Continuous Random Variables Lecture 25 Section 7.5.4 Wed, Oct 10, 2007
Uniform Distributions The uniform distribution from a to b is denoted U(a, b). 1/(b – a) a b
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). H0: X is U(0, 1). H1: X is U(0.5, 1.5). One value of X is sampled (n = 1).
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). H0: X is U(0, 1). H1: X is U(0.5, 1.5). One value of X is sampled (n = 1). If X is more than 0.75, then H0 will be rejected.
Hypothesis Testing (n = 1) Distribution of X under H0: Distribution of X under H1: 0.5 1 1.5 0.5 1 1.5
Hypothesis Testing (n = 1) What are and ? 1 0.5 1 1.5 1 0.5 1 1.5
Hypothesis Testing (n = 1) What are and ? 1 0.5 0.75 1 1.5 1 0.5 0.75 1 1.5
Hypothesis Testing (n = 1) What are and ? 1 0.5 0.75 1 1.5 Acceptance Region Rejection Region 1 0.5 0.75 1 1.5
Hypothesis Testing (n = 1) What are and ? 1 0.5 0.75 1 1.5 1 0.5 0.75 1 1.5
Hypothesis Testing (n = 1) What are and ? = ¼ = 0.25 1 0.5 0.75 1 1.5 1 0.5 0.75 1 1.5
Hypothesis Testing (n = 1) What are and ? = ¼ = 0.25 1 0.5 0.75 1 1.5 = ¼ = 0.25 1 0.5 0.75 1 1.5
Example Now suppose we use the TI-83 to get two random numbers from 0 to 1, and then add them together. Let X2 = the average of the two random numbers. What is the pdf of X2?
Example The graph of the pdf of X2. f(y) ? y 0.5 1
Example The graph of the pdf of X2. f(y) 2 Area = 1 y 0.5 1
Example What is the probability that X2 is between 0.25 and 0.75? f(y) 0.25 0.5 0.75 1
Example What is the probability that X2 is between 0.25 and 0.75? f(y) 0.25 0.5 0.75 1
Example The probability equals the area under the graph from 0.25 to 0.75. f(y) 2 y 0.25 0.5 0.75 1
Example Cut it into two simple shapes, with areas 0.25 and 0.5. f(y) 2 0.25 0.5 0.75 1
Example The total area is 0.75. f(y) 2 Area = 0.75 y 0.25 0.5 0.75 1
Verification Use Avg2.xls to generate 10000 pairs of values of X. See whether about 75% of them have an average between 0.25 and 0.75.
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). H0: X is U(0, 1). H1: X is U(0.5, 1.5). Two values of X are sampled (n = 2). Let X2 be the average. If X2 is more than 0.75, then H0 will be rejected.
Hypothesis Testing (n = 2) Distribution of X2 under H0: Distribution of X2 under H1: 0.5 1 1.5 2 0.5 1 1.5 2
Hypothesis Testing (n = 2) What are and ? 0.5 1 1.5 2 0.5 1 1.5 2
Hypothesis Testing (n = 2) What are and ? 0.5 1 1.5 2 0.75 0.5 1 1.5 2 0.75
Hypothesis Testing (n = 2) What are and ? 0.5 1 1.5 2 0.75 0.5 1 1.5 2 0.75
Hypothesis Testing (n = 2) What are and ? 0.5 1 1.5 2 = 1/8 = 0.125 0.75 0.5 1 1.5 2 0.75
Hypothesis Testing (n = 2) What are and ? 0.5 1 1.5 2 = 1/8 = 0.125 0.75 0.5 1 1.5 2 = 1/8 = 0.125 0.75
Conclusion By increasing the sample size, we can lower both and simultaneously.
Example Now suppose we use the TI-83 to get three random numbers from 0 to 1, and then average them. Let X3 = the average of the three random numbers. What is the pdf of X3?
Example The graph of the pdf of X3. 3 y 0.25 0.5 0.75 1
Example The graph of the pdf of X3. 3 Area = 1 y 0.25 0.5 0.75 1
Example What is the probability that X3 is between 0.25 and 0.75? 3 y 0.25 0.5 0.75 1
Example What is the probability that X3 is between 0.25 and 0.75? 3 y 0.25 0.5 0.75 1
Example The probability equals the area under the graph from 0.25 and 0.75. 3 Area = 0.8594 y 0.25 0.5 0.75 1
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). H0: X is U(0, 1). H1: X is U(0.5, 1.5). Three values of X3 are sampled (n = 3). Let X3 be the average. If X3 is more than 0.75, then H0 will be rejected.
Hypothesis Testing (n = 3) Distribution of X3 under H0: Distribution of X3 under H1: 0.5 1.5 1 1.5 0.5 1
Hypothesis Testing (n = 3) Distribution of X3 under H0: Distribution of X3 under H1: = 0.0703 0.5 1.5 1 = 0.0703 1.5 0.5 1
Example Suppose we get 12 random numbers, uniformly distributed between 0 and 1, from the TI-83 and get their average. Let X12 = average of 12 random numbers from 0 to 1. What is the pdf of X12?
Example It turns out that the pdf of X12 is nearly exactly normal with a mean of 1/2 and a standard deviation of 1/12. N(1/2, 1/12) x 1/3 1/2 2/3
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.
Experiment Use the TI-83 to generate 100 averages of 12 random numbers each. Use rand(100) L1 L1 + L2 L2 L2/12 L2 Test the 68-95-99.7 Rule. 12 times