The Binomial
Distribution
Karl L. Wuensch
Department of Psychology
East 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  
n!
y
p q
n  y 
y ! (n - y) !
• P(Y = y) may also be determined by
reference to a binomial table.
• The binomial distribution has:
  np

2
 npq
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
•
•
•
•
•
•
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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 = 2P(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).
25 (. 5 )  2 25 (. 5 )(. 5 )  12 . 5  2 ( 2 . 5 )  7 . 5  17 . 5
• which is contained within 0  25, so
approximation should be good.
Correction for Continuity
• When computing the z, move the
observed value of Y one-half point towards
the mean under the null.
z 
17 . 5  12 . 5
2 .5
• psig = .0228
 2
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
• 2P(Y  9  n = 10, p = .5) =
• 1-(2P(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.