BIOL 570 1nd Edition Lecture 12 Outline of Last 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. CIOutline of Current Lecture I. Chi square goodness of fit testsa. test statisticb. determining P valuec. Table A (p 669-671)II. assumptions of testsCurrent LectureGoodness of fit- method for comparing observed frequency distribution with an expected frequency distribution - Binomial test is a G.O.F. testx2 goodness of fit examplePopulation: 16% < 25 years, 44% 25 – 44, 27% 45 – 64, 13% 64> Step 1: Q1 – Does age affect the likelihood of a car crash?Step 2: HO: Pr (accident) is equal for all ages [pr (accident <25) = pr (accident 25-44) = pr (accident 45-64) = pr (accident 64>)]HA: at least one probability differs from aboveThese 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.Age Observed(Obs)Expected (Exp)<25 36 88 x .16 = 14.0825-44 21 88 x .44 = 38.7245-64 12 88 x .27 = 23.7664> 19 88 x .13 = 11.4488 total*okay to have decimalsTest statistic: ∑ = # of categoriesx2 = (36 – 14.08)2 + (21 – 38.72)214.08 38.72 ….. x2 = 53.05 (test statistic) P-value= shaded region- will be skewed to the right: because sampling errorDegrees of freedom:DF = # of categories – 1- # of parameters estimated from dataHandout:Reject HO, state how small P is, P < 0.01Not reject HO, state how big P is, p > 0.05Based on a random sample of 88, the age of drivers does affect the probability of a crash. (x2= 53.05, df = 3, P < 0.01)Assumptions1) Random sample from population2) None of the expected numbers should be less than 1; no more than 20% should be less than 5 |53.05Probability densityTest
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