BIOL 570 1nd Edition Lecture 15 Outline of Last Lecture I. Normal distributionII. Z testIII. Central limit theoremOutline of Current Lecture I. Null distribution II. Z testIII. Central Limit TheoremCurrent LectureNull distribution- Sampling distribution of the test statistic if the null hypothesis is true- When the null is true, Y-bar ~ normal - Sampling distribution depends on sample size*X = Ῡµ1985 = 0 .0625gσ1985= 0.0456gῩ2009= 0.7gn = 1 n = 9Z = (0.7 – 0.625)/ 0.0456 = 1.64 Z = (0.7 – 0.625)/ (0.0456/sqrt 9) = 4.92 3 times as largePr (Z > 4.09) = 0.0002 ← largest value on the table!P < 0.0004 (P-value)Two sided (multiply P-value by 2)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.Binomial coefficient: 10! = (10!)(9!)(8!)2! 8! 2! 8! = 45When plotted, the binomial coefficient looks like a normal distribution!!People between 60 and 64 with certain risk factors have a 16% chance of having a heart attack with a time window. In a study, 15 of 300 subjects using the drug have a heart attack. Does the drug alter the probability of a heart attack?n = 300 x = 15Ho: p = 0.16HA: p ≠ 0.16E [Heart Attack] = np = (300)(0.16) = 4848 – 15 = 3348 + 33 = 81When n is large, binomial for n trials and p (probability of successes) ~ Normal [µ = np, σ = sqrt (np)(1-p)]Pr (x ≤ 15) if x ~ normalZ = 15 – 48 sqrt [300(.16)(.84)] = -5.2**In order to calculate the Z statistic, have to know the mean and st
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