuenschDepartment of PsychologyEast Carolina University

A Binomial Experiment• consists of n identical trials.• each trial results in one of two outcomes, a

“success” or a “failure.”• the probabilities of success ( p ) and of

failure ( q = 1 ‑ p ) are constant across trials.

• trials are independent, not affected by the outcome of other trials.

• Y is the number of successes in n trials.

• P(Y = y) may also be determined by reference to a binomial table.

• The binomial distribution has:

ynyqpy

nyYP !y)-(n !

!

np npq2

Binomial Hypotheses, Directional

• H0: Mothers cannot identify their babies by scent alone, binomial p .5

• H1: Yes they can, binomial p > .5• The data: 18 of 25 mothers correctly

identified their baby.• P(Y 18 | n = 25, p = .5) =• 1 - P(Y < 17 | n = 25, p = .5) = .022

Mothers were allowed to smell two articles of infant’s clothing and asked to pick the one which was their infant’s. They were successful in doing so 72% of the time, significantly more often than would be expected by chance, exact binomial p (one-tailed) = .022.

The Basenji is fearful of strangers.

The cocker spaniel is not.

What About A Cockenji?

Inheritance of Fearfulness

• John Paul Scott and John Fuller• Basenji x Basenji fearful pups• Cocker x Cocker fearless pups• Basenji x Cocker fearful pups• Dominant F gene codes for Fearfulness• Recessive f gene codes for fearlessness• F1 dogs are heterozygous, Ff

Breed F1 Dogs With Each Other

Mother

Father F f

F FF Ff

f fF ff

Binomial Hypotheses: Nondirectional

• H0: 75% of the babies will fear strangers, binomial p = .75.

• H1: binomial p .75• The data: 18 of 25 puppies were fearful of

strangers.• Under the null, we expect 75% of pups to

be fearful. 18/25 = 72% were.• psig = 2P(Y 18 | n = 25, p = .75)

• “p = 2*PROBBNML(.75, 25, 18);”• p = .8778• The high value of p indicates very good fit

between the null hypothesis and the data.

Eighteen of 25 pups (72%) born to F1 parents were fearful of strangers. The obtained proportion was not significantly different from the expected .75, p = .88

Normal Approximation• If falls within 0 to n, then the

binomial approximation should be good.• We want P(Y ≥ 18 | n = 25, p = .5).

• which is contained within 0 25, so approximation should be good.

5.175.7)5.2(25.12)5)(.5(.252)5(.25

Correction for Continuity• When computing the z, move the

observed value of Y one-half point towards the mean under the null.

• psig = .0228

25.2

5.125.17

z

The Binomial Sign Test• Design = Matched Pairs• Pre and post data for patients given a

blood pressure treatment• Of 10 patients, 9 had lower pressure at

post-test.• Under the null of no effect of treatment, we

expect .5(10) = 5 lower and 5 higher.

• H0: The treatment has no effect on blood pressure, binomial p = .5

• H1: The treatment does affect blood pressure, binomial p .5

• 2P(Y 9 n = 10, p = .5) =• 1-(2P(Y < 8 n = 10, p = .5)) = .0215

An exact binomial sign test indicated that the treatment significantly lowered blood pressure, 9 of 10 patients having post-treatment pressure lower than their pre-treatment pressure, p = .021.