1. 1 Counting in probability Permutations The number of orderings of different events Combinations The number ways that outcomes can be grouped

2. 2 Permutations The number of orderings of different events. Three cards: AKQ AKQ, AQK, KQA, KAQ, QAK QKA = 6 Four digits: 7412 24 Five symbols: # $ % & @ 120

3. 3 Three different cards: AKQ AKQ, AQK, KQA, KAQ, QAK QKA = 6 Complete permutations

4. 4 Partial Permutations Four Different Digits (of 10) In the first spot: 10 different digits possible In the second spot: 9 possible left In the third spot: 8 possible left In the last spot: 7 possible left 10*9*8*7 = 5040

5. 5 Combinations The number ways that outcomes can be grouped Calculate the number of combinations of N objects taken r (r <= N) at a time. Example: 13 cards (AKQ…2) taken 5 at a time = 1287

6. 6 Review the problem We want to be able to say something about the population from a single sample that we have drawn. This is the problem of statistical inference What can we infer from our sample?

7. 7 Our logic for proceeding If the population has certain characteristics, then our sample will probably include certain outcomes and probably not include other outcomes If our sample has outcomes that are unlikely to come from that population, it probably did not come from that population

8. 8 Our logic (continued) If our sample has outcomes that are unlikely to come from that population, it probably did not come from that population Our conclusion must be either that the population is as hypothesized or it is not as hypothesized. We reject or fail to reject the hypothesized population characteristic

9. 9 Once More We reason from population characteristics to probability distribution of all possible samples -- the sampling distribution. We calculated this for the population of 7 black and 3 red marbles We noted that the sampling distribution began to look like a normal distribution

10. 10 Sampling distributions There are a few sampling distributions that occur very frequently binomial normal student’s t statistic chi-square statistic F statistic

11. 11 Binomial This is the distribution we have already calculated for the red (p=.3) and black (p=.7) marbles. It is also the distribution for coin tosses where p=.5 for heads and p=.5 for tails It is the distribution for all binary outcomes.

12. 12 Normal This is the distribution for many empirical distribution. This is the distribution which many other distributions approach when working with large numbers: binomial, t, chi- square, etc. This is the distribution for the sum of a set of random variables.

13. 13 Normal distribution characteristics The formula

14. 14 Normal distribution characteristics The shape 68%

15. 15 Normal distribution characteristics The shape 95%

16. 16 Normal distribution characteristics The shape Mean One unit is one std. dev.

17. 17 Why the normal distribution? Central Limit Theorem Sampling distribution of the mean approaches normal distribution Mean of sampling distribution approaches mean of population sd of sampling distribution approaches population sd / sqrt(n) as n becomes large

18. 18 Thus pop mean = 5, pop sd = 1 If n=4, sampling dist mean = 5, sd =.5 If n=9, sampling dist mean = 5, sd =.333 If n=25, samp dist mean = 5, sd =.20 If n=900, samp dist mean = 5, sd =.033

19. 19 N=4 N=9 N=900 Pop