University:
McMaster University
Course:
STATS 1LL3 | Introduction to Probability and Statistics
Academic year:

2023
Views:

52

Pages:

4
Author:

Sabrina S.

Video 12: The Poisson and Exponential Distributions Excel spreadsheet containing pothole example 1000 kms of road 850 potholes on this stretch of road on any repetition of a random experment, we can't know what the outcome will be. theoretical model to give idea of what to expect us because its random We can use some a probability, so we have on observed pattern gets closer to theoretical pattern with more repetitions. X the number of potholes that road (5km) occur in a given stretch of Discrete probability moss function the function will depend on only 1 parameter the average rate of occurence of potholes average rate of occurence 850 1000 per km 2 50 1000 x S 4.25 potholes per 5 km of road X~P(2=4.25) p(x). X= 0, 1, 2, 3 ... 0 x else where this distribution depends on events with an average rate of occurence occ occuring randomly but POISSON PROCESS Assuming no association between ownership and outcome Outcome of toss of die (6 We reject Ho if D² > 11.07 1 2 3 4 S 6 You friend 11 6 15 10 60 3 5 S 4 6 17 14 14 11 19 15 27 100 Pr (Your die ralling a 0.084 8.4 expected value of rolls of your die being 1 OR 100 60 14 X 100 100 0.084 60% of the die rolls were with your die. Theres a total of 14 1's rolled. We would expect your die to account for 60% of is rolled 60% of 14 - 8.4 Eij = row; total x Coloumn; total grand total 1 3 4 S 6 You friend 2.4 8.4 6.6 11.4 16.2 60 5.6 5.6 4.4 7.6 6 10.8 40 Expected values 14 14 ۱۹ IS 27 100 -Σ()- D n 11.03 this summation is taken over all cells 11.03 At 5% level we do not have evidence to reject Ho and so must conclude that the outcome of the die roll does not depend on whose die is rolled. the modified approach p-value: The probability of observing a Test Statistic 11.03 if Ho is true... Chi-squared table 5 af and less than 0.05 0.1 and 9.236 (0) Xs 9.236 therefore p-value < 0.1 In excel calculating p-value = CHISQ TEST (actual range, expected range) 0.050796 Inverse of Chi-squared ie observed right-tailed Critical value in excel = CHISQ INV.RT (significance, df) CHISG INV. ET (0.05, 5) 11.075 test statistic in excel CHISQ INV. RT (p-value, degrees of f) CHISQ.INV.RT (0.050796, 5) 11.02962 The Chi-Squared test of independence categorical variables are associated and is used to test two The test statistic measures the difference between what we observe and what we would expect under Ho (no association) The larger the test statistic, the more the evidence will be against Ho Note with these chi-squared tests we always do a one-sided test