Question

For each question, assume you're using an alpha of 05, and a two-tailed test, to make your decision about whether to reject the null hypothesis. 4. Assume that Data Set C depicts scores for 30 individuals who participated in a two-way independent groups design Analyze the data using a two-way ANOVA and complete an ANOVA table reporting the relevant DD, df, MS, and Fs Make sure to also report each of your critical values and indicate whether you would reject each of your three null hypotheses (i e, main effect of A, main effect of B, interaction of A x B) (25 points) Data Set C Variable B Group 2 Variable A Group 1 Group 3 10 12 1.4 18 17 15 17 19 23 15 19 20 25 19 26 30 31 Group 1 17 16 19 27 Group 2 21 29

please use manual calculations do not use excel, please

Answer 1

4)

Variable B Summary Group |Statistics Group Group 15 20 25 Group 1 14 18 Variable A 26 30 Group 2 36 29 23 27 20 Summary Statistics The summarized statistics is as shown below: Summary Statistics в, x = 98 mean 19.6 ΣΧ2= 1972 mean -16.6 mean 16.8 mean = 1 4.2 x2=1426 | 2x-19724448 Variance 11.2Variance 11.3Variance 12.8 Variance 15.31 Std.dev 3.91 stderr= 1.01 Std.dev- 3.35 Std.dev 336 Std.dev 3.58 stderr = 1.6 std.err-1.5 std.em 1.5 yx=353 mean 30.4 mean 23.5333 ΣΧ2-18041 ΣΧ2-23391 ΣΧ2-4674| ΣΧ2-8817 Variance 9.2 Variance 12.3Variance 13.3 Variance 36.41 Std.dev- 6.03 std.err 1.56 N 30 r605 mean 25mean 20.1667 Summary Statistics | Tx2-28 5 7 | Jr-3762 | Уґ-6646 | Ух2-13265 Variance 14.94Variance 16.89 Variance 44 Variance 36.7 Std.dev-6.06 std err - 1.11 x 107 x 152 mean -18.8 mean 21.4 .4 Std.dev- 3.03Std.dev 3.5Std.dev 3.65 std.err 1.57 std.err 1.63 std.err 1.36 N 10 N 10 x 190 mean -19 x=250 mean 16.5 Std.dev- 3.87 Std.dev 4.11 Std.dev-6.63 std.err-1.3 | std.err = 2.1 std.err 122

SS (4) = nb Σ 05,--. ..) = 15「(16.8-20. 1667)' + (23.5333-20.1667)2 340.03 (5x2) (16.5-20. 166が4(19-20, 1667)'t(25-20, 1667) = 381.667 (10-14.2) +. +(17-14.2) +(13-16.6)(16-16.6) +(15-19.6)(19-19.6) +(15-16.5) + + (20-16.5) +(18-19)(21-19) +(26-25) (29- 25) 280.4 SS Total-ΣΣ (10-20.1667)-+ -1064.17 (29-20.1667) SS (AB) = SS Total-SS Error-SS (4)-SS (B) = 1064, 17-280.4-340.03-381.667 = 62.0734 The resultant ANOVA table is as follows df 2-1=1 3-1=2 Source SS MS 340.03 29. 1 | 0.0001 A 1340.03 -340.3 | -4-340.3 = MS 11.68 MS8 = 190:84 1633 | 0.0001 MSE 11.68 381.67 11681633 MSLA = 31.04 = 2.66 | 0.0905 MSg 11.68 B 381.67 | 62.07 | (a-1)(b-1)=2 | 280.4 | n-ab-30-6=241 | 1064.17 | n-1=30-1=29 ー= 190.84 |ー2-1 31.04 Ax8 011.68 Error ! 24 Total