Analysis of Variance II: ANOVA and Multiple Comparisons - Prof. Michae Hudgens, Study notes of Data Analysis & Statistical Methods

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Analysis of Variance II Bios 662

Michael G. Hudgens, Ph.D. [email protected] http://www.bios.unc.edu/∼mhudgens 2006-11-07 15:

Outline

• ANOVA Example: SAS/R
• Multiple Comparisons
  • Scheffe
  • Tukey
  • Bonferroni

ANOVA: Example

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Group

Age

Active Passive None Control

ANOVA: SAS

proc glm data=one; * proc anova data=one; class group; model age=group; run;

Sum of Source DF Squares Mean Square F Value Pr > F Model 3 14.77780797 4.92593599 2.14 0. Error 19 43.68958333 2. Corrected Total 22 58.

ANOVA: R

av <- aov(age ~ group) anova(av)

Analysis of Variance Table

Response: age Df Sum Sq Mean Sq F value Pr(>F) group 3 14.778 4.926 2.1422 0. Residuals 19 43.690 2.

Multiple Comparisons

• Suppose we do n independent tests, each with probabil- ity α of making a type I error
• Suppose all n null hyp are true
• What is the probabilty of making at least one type I error? 1 − (1 − α)n

ANOVA and Multiple Comparisons

• Rejection of H 0 : μ 1 = μ 2 = · · · μK does not indicate where the 6 =’s exists
• E.g. HA : μ 1 = μ 2 = · · · = μK− 1 6 = μK HA : μ 1 6 = μ 2 6 = · · · 6 = μK− 1 6 = μK
• Usually we want to identify where the 6 =’s exist

ANOVA

• If H 0 is not rejected, no need for further analysis
• Otherwise, need multiple comparisons method to test the

(K

2

null hyps H 0 : μi = μj (i 6 = j)

• Popular methods:
  • Scheffe
  • Tukey
  • Bonferroni (Sidak, Holm, Hochberg)

Scheffe: Passive smoking example

Comparison tij Significant NS-PS 6.01 yes NS-NI 3.65 yes NS-LS 6.27 yes NS-MS 13.17 yes NS-HS 14.92 yes PS-NI -0.16 no PS-LS 0.88 no PS-MS 7.14 yes PS-HS 8.90 yes NI-LS 0.71 no NI-MS 4.67 yes NI-HS 5.79 yes LS-MS 6.27 yes LS-HS 8.03 yes MS-HS 1.76 no

Scheffe: Example

• Overall conclusions about similarities and difference among the population means using schematic diagram
• Use overbars to connect means that do not significantly differ

2.59 2.73 3.23 3.30 3.

HS MS LS P NI NS

Scheffe: Example II

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Substance

Dosage at death

1 2 3 4

Scheffe: Example II

• Global F-test strongly rejects null of equality of four population means (p = 0.0002)
• Critical region

C 0. 05 = {|tij| >

3 F 3 , 36 , 0. 95 =

• Note denominator of tij is always

M SE/5 = 1. 396

• So could also write critical region in terms of minimum signficant difference C 0. 05 = {| Y¯i· − Y¯j·| > 2. 93 ∗ 1 .396 = 4. 09 }

ANOVA: Scheffe

• For each pair of means, can compute multiplicity ad- justed confidence intervals using Scheffe’s method also

Y¯i− Y¯j±

M SE

ni

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×

(K − 1)FK− 1 ,N −K, 1 −α

• The probability is 1 − α these intervals simultaneously straddle the associated population mean differences
• What happens when K = 2?
• For cardiac substance example,

Y¯i − Y¯j ± 4. 09

Scheffe: SAS

proc glm; class group; model dose=group; means group/scheffe cldiff;

Scheffe’s Test for dose Comparisons significant at the 0.05 level are indicated by ***. Difference Simultaneous group Between 95% Confidence Comparison Means Limits 1 - 2 3.700 -0.394 7. 1 - 3 5.900 1.806 9.994 *** 1 - 4 6.300 2.206 10.394 *** 2 - 1 -3.700 -7.794 0. 2 - 3 2.200 -1.894 6. 2 - 4 2.600 -1.494 6. 3 - 1 -5.900 -9.994 -1.806 *** 3 - 2 -2.200 -6.294 1. 3 - 4 0.400 -3.694 4. 4 - 1 -6.300 -10.394 -2.206 *** 4 - 2 -2.600 -6.694 1. 4 - 3 -0.400 -4.494 3.