1. Continuous Random Variables Lecture 26 Section 7.5.4 Mon, Mar 5, 2007

2. Uniform Distributions The uniform distribution from a to b is denoted U(a, b). ab 1/(b – a)

3. Hypothesis Testing (n = 1) An experiment is designed to determine whether a random variable X has the distribution U(0, 1) or U(0.5, 1.5).  H 0 : X is U(0, 1).  H 1 : X is U(0.5, 1.5). One value of X is sampled (n = 1).

4. Hypothesis Testing (n = 1) An experiment is designed to determine whether a random variable X has the distribution U(0, 1) or U(0.5, 1.5).  H 0 : X is U(0, 1).  H 1 : X is U(0.5, 1.5). One value of X is sampled (n = 1). If X is more than 0.75, then H 0 will be rejected.

5. Hypothesis Testing (n = 1) Distribution of X under H 0 : Distribution of X under H 1 : 00.511.5 1 00.511.5 1

6. Hypothesis Testing (n = 1) What are  and  ? 00.511.5 1 00.511.5 1

7. Hypothesis Testing (n = 1) What are  and  ? 0.75 00.511.5 1 00.511.5 1

8. Hypothesis Testing (n = 1) What are  and  ? 0.75 00.511.5 1 00.511.5 1 Acceptance RegionRejection Region

9. Hypothesis Testing (n = 1) What are  and  ? 0.75 00.511.5 1 00.511.5 1

10. Hypothesis Testing (n = 1) What are  and  ? 0.75 00.511.5 1 00.511.5 1   = ¼ = 0.25

11. Hypothesis Testing (n = 1) What are  and  ? 0.75 00.511.5 1 00.511.5 1   = ¼ = 0.25   = ¼ = 0.25 0.75

12. Example Now suppose we use the TI-83 to get two random numbers from 0 to 1, and then add them together. Let X 2 = the average of the two random numbers. What is the pdf of X 2 ?

13. Example The graph of the pdf of X 2. y f(y)f(y) 00.51 ?

14. Example The graph of the pdf of X 2. y f(y)f(y) 00.51 2 Area = 1

15. Example What is the probability that X 2 is between 0.25 and 0.75? y f(y)f(y) 00.510.250.75 2

16. Example What is the probability that X 2 is between 0.25 and 0.75? y f(y)f(y) 00.510.250.75 2

17. Example The probability equals the area under the graph from 0.25 to 0.75. y f(y)f(y) 00.5 2 10.250.75

18. Example Cut it into two simple shapes, with areas 0.25 and 0.5. y f(y)f(y) 00.510.250.75 2 Area = 0.5 Area = 0.25 0.5

19. Example The total area is 0.75. y f(y)f(y) 00.510.250.75 2 Area = 0.75

20. Verification Use Avg2.xls to generate 10000 pairs of values of X.Avg2.xls See whether about 75% of them have an average between 0.25 and 0.75.

21. Hypothesis Testing (n = 2) An experiment is designed to determine whether a random variable X has the distribution U(0, 1) or U(0.5, 1.5).  H 0 : X is U(0, 1).  H 1 : X is U(0.5, 1.5). Two values of X are sampled (n = 2). Let X 2 be the average. If X 2 is more than 0.75, then H 0 will be rejected.

22. Hypothesis Testing (n = 2) Distribution of X 2 under H 0 : Distribution of X 2 under H 1 : 00.511.5 2 00.511.5 2

23. Hypothesis Testing (n = 2) What are  and  ? 00.511.5 2 00.511.5 2

24. Hypothesis Testing (n = 2) What are  and  ? 00.511.5 2 00.511.5 2 0.75

25. Hypothesis Testing (n = 2) What are  and  ? 00.511.5 2 00.511.5 2 0.75

26. Hypothesis Testing (n = 2) What are  and  ? 00.511.5 2 00.511.5 2 0.75   = 1/8 = 0.125

27. Hypothesis Testing (n = 2) What are  and  ? 00.511.5 2 00.511.5 2 0.75   = 1/8 = 0.125   = 1/8 = 0.125

28. Conclusion By increasing the sample size, we can lower both  and  simultaneously.

29. Example Now suppose we use the TI-83 to get three random numbers from 0 to 1, and then average them. Let X 3 = the average of the three random numbers. What is the pdf of X 3 ?

30. Example The graph of the pdf of X 3. y 01/32/3 3 1

31. Example The graph of the pdf of X 3. y 01/32/31 Area = 1 3

32. Example What is the probability that X 3 is between 1/3 and 2/3? y 01/32/31 3

33. Example What is the probability that X 3 is between 1/3 and 2/3? y 01/32/31 3

34. Example The probability equals the area under the graph from 1/3 to 2/3. y 01/32/31 Area = 2/3 3

35. Verification Use Avg3.xls to generate 10000 triples of numbers.Avg3.xls See if about 2/3 of the averages lie between 1/3 and 2/3.

36. Hypothesis Testing (n = 3) An experiment is designed to determine whether a random variable X has the distribution U(0, 1) or U(0.5, 1.5).  H 0 : X is U(0, 1).  H 1 : X is U(0.5, 1.5). Three values of X 3 are sampled (n = 3). Let X 3 be the average. If X 3 is more than 0.75, then H 0 will be rejected.

37. Hypothesis Testing (n = 3) Distribution of X 3 under H 0 : Distribution of X 3 under H 1 : 01/32/31 4/3 01/32/31 1 4/3

38. Hypothesis Testing (n = 3) Distribution of X 3 under H 0 : Distribution of X 3 under H 1 : 01/32/31 4/3  = 0.07 01/32/31 1 4/3  = 0.07

39. Example Suppose we get 12 random numbers, uniformly distributed between 0 and 1, from the TI-83 and get their average. Let X 12 = average of 12 random numbers from 0 to 1. What is the pdf of X 12 ?

40. Example It turns out that the pdf of X 12 is nearly exactly normal with a mean of 1/2 and a standard deviation of 1/12. x 1/22/31/3 N(1/2, 1/12)

41. Example What is the probability that the average will be between 0.45 and 0.55? Compute normalcdf(0.45, 0.55, 1/2, 1/12). We get 0.4515.

42. Experiment Use the Excel spreadsheet Avg12.xls to generate 10000 values of X, where X is the average of 12 random numbers from U(0, 1).Avg12.xls Test the 68-95-99.7 Rule.