BIOL 570 1nd Edition Lecture 11 Outline of Last Lecture I. Hypothesis tests and scientific researchII. Binomial formula and distributionIII. Using binomial to create null distributionIV. Binomial TestOutline of Current Lecture I. Estimation with proportionsII. Genetics exampleIII. Review concepts (point estimate, interval estimate (=CI))IV. SE of p hatV. CI for pVI. Hypothesis tests vs. CICurrent LectureTrue breeding linesGreen YellowParent (G-G) (g-g)F1 (Gg) – GreenF2 428 Green, 152 YellowStatistic:p-hat = x/n → 428/580 = .738Population: Parameter- true proportion that are green (P)Point estimate = Y-bar (sample statistic, µ parameter) p-hat (sample statistic, P parameter)Interval estimate = 95% Confidence Interval (CI)These 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.95% CI for µ =Y −¯± 2(S√n)Point estimate ± (appropriate value from a x (standard error of of a parameter statistical distribution) point estimate) margin of errorp-hat ± ( )(SE of p-hat)Sampling distribution: how much spread- Large spread = Large SE- SE decreases with increasing sample sizeStandard error:- SD of sampling distributiono SEY-bar = S√nSE (dealing with proportions): P157 Law of Large Numbers- Large samples yield more precise estimatesWald Methodp-hat ± (value from statistical distribution) (SEp-hat)- value from statistical distribution:o 95% CI = 1.96o 99% CI = 2.58Agresti-Coull Methodp’ = x + 2 p’ = 428 + 2p'≈ p−^¿ n + 4 580 + 4 = .741 p-hat = .738 (point estimate)95% CI for pp'± Z √p'(1−p')n+4(Z = 1.96).741 ± 1.96 sqrt [.741 (1 – .741)/ 580 +4].7058 < p < .776895% confident p falls between .7058 and .7768- To narrow the CI, increase the sample size99% CI- Larger than 95% CI (defining larger area).6941 < p < .7877p population proportionp-hat sample proportion (point estimate)p’ Agresti-Coull (CI)P-value hypothesis testingHypothesis Test: HO: proportion with free lobes = .65HA: p ≠ .65Reject HO p-value = 0.03 α = 0.05n = 20x = 8 p’ = (8 + 2)/ (20+4) = .417 CI = .417 ± 1.96 sqrt [.417(.583)/ 20 + 4].219 < p < .614HO: p = .65, not included in CIRejectHO: parameter = value xa) If you reject HO, α = 0.05, then a 95% CI should not include xb) If you do not reject HO, α= 0.05, then a 95% CI should include