BIOL 570 1nd Edition Lecture 14 Outline of Last Lecture I. Two categories: binomial test vs. chi square testII. chi square test of associationIII. odds ratiosOutline of Current Lecture I. Normal distributionII. Z testsIII. Central limit theorem.Current LectureHas fish size changed in a Salmon hatchery?Consider a case in which very extensive sampling (huge n) in 1985 gives us essentially error – free estimates of fish sizes. In 2009 we measure a hatchling and a 3-week old fish. HO: µ2009 = µ1985, size is normally distributed, σh 2009 = σh 1985HA: µ2009 ≠ µ1985, or σ2009 ≠ σ1985Mass of salmon hatchlings:µh 1985 = 0.265gYh 2009 = 0.190gσh 1985 = 0.02gTest statistic under the null hypothesis:D = Ybar - µOE[D] = 0 E=expected valueD= -0.075gHypothetical mass at 3 weeks:µ 1985 = 0.625gY 2009 = 0.7gσh 1985 = 0.0456gD= 0.075gThese 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.HO: 3 week size = 1985 sizeD= Y-µZ statistic (what we actually use): Z= (Y-µ)/σ Z~ Normal (µ=0, σ=1)Z (test statistic): how many SD away from the meanZh 2009 = -0.075/ 0.02 = -3.75 Z2009= 0.075/0.0456 = 1.64(3.75 SD smaller than/away from the mean) (1.64 SD away from mean)- unusual, not likelyIf there is very little sampling error we will be close to what’s expected. Z = 1.64P-value = 2(0.0505) 0.101 **table gives half P-values so multiply by 2Do Not RejectNormal is symmetrical!Z= -3.75 → 3.75 (table does not have negative numbers)P-value = 2(0.00009) = 0.00018Reject Z-test assumptions: 1) Random sample2) Null hypothesis:a. Normal distributionb. Continuous variablec. With a specific meand. With a specific standard deviation Critical point: when test statistic is too largeThe Central Limit Theory: if you add a large number of independent (random) variables, then the sum will be normally