This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 05/2 = 0.025 the critical value is: t c = 2.093 It is clear that t = 0.556 is smaller than the critical value and so there is no evidence to reject the null hypothesis. 7 Chapter 11 The graph below shows both the p‐value calculation and the rejection region for the stock market returns example. PDF of t (19 ) p-value / 2 f(t) p-value / 2 area = 0.025 -tc=-2.093 -0.556 0
t 8 0.556 tc=2.093 Chapter 11 Comparing Two Means from Independent Samples The construction of a test statistic for the comparison of two means from independent samples depends on the assumptions made about the population variances. Different variance assumptions lead to different test statistics. The lecture notes for Chapter 9.2 presented one set‐up that will be presented here. The key assumption is that the two populations have a common variance σ that is estimated from the sample observations. 2 Example: The Chapter 9.2 example analyzed daily closing prices for the company Johnson & Johnson for the sample period January to June and the sample period July to December of the year 1999. The data set contained n x = 124 observations for the first sample period and n y = 128 observations for the second sample period. 9 Chapter 11 Deno...
View Full Document
This note was uploaded on 02/06/2014 for the course ECON ECON 325 taught by Professor Whistler during the Spring '10 term at The University of British Columbia.
- Spring '10