Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Continuous Random Variables Lecture 26 Section 7.5.4 Mon, Mar 5, 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: 1 0 0.5 1 1.5 1 1.5 Distribution of X under H1: 1 0 0.5 Hypothesis Testing (n = 1) What are and ? 1 0 0.5 1 1.5 0 0.5 1 1.5 1 Hypothesis Testing (n = 1) What are and ? 1 0 0.5 0.75 1 1.5 0 0.5 0.75 1 1.5 1 Hypothesis Testing (n = 1) What are and ? 1 0 0.5 0.75 Acceptance Region 1 1.5 Rejection Region 1 0 0.5 0.75 1 1.5 Hypothesis Testing (n = 1) What are and ? 1 0 0.5 0.75 1 1.5 0 0.5 0.75 1 1.5 1 Hypothesis Testing (n = 1) What are and ? = ¼ = 0.25 1 0 0.5 0.75 1 1.5 0 0.5 0.75 1 1.5 1 Hypothesis Testing (n = 1) What are and ? = ¼ = 0.25 1 0 1 0.5 0.75 1 1.5 0.5 0.75 1 1.5 = ¼ = 0.25 0 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 0.5 1 Example The graph of the pdf of X2. f(y) 2 Area = 1 y 0 0.5 1 Example What is the probability that X2 is between 0.25 and 0.75? f(y) 2 y 0 0.25 0.5 0.75 1 Example What is the probability that X2 is between 0.25 and 0.75? f(y) 2 y 0 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 0.25 0.5 0.75 1 Example Cut it into two simple shapes, with areas 0.25 and 0.5. f(y) 2 Area = 0.25 0.5 Area = 0.5 y 0 0.25 0.5 0.75 1 Example The total area is 0.75. f(y) 2 Area = 0.75 y 0 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: 2 0 0.5 1 1.5 Distribution of X2 under H1: 2 0 0.5 1 1.5 Hypothesis Testing (n = 2) What are and ? 2 0 0.5 1 1.5 0 0.5 1 1.5 2 Hypothesis Testing (n = 2) What are and ? 2 0 0.5 0.75 1 1.5 0 0.5 0.75 1 1.5 2 Hypothesis Testing (n = 2) What are and ? 2 0 0.5 0.75 1 1.5 0 0.5 0.75 1 1.5 2 Hypothesis Testing (n = 2) What are and ? 2 = 1/8 = 0.125 0 0.5 0.75 1 1.5 0 0.5 0.75 1 1.5 2 Hypothesis Testing (n = 2) What are and ? 2 = 1/8 = 0.125 0 2 0.5 0.75 1 1.5 0.75 1 1.5 = 1/8 = 0.125 0 0.5 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 1/3 2/3 1 Example The graph of the pdf of X3. 3 Area = 1 y 0 1/3 2/3 1 Example What is the probability that X3 is between 1/3 and 2/3? 3 y 0 1/3 2/3 1 Example What is the probability that X3 is between 1/3 and 2/3? 3 y 0 1/3 2/3 1 Example The probability equals the area under the graph from 1/3 to 2/3. 3 Area = 2/3 y 0 1/3 2/3 1 Verification Use Avg3.xls to generate 10000 triples of numbers. See if about 2/3 of the averages lie between 1/3 and 2/3. 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: 0 1/3 2/3 1 4/3 1 4/3 Distribution of X3 under H1: 1 0 1/3 2/3 Hypothesis Testing (n = 3) Distribution of X3 under H0: = 0.07 0 1/3 2/3 1 4/3 1 4/3 Distribution of X3 under H1: 1 = 0.07 0 1/3 2/3 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 Excel spreadsheet Avg12.xls to generate 10000 values of X, where X is the average of 12 random numbers from U(0, 1). Test the 68-95-99.7 Rule.