Four were selected at random from among the 11

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the experiment. Four were selected at random from among the 11 problem cases and treated with method 1. Four of the remaining seven students were selected at random and treated with method 2. The remaining three students were treated with method 3. All treatments were continued for a one-semester period. Each student was given the HLT test at the end of the semester, with the results shown in Table 13.8. Use these data to perform an analysis of variance to determine if there are differences among mean scores for the three methods. Use α = . 05. Method Test Scores Totals 1 80 92 87 83 342 2 70 81 78 74 303 3 63 76 70 209 Total 854 Solution: The null and alternative hypotheses are H 0 : μ 1 = μ 2 = μ 3 H a : At least one of the population means differes from the rest. For n 1 = 4 , n 2 = 4 , and n 3 = 3, we have a total sample size of n T = 11. The totals from Table are y 1 · = 342 , y 2 · = 303 , y 3 · = 209 , y ·· = 854 . Substituting into the computational formulas for TSS and SSB, we have TSS = X i,j y 2 ij - y 2 ·· n T = (80) 2 + (92) 2 + · · · + (70) 2 - (854) 2 11 = 66 , 988 - 66 , 301 . 45 = 686 . 55 SSB = X y 2 i · n i - y 2 ·· n T = (342) 2 4 + (303) 2 4 + (209) 2 3 - 66 , 301 . 45 = 66 , 753 . 58 - 66 , 301 . 45 = 452 . 13 . Then SSW = 686 . 55 - 452 . 13 = 234 . 42 . The AOV table for these data is shown in Table 13.9. 3
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Source SS df MS F Between samples 452.13 2 226.07 226.07/29.3=7.72 Within samples 234.42 8 29.30 Totals 686.55 10 The critical value of F is obtained from Table 6 in the Appendix for α = . 5, df 1 = 2, and df 2 = 8; this value is 4.46. Since the computed value of F, 7.72, exceeds the tabulated value, 4.46, we reject the null hypothesis of
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  • Fall '10
  • WeiZhu
  • Yij, total sample size, Sample observations, SSB SSW, SS SSB SSW

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