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(c) 2007 IUPUI SPEA K300 (4392) Outline Comparing Group Means Data arrangement Linear Models and Factor Analysis of Variance (ANOVA) Partitioning Variance.

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Presentation on theme: "(c) 2007 IUPUI SPEA K300 (4392) Outline Comparing Group Means Data arrangement Linear Models and Factor Analysis of Variance (ANOVA) Partitioning Variance."— Presentation transcript:

1 (c) 2007 IUPUI SPEA K300 (4392) Outline Comparing Group Means Data arrangement Linear Models and Factor Analysis of Variance (ANOVA) Partitioning Variance F-test (Computation) T-test and ANOVA Conclusion

2 (c) 2007 IUPUI SPEA K300 (4392) Comparing group means Compares the means of independent variables x1 and x2 (independent sample t-test) Compares two group means of a variable x; examines how group makes difference in x. Data arrangements: SPSS requires the second arrangement for independent sample t-test

3 (c) 2007 IUPUI SPEA K300 (4392) Data Arrangement xgroup 10 12 9 15 13 6 8 3 0 2 00000111110000011111 x1x2 10 12 9 15 13 6830268302

4 (c) 2007 IUPUI SPEA K300 (4392) Linear Model and Factor Examines how group makes difference in the variable x. X = µ + g i + e i µ is the overall mean, g i is group i’s mean difference from the overall mean G is called factor (categorical independent variable) that makes difference in the left-hand side variable (dependent variable) x.

5 (c) 2007 IUPUI SPEA K300 (4392) Comparing More Variables What if you need to compare x1, x2, and x3? Or compares means of three groups? Compares x1 and x2, x1 and x3, and x2 and x3? How about comparing 4 and 5 groups? Any way to make comparison easy? ANOVA is the answer

6 (c) 2007 IUPUI SPEA K300 (4392) T-test and ANOVA 1 T-test directly compares means of two variables (groups) ANOVA (Analysis of Variance) partitions overall variance and examines the impact of factors on mean difference So, t-test is a special case of ANOVA that considers only TWO GROUPS

7 (c) 2007 IUPUI SPEA K300 (4392) Data Arrangement: x ij An observation of i group and jth data point x ij xgroup x 11 x 12 x 13 x 14 x 15 x 21 x 22 x 23 x 24 x 25 00000111110000011111 x1x2 x 11 x 12 x 13 x 14 x 15 x 21 x 22 x 23 x 24 x 25

8 (c) 2007 IUPUI SPEA K300 (4392) Partitioning of Variance 1 Overall mean x ** ; group i mean x i* Sum of squares between group (treatment) Sum of squares within group SST = SSM (between) + SSE (within)

9 (c) 2007 IUPUI SPEA K300 (4392) Partitioning of Variance 2: µ=7.7 Treatmentx(x-xbar)^2xibar(xibar-xbar)^2(x-xibar)^2 1105.13811.816.5383.24 11218.20411.816.5380.04 191.60411.816.5387.84 11552.80411.816.53810.24 11327.73811.816.5381.44 263.0043.815.4714.84 280.0713.815.47117.64 2322.4043.815.4710.64 2059.8043.815.47114.44 2232.8713.815.4713.24 357.4717.60.0186.76 391.6047.60.0181.96 31218.2047.60.01819.36 380.0717.60.0180.16 3413.9387.60.01812.96 Total 264.933 160.133104.800

10 (c) 2007 IUPUI SPEA K300 (4392) ANOVA: F test 1 H 0 : all group have the same mean Ratio of MSM to MSE Degrees of freedom 1 is t-1 (t is the number of groups) Degrees of freedom 2 is N-t (N is the number of overall observation) SourcesSum of SquaresDFMean SquaresF Model (treatment)SSMt-1MSM=SSM/(t-1)MSM/MSE Error (residual)SSEN-tMSE=SSE/(N-t) TotalSSTN-1

11 (c) 2007 IUPUI SPEA K300 (4392) ANOVA: F test 2 T=3 (three groups); N=15 (5 in each group) SST=264.9=160.1+104.8 SSM=160.1, MSM=80.1=160/(3-1) SSE=104.8, MSE=8.7=104.8/(15-3) F=9.2=80.1/8.7 (CV=5.10 p.639) df1=2=3-1, df2=12=15-3 SourcesSum of SquaresDFMean SquaresF Model (treatment)160.1280.19.2 Error (residual)104.8128.7 Total264.914

12 (c) 2007 IUPUI SPEA K300 (4392) ANOVA: F-test 3 If F score is larger than the critical value or the p-value is smaller than the significance level, reject the null hypothesis Rejection of H 0 is interpreted as there is at least one group that has a mean different from other group means. Rejection of H 0 does not, however, say which groups have different means.

13 (c) 2007 IUPUI SPEA K300 (4392) ANOVA: Short Cuts Compute sum of observations (overall and individual groups) or overall mean x ** Compute variances of groups (s i ) 2

14 (c) 2007 IUPUI SPEA K300 (4392) Independent sample t-test X1bar=$26,800, s1=$600, n1=10 X2bar=$25,400, s2=$450, n2=8 Since 5.47>2.58 and p-value <.01, reject the H 0 at the.01 level.

15 (c) 2007 IUPUI SPEA K300 (4392) ANOVA for two group means sum1=268,000=26,800*10, s1=$600, n1=10 sum2=203,200=24,400*8, s2=$450, n2=8 Overall sum=471,200=268,000+203,200, N=18 SST=SSM+SSE=13,368,611=8,711,111+4,657,500 F=MSM/MSE=29.9254; sqrt(29.9254)=5.4704=t

16 (c) 2007 IUPUI SPEA K300 (4392) T-test versus ANOVA T-test examines mean difference using t distribution (mean difference/standard deviation) ANOVA examines the mean difference by partitioning variance components of between and within group (F-test) T-test is a special case of ANOVA T score is the square root of F score when df 1 is 1 (comparing two groups)

17 (c) 2007 IUPUI SPEA K300 (4392) Summary of Comparing Means


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