hw13sol - Solutions to Homework 13 Problem 13.5 First, SSE...

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Solutions to Homework 13 Problem 13.5 First, SSE = 1800 - 1200 = 600. The degrees of freedom for the Treatments is k - 1 = 3 - 1 = 2. The degrees of freedom for the Error is n T - k = 47 - 3 = 44. The degrees of freedom for the Total is n T - 1 = 47 - 1 = 46. We then have MSTR = 1200 2 = 600 and MSE = 600 44 = 13 . 64 . The F statistic is MSTR MSE = 600 13 . 64 = 43 . 99 The p -value is the area to the right of 43.99 for an F distribution with 2 and 44 degrees of freedom. This area is less than 0.01. The complete ANOVA table is SS df MS F p -value Treatments 1200 2 600 43.99 < 0 . 01 Error 600 44 13.64 Total 1800 46 Problem 13.9 We first compute the sample means and sample variances for each group. We get ¯ x 1 = 33, ¯ x 2 = 29, and ¯ x 3 = 28. We also have s 1 = 5 . 66, s 2 = 4 . 18, and s 3 = 3 . 08. The overall mean is ¯ ¯ x = 30. Now, SSTR = n 1 x 1 - ¯ ¯ x ) 2 + n 2 x 2 - ¯ ¯ x ) 2 + n 3 x 3 - ¯ ¯ x ) 2 = 70 MSTR = SSTR k - 1 = 70 3 - 1 = 35 SSE = ( n 1 - 1) s 2 1 + ( n 2 - 1)
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This note was uploaded on 01/24/2011 for the course STATISTICS 19897 taught by Professor Jager,abigaill during the Spring '10 term at Kansas State University.

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hw13sol - Solutions to Homework 13 Problem 13.5 First, SSE...

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