Download presentation

Presentation is loading. Please wait.

Published byAmberlynn Kelley Modified over 2 years ago

1
The Binomial Distribution Karl L. Wuensch Department of Psychology East Carolina University

2
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.

3
P(Y = y) may also be determined by reference to a binomial table. The binomial distribution has:

4
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

5
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.

6
The Basenji is fearful of strangers.

7
The cocker spaniel is not.

8
What About A Cockenji?

9
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

10
Breed F 1 Dogs With Each Other Mother FatherFf FFFFf ffFff

11
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)

12
“p = 2*PROBBNML(.75, 25, 18);” p =.8778 The high value of p indicates very good fit between the null hypothesis and the data.

13
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

14
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.

15
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

16
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.

17
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

18
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.

Similar presentations

OK

Section 10.1 Estimating with Confidence AP Statistics February 11 th, 2011.

Section 10.1 Estimating with Confidence AP Statistics February 11 th, 2011.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Free ppt on american war of independence Ppt on library management system in java Ppt on nature of human resource management Ppt on phonetic transcription of words Ppt on complex numbers class 11th biology Ppt on msme in india Ppt on social contract theory ethics Ppt on chromosomes and genes model Ppt on duty roster spreadsheet Ppt on kindness is contagious