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Stat 301 – Day 20 Fisher’s Exact Test. Announcements HW 4 (45 pts)  Classify variables (categorical vs. quantitative)  Round vs. truncating to produce.

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Presentation on theme: "Stat 301 – Day 20 Fisher’s Exact Test. Announcements HW 4 (45 pts)  Classify variables (categorical vs. quantitative)  Round vs. truncating to produce."— Presentation transcript:

1 Stat 301 – Day 20 Fisher’s Exact Test

2 Announcements HW 4 (45 pts)  Classify variables (categorical vs. quantitative)  Round vs. truncating to produce counts  Defining the parameter  Simulation vs. theory-based One vs. two-sided hypotheses Two-sided p-values Interpretation of p-value, random chance = sampling  Some times one sample but will to consider the two sub-samples independent (male, female) Project 2 proposal

3 Extra credit possibility? Taysseer Sharaf  22-212  4:10-5pm  Teaching demonstration (paired t-test)

4 Last Time – Comparing treatments Random chance = random assignment Null hypothesis: Equal treatment probabilities  No association between explanatory and response variables Simulation: Assume the successes/failures were going to be successes/failures no matter which group they were assigned to, see how often “shuffling” creates a difference in the groups at least as extreme as observed

5 PP 2.6 (p. 151) sacrificeno reminder total Gifts ok571370 Not ok6347110 12060180 180 cards, 70 blue and 110 green Deal out 120 and 60 Count the number of blue cards in sacrifice group How often 57 or more blue cards in sacrifice group? How often 13 or fewer in no reminder?

6 Quiz 9

7 Investigation 2.7 (p. 152) (a) EV: yawn seed or no RV: whether or not yawned (b) Parameter: difference in probability of yawning (yawn seed – no yawn seed) H 0 : the two probabilities are the same H a : the probability of yawning is larger if given a yawn seed

8 Hypergeometric Distribution X counts the number of successes in a sample of n objects selected from a population of N objects consisting of M successes E(X) = n(M/N) Two-way table applet, R  Fisher’s Exact Test

9 Let X represent the number of yawners in the yawn seed group p-value = P(X> 10) with M = 14, n = 34, N = 50 (p) Success = “not yawning” p-value = P(X < 24) Yawning study M N n M N n

10 (q) Let X represent the number that didn’t yawn in the no yawn seed group p-value = P(X > 12) with M = 36, n = 16, N = 50

11 To Do PP 2.7A and 2.7B(a)(b) (p. 157) Investigation 2.8 through part (d) Have one group member submit project 2 proposal


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