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Unformatted text preview: 8 of 14 1.3 One-Way Analysis of Variance Test statistic To test the hypothesis H : 1 = 2 = = k , the test statistic is: F = MS (treatment) MS (error) (9) having an F distribution with two degrees of freedom: (numerator) df 1 = k- 1 (10) (denominator) df 2 = N- k (11) Like the 2 distribution, we always find the area to the RIGHT of the test statistic. F distribution CDF: pf(F, df1, df2) Finds the area to the left of F on the F density, p = P ( F < F ) = F ( F ), given the numerator and denominator degrees of freedom df1 and df2 . R Command The F distribution 1 2 3 4 5 0.0 0.2 0.4 0.6 0.8 1.0 F distribution for various degrees of freedom F df1=2, df2=10 df1=3, df2=33 df1=5, df2=50 A WORKED OUT EXAMPLE: MANUAL Example 3 . A clinical psychologist wishes to test three methods (A, B, C) for reducing hostility levels in university students to see if there is any real difference between the methods. A certain psychological test (HLT) was used to measure the degree of hostility (higher scores indicate greater hostility). Eleven students participated in the experiment and the results are shown in the table below: Anthony Tanbakuchi MAT167 Analysis of Variance 9 of 14 score method gender 1 75.00 A MALE 2 83.00 A FEMALE 3 78.00 A MALE 4 68.00 A FEMALE 5 83.00 A MALE 6 54.00 B FEMALE 7 78.00 B MALE 8 71.00 B FEMALE 9 82.00 C MALE 10 95.00 C FEMALE 11 88.00 C MALE Step 0: known info : Enter data: R: A = c (75 , 83 , 78 , 68 , 83) R: B = c (54 , 78 , 71) R: C = c (82 , 95 , 88) Enter number of levels k , level sample sizes n j , and level means x j : R: k = 3 R: n = c ( length (A) , length (B) , length (C) ) R: n [ 1 ] 5 3 3 R: x . bar = c (mean(A) , mean(B) , mean(C) ) R: x . bar [ 1 ] 77.400 67.667 88.333 Make one more vector x i with all data and find total sample size N and overall mean x R: x = c (A, B, C) R: N = length (x) R: N [ 1 ] 11 R: x . bar . bar = mean(x) R: x . bar . barR: x ....
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