Presentation on theme: "The Binomial Distribution Karl L. Wuensch Department of Psychology East Carolina University."— Presentation transcript:
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) may also be determined by reference to a binomial table. The binomial distribution has:
Binomial Hypotheses, Directional H 0 : Mothers cannot identify their babies by scent alone, binomial p .5 H 1 : 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 F 1 dogs are heterozygous, Ff
Breed F 1 Dogs With Each Other Mother FatherFf FFFFf ffFff
Binomial Hypotheses: Nondirectional H 0 : 75% of the babies will fear strangers, binomial p =.75. H 1 : 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. p sig = 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 F 1 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.
Correction for Continuity When computing the z, move the observed value of Y one-half point towards the mean under the null. p sig =.0228
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.
H 0 : The treatment has no effect on blood pressure, binomial p =.5 H 1 : 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.