Download Lecture XXV: Hypothesis Testing in Mathematical Statistics - Prof. Charles Britt Moss and more Study notes Introduction to Macroeconomics in PDF only on Docsity! 1 Examples and Multivariate Testing Lecture XXV I. Examples. A. Example 9.6.1 (mean of a binomial distribution) Assume that we want to know whether a coin toss is biased based on a sample of ten tosses. B. Our null hypothesis is that the coin is fair ( 0 1: 2 H p ) versus an alternative hypothesis that the coin toss is biased towards heads ( 1 1: 2 H p ). 1. Assume that you tossed the coin ten times and observed eight heads. What is the probability of drawing eight heads from ten tosses of a fair coin? 0 1 210 9 8 10 10 10 8 1 1 1 10 9 8 P n p p p p p p If 1 2 p , 8 0.054688P n . Thus, we reject 0 H at a confidence level of 0.10 and fail to reject 0 H at a 0.05 confidence level. 2. Moving to the Likelihood ratio test: 28 28 .5.5 1 .5 .1455 ˆ .8.8 1 .8 MLE p p Given that 2 1 2 ln ~ We reject the hypothesis of a fair coin toss at a 0.05 confidence level. ( 2ln 3.854 ) and the critical region for a chi-squared distribution at one degree of freedom is 3.84. C. Example 9.6.2. Suppose the heights of male Stanford students is distributed 2,N with a known variance of 0.16. 1. Assume that we want to test whether the mean of this distribution is 5.8 against the hypothesis that the mean of the distribution is 6. What is the test statistic for a 5 percent level of confidence and a 10 percent level of confidence? 2. Under the null hypothesis .16 ~ 5.8, 10 X N AEB 6933 – Mathematical Statistics for Food and Resource Economics Lecture XXV Professor Charles Moss Fall 2005 2 The test statistic then becomes 6 5.8 1.58 ~ 0,1 .1265 Z N Given that 1.58 0.0571P Z , we have the same decisions as above, namely that we reject the hypothesis at a confidence level of 0.10 and fail to reject the hypothesis at a confidence level of 0.05. D. Example 9.6.3 (mean of normal with variance unknown) Assume the same scenario as above, but that the variance is unknown, but estimated to be 0.16. The test then becomes: 9 6 ~ .16 10 X t The computed statistic becomes 9 1.58 0.074P t . E. Example 9.6.7 (differences in variances). In lecture XVIII, we discussed the chi-squared distribution as a distribution of the sample variance: 1. Theorem 5.4.1: Let 1 2 , , n X X X be a random sample from a 2 ,N distribution, and let 1 1 n ii X X n and 22 1 1 1 n ii S X X n Then a) X and 2S are independent random variables b) n NX 2 ,~ c) 2 2 1N S has a chi-squared distribution with 1n degrees of freedom. 2. Given the distribution of the sample variance, we may want to compare two sample variances: 2 2 2 2 1 12 2 ~ and ~ X Y X X Y Y n n X Y n S n S
