CJUS K300 1nd Edition Lecture 13 Outline of Last Lecture I Hypothesis testing 1 tailed directional tests Outline of Current Lecture II Chi Square Goodness of Fit Test Current Lecture Time of day Observed frequency Expected frequency 12am 6am 20 25 6am 12pm 7 25 12pm 6pm 31 25 6pm 12am 42 25 Total 100 100 H o x 2 0 In the population of all ambulance calls at Green College the calls are distributed evenly across the 4 time of day categories H i x 2 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 7 815 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 2 X o e e 2 obt This is the equation in theory and we have to understand this equation in order to understand what our answer means but we will be using a somewhat more simple equation to work with day to day X 2obt 26 96 o Time of day o e e o e 2 o e 2 e 12am 6am 20 25 5 25 1 6am Noon 7 25 18 324 12 96 Noon 6pm 31 25 7 49 1 44 6pm 12am 42 25 17 289 11 56 Total 100 26 96 Is 26 96 7 815 Yes so we Reject Ho Conclusion We are 95 certain that in the population of all Green College ambulance calls the ambulance 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 2 o X n e 2 obt Using the same example with this equation we would do 20 2 25 72 312 25 25 422 100 26 96 25
View Full Document
Unlocking...