Practice

2. Practice • Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as DeMere's problem, named after Chevalier De Mere. • Blaise Pascal later solved this problem.

3. Binomial Distribution p = .482 of zero aces 1 - .482 = .518 at least one ace will occur

4. Binomial Distribution p = .508 of zero double aces 1 - .508 = .492 at least one double ace will occur

5. Practice • Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as DeMere's problem, named after Chevalier De Mere. • More likely at least one ace with 4 throws will occur

6. Example • You give 100 random students a questionnaire designed to measure attitudes toward living in dormitories • Scores range from 1 to 7 • (1 = unfavorable; 4 = neutral; 7 = favorable) • You wonder if the mean score of the population is different then 4

7. Hypothesis • Alternative hypothesis • H1: sample = 4 • In other words, the population mean will be different than 4

8. Hypothesis • Alternative hypothesis • H1: sample = 4 • Null hypothesis • H0: sample = 4 • In other words, the population mean will not be different than 4

9. Results • N = 100 • X = 4.51 • s = 1.94 • Notice, your sample mean is consistent with H1, but you must determine if this difference is simply due to chance

10. Results • N = 100 • X = 4.51 • s = 1.94 • To determine if this difference is due to chance you must calculate an observed t value

11. Observed t-value tobs = (X - ) / Sx

12. Observed t-value tobs = (X - ) / Sx • This will test if the null hypothesis H0:  sample = 4 is true • The bigger the tobs the more likely that H1:  sample = 4 is true

13. Observed t-value tobs = (X - ) / Sx Sx = S / N

14. Observed t-value tobs = (X - ) / .194 .194 = 1.94/ 100

15. Observed t-value tobs = (4.51 – 4.0) / .194

16. Observed t-value 2.63 = (4.51 – 4.0) / .194

17. t distribution

18. t distribution tobs = 2.63

19. t distribution tobs = 2.63 Next, must determine if this t value happened due to chance or if represent a real difference in means. Usually, we want to be 95% certain.

20. t critical • To find out how big the tobs must be to be significantly different than 0 you find a tcrit value. • Calculate df = N - 1 • Page 747 • First Column are df • Look at an alpha of .05 with two-tails

21. t distribution tobs = 2.63

22. t distribution tcrit = -1.98 tcrit = 1.98 tobs = 2.63

23. t distribution tcrit = -1.98 tcrit = 1.98 tobs = 2.63

24. t distribution tcrit = -1.98 tcrit = 1.98 tobs = 2.63 If tobs fall in critical area reject the null hypothesis Reject H0:  sample = 4

25. t distribution tcrit = -1.98 tcrit = 1.98 tobs = 2.63 If tobs does not fall in critical area do not reject the null hypothesis Do not reject H0:  sample = 4

26. Decision • Since tobs falls in the critical region we reject Ho and accept H1 • It is statistically significant, students ratings of the dorms is different than 4. • p < .05

27. Example • You wonder if the average IQ score of students at Villanova significantly different (at alpha = .05)than the average IQ of the population (which is 100). You sample the students in this room. • N = 54 • X = 130 • s = 18.4

28. The Steps • Try to always follow these steps!

29. Step 1: Write out Hypotheses • Alternative hypothesis • H1: sample = 100 • Null hypothesis • H0: sample = 100

30. Step 2: Calculate the Critical t • N = 54 • df = 53 •  = .05 • tcrit = 2.0

31. Step 3: Draw Critical Region tcrit = -2.00 tcrit = 2.00

32. Step 4: Calculate t observed tobs = (X - ) / Sx

33. Step 4: Calculate t observed tobs = (X - ) / Sx Sx = S / N

34. Step 4: Calculate t observed tobs = (X - ) / Sx 2.5 = 18.4 / 54

35. Step 4: Calculate t observed tobs = (X - ) / Sx 12 = (130 - 100) / 2.5 2.5 = 18.4 / 54

36. Step 5: See if tobs falls in the critical region tcrit = -2.00 tcrit = 2.00

37. Step 5: See if tobs falls in the critical region tcrit = -2.00 tcrit = 2.00 tobs = 12

38. Step 6: Decision • If tobs falls in the critical region: • Reject H0, and accept H1 • If tobs does not fall in the critical region: • Fail to reject H0

39. Step 7: Put answer into words • We reject H0 and accept H1. • The average IQ of students at Villanova is statistically different ( = .05) than the average IQ of the population.

40. Practice • You recently finished giving 5 of your friends the MMPI paranoia measure. Is your friends average average paranoia score significantly ( = .10) different than the average paranoia of the population ( = 56.1)?

41. Scores

42. Step 1: Write out Hypotheses • Alternative hypothesis • H1: sample = 56.1 • Null hypothesis • H0: sample = 56.1

43. Step 2: Calculate the Critical t • N = 5 • df =4 •  = .10 • tcrit = 2.132

44. Step 3: Draw Critical Region tcrit = -2.132 tcrit = 2.132

45. Step 4: Calculate t observed tobs = (X - ) / Sx -.48 = (55.2 - 56.1) / 1.88 1.88 = 4.21/ 5

46. Step 5: See if tobs falls in the critical region tcrit = -2.132 tcrit = 2.132 tobs = -.48

47. Step 6: Decision • If tobs falls in the critical region: • Reject H0, and accept H1 • If tobs does not fall in the critical region: • Fail to reject H0

48. Step 7: Put answer into words • We fail to reject H0 • The average paranoia of your friends is not statistically different ( = .10) than the average paranoia of the population.

49. SPSS