BIOL 570 1nd Edition Lecture 10 Outline of Last Lecture I. Review of hypothesis testingII. Type I and Type II errorsIII. PowerOutline of Current Lecture IV. Hypothesis tests and scientific researchV. Binomial formula and distributionVI. Using binomial to create null distributionVII. Binomial TestVIII. Began discussion of estimationCurrent LectureScientific Paper- Title- Abstract- Introduction- Methods- Results- Discussion/ ConclusionP (H) = 0.25 P (N) = 0.75n = 8P (all have H) = .25 x .25 x .25 ….= (0.25) ^81) 2 outcomes – success, failure2) Fixed number of trials – sample size n = 83) Trials are independent4) Constant probability of success – p = .25Binomial distributionThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- Describes the probability of a given number of “successes” from a fixed number of independent trials when the probability of a success is the same in all trials Binomial Coefficient = number of waysProbabilitiesPr (4 of 8 have H) = Pr (x=4) = 8! 4! 4! (.25) ^4 (.75) ^4Pr (x = 4) = 8 x 7 x 6 x 5 x 44! 4! = 70Pr (4) = 70 (.25) ^4 (.75) ^4 = 0.08651730 ! = 11 ! = 1What test should I use (p 315)Binomial Test - Use data to test whether a population proportion (p) matches the null expectation (pO) for the proportion1) Efficient method: null distribution (binomial formula)2) HO can have p = any fractionStep 1: Do KU student have hitch hikers thumb (HT) in proportions expected given the literature?Step 2: HO: p = .25 HA: p ≠ .25Step 3: 8 studentsStep 4: 4 HTStep 5: test statistic = 4Mean of binomial distribution: n x p8 (.25) =20 1 2 3 4 5 6 7 8X ↑ X X X X X MEAN (n x p)Two tailed: what would be shaded to determine P-valueP 160 Do not multiply by 2What we actually got or more extreme: 4>0 1 2 3 4 5 6 7 8↑ X X X XGg x GgGG: green P (green) = .75Gg: greengg: yellow P (yellow) = .252 outcomes: green, yellown = 3independence of outcomesfixed p = .75n = 3 (green)Pr (x = 3) = 3! 3! 0! (.75) ^3 (.25)