# 2 is obtained by selecting insert function tdist0556

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( > - 2 is obtained by selecting Insert Function: TDIST(0.556, 19, 2) This returns a p-value of 0.58. That is, for the t-distribution with 19 degrees of freedom, the area under the probability density function to the right of the value 0.556 is 0.58/2 = 0.29, and, by symmetry, the area to the left of the value - 0.556 is also 0.29. The calculated p-value of 0.58 exceeds any reasonable significance level and therefore, with the given data set, there is no evidence to reject the null hypothesis of equal population means for the company returns and the overall market returns. The same conclusion is obtained by comparing the test statistic with a t-distribution critical value. A significance level must be chosen. A sensible choice is a 5% significance level. From the Appendix Table for the t-distribution, using 19 degrees of freedom and a upper tail area of 0.05/2 = 0.025 the critical value is: = c t 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. Econ 325 – Chapter 10 8 The graph below shows both the p-value calculation and the rejection region for the stock market returns example. PDF of ) 19 ( t 0.556 -0.556 0 tc=2.093 -tc=-2.093 f(t) t p-value / 2 p-value / 2 area = 0.025

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Econ 325 – Chapter 10 9 Chapter 10.2 Tests of the Difference Between 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 8.2 introduced one set-up that will be presented here. The key assumption is that the two populations have a common variance 2 σ that is estimated from the sample observations. Example: The Chapter 8.2 example looked at daily closing prices for the company Johnson & Johnson in the year 1999. The data set was split into two sample periods: January to June, 1999, 124 = x n observations.

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