CJUS-K300 1nd Edition Lecture 13Outline of Last Lecture I. Hypothesis testing: 1 tailed, directional testsOutline of Current Lecture II. Chi Square Goodness of Fit TestCurrent LectureTime of day Observed frequencyExpected frequency12am-6am 20 256am-12pm 7 2512pm-6pm 31 256pm=12am 42 25Total 100 100Ho: x2 = 0 (In the population of all ambulance calls at Green College, the calls are distributed evenly across the 4 time of day categories)Hi: x2 > 0 (In the population of all ambulance calls at Green College, the calls are not distributed evenly across the 4 time of day categories)alpha = .05 Degrees of freedom = K-1 (K most likely stands for categories in german) d.f = 4-1 = 3 at an alpha of .05 X^2crit = 7.815These 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.Xobt2=∑(ƒo− ƒe)2ƒe This is the equation in theory and we have to understand this equation inorder to understand what our answer means but we will be using a somewhat more simple equation to work with day to day. Xobt2= 26.96Time of dayƒoƒeƒo−ƒe(ƒo−ƒe)2(ƒo−ƒe)ƒe212am -6am 20 25 -5 25 16am-Noon 7 25 -18 324 12.96Noon-6pm 31 25 7 49 1.446pm-12am 42 25 17 289 11.56Total 100 26.96Is 26.96>7.815? Yes so we Reject HoConclusion: We are 95% certain that, in the population of all Green College ambulance calls, theambulance calls are not evenly distributed across the 4 times of day categories. (You will absolutely need to know how to write this out.)Chi-Squared Computational Formula (aka the formula we will actually be using)Xobt2=∑ƒo2ƒe−n Using the same example with this equation we would do:20225 + 7225+31225 +
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