1. Today: lab 2 due Monday: Quizz 4 Wed: A3 due Friday: Lab 3 due Mon Oct 1: Exam I  this room, 12 pm

2. Recap last lecture Ch 6.1 Empirical frequency distributions Discrete Continuous Four forms F(Q=k), F(Q=k)/n, F(Qqk), F(Qqk)/n Four uses Summarization gives clue to process Summarization useful for comparisons Used to make statistical decisions Reliability evaluation

3. Today Read lecture notes!

4. Distribution of ages of mothers Sample: students that attended class in 1997 Population: MUN students Unknown distribution

5. Distribution of ages of mothers Sample: students that attended class in 1997 Population: MUN students Unknown distribution Solution: use theoretical frequency dist to characterize pop Assumption: observations are distributed in the same way as theoretical dist Theoretical distribution is a model of a frequency distribution

6. Commonly used theoretical dist: Discrete Binomial Multinomial Poisson Negative binomial Hypergeometric Uniform Continuous Normal Chi-square (  2) t F Log-normal Gamma Cauchy Weibull Uniform

7. Commonly used theoretical dist: Discrete Binomial Multinomial Poisson Negative binomial Hypergeometric Uniform Continuous Normal Chi-square (  2) t F Log-normal Gamma Cauchy Weibull Uniform

8. Theoretical frequency distributions 4 forms Empirical (n=sample) Theoretical (N=pop discrete) Theoretical (N=pop continuous)

9. Theoretical frequency distributions - 4 uses 1. Clue to underlying process If an empirical dist fits one of the following, this suggests the kind of mechanism that generated the data a)Uniform dist e.g. # of people per table  mechanism: all outcomes have equal prob b)Normal dist e.g. oxygen intake per day  mechanism: several independent factors, no prevailing factor

10. Theoretical frequency distributions - 4 uses 1. Clue to underlying process c)Poisson dist e.g. # of deaths by horsekick in the Prussian army, per year  mechanism: rare & random event c)Binomial dist e.g. # of heads/tails on coin toss  mechanism: yes/no outcome

11. Theoretical frequency distributions - 4 uses 2. Summarize data  dist info contained in parameters e.g. number of events per unit space or time can be summarized as the expected value of a Poisson dist

12. Theoretical frequency distributions - 4 uses 2. Summarize data e.g. number of events per unit space or time can be summarized as the expected value of a Poisson dist Can make comparisons

13. Theoretical frequency distributions - 4 uses 3. Decision making. Use theoretical dist to calculate p-value

14. Theoretical frequency distributions - 4 uses 3. Decision making. Use theoretical dist to calculate p-value p(X 1 qx) p(X 2 >x)

15. Theoretical frequency distributions - 4 uses 3. Decision making. Use theoretical dist to calculate p-value p(X 1 qx) MiniTab: cdf R: pnorm()

16. Theoretical frequency distributions - 4 uses 4. Reliability. Put probability range around outcome

17. Theoretical frequency distributions - 4 uses 4. Reliability. Put probability range around outcome MiniTab: invcdf R: qnorm()

18. Computing probabilities from observed vs theoretical dist Theoretical AdvantagesDisadvantages Easy Assumptions may not apply  wrong p-values FamiliarChecking assumptions is laborious Recipes, known performance Empirical AdvantagesDisadvantages No assumptionsComputation Easy to defendNot always easy to carry out

19. Ch 6.3 Fit of Observed to Theoretical Will present 2 examples: 1 continuous, 1 discrete More examples in lecture notes

20. Ch 6.3 Fit of Observed to Theoretical Example 1 (Poisson) Number of coal mining disasters, 1851-1866 (England) NDisaster = [4 5 4 1 0 4 3 4 0 6 3 3 4 0 2 4] sum(N)=47 k = [0 1 2 3 4 5 6] = outcomes(N) n = 16 observations kF(N=k) 0 1 2 3 4 5 6

21. Example 1 (Poisson) Number of coal mining disasters, 1851-1866 (England) kF(N=k)F(N=k)/n 030.1875 110.0625 21 330.1875 460.3750 510.0625 61

22. Example 1 (Poisson) Number of coal mining disasters, 1851-1866 (England) kF(N=k)F(N=k)/nPr(N=k) 030.1875 110.0625 21 330.1875 460.3750 510.0625 61

23. Example 1 (Poisson) Number of coal mining disasters, 1851-1866 (England) kF(N=k)F(N=k)/nPr(N=k) 030.18750.053 110.06250.1557 210.06250.2287 330.18750.2239 460.37500.1644 510.06250.0966 610.06250.0473

24. Example 1 (Poisson) Number of coal mining disasters, 1851-1866 (England) kF(N=k)F(N=k)/nPr(N=k)Obs-Exp 030.18750.053 110.06250.1557 210.06250.2287 330.18750.2239 460.37500.1644 510.06250.0966 610.06250.0473

25. Example 1 (Poisson) Number of coal mining disasters, 1851-1866 (England) kF(N=k)F(N=k)/nPr(N=k)Obs-Exp 030.18750.0530.1345 110.06250.1557-0.0932 210.06250.2287-0.1662 330.18750.2239-0.0364 460.37500.16440.2106 510.06250.0966-0.0341 610.06250.04730.0152

26. Example 2 (Normal) Age of mothers of students in quant 1997 Are the ages normally distributed?

27. Example 2 (Normal) Age of mothers of students in quant 1997 Are the ages normally distributed?

28. Example 2 (Normal) Age of mothers of students in quant 1997 Are the ages normally distributed? Strategy  work with probability plots  compute cdf

29. Example 2 (Normal) Age of mothers of students in quant 1997 Are the ages normally distributed? Strategy  work with probability plots  compute cdf Expected distribution:

30. Example 2 (Normal) Age of mothers of students in quant 1997 Are the ages normally distributed? Strategy  work with probability plots  compute cdf Expected distribution:

31. Example 2 (Normal) Age of mothers of students in quant 1997 Are the ages normally distributed? Strategy  work with probability plots  compute cdf