582029andbysymmetrytheareatotheleftofthevalue

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Unformatted text preview: − 0.173 and s d = 1.391 To test the null hypothesis of equal population means for the two samples against a two‐sided alternative the t‐test statistic is: 6 t= d sd = n − 0.173 = − 0.556 1.391 20 Chapter 11 With Microsoft Excel, the p‐value calculation 2 ⋅ P(t ( n − 1 ) > t ) 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....
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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.

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