Question

The times (in seconds) for a sample of New York Marathon runners were as follows:

Gender	Age Class
Male	20-29	30-39	40+
	13615	14677	14528
	18784	16090	17034
	14256	14086	14935
	10905	16461	14996
	12077	20808	22146
Female	16401	15357	17260
	14216	16771	25399
	15402	15036	18647
	15326	16297	15077
	12047	17636	25898

Conduct a two-way analysis of variance including interactions to examine these data. Perform model criticism. What do you conclude? Note that R users do not need to invoke the “car” package since the equal replication means that Type I = Type III SS (using car package can be tricky when fitting models with interactions since the contrasts also need to be altered to make them comparable).

Answer 1

R code:

Input =("
AgeClass Gender Time
20-29 Male 13615
20-29 Male 18784
20-29 Male 14256   
20-29 Male 10905
20-29 Male 12077
30-39 Male 14677
30-39 Male 16090
30-39 Male 14086   
30-39 Male 16461
30-39 Male 20808
40+ Male 14528
40+ Male 17034
40+ Male 14935   
40+ Male 14996
40+ Male 22146
20-29 Female 16401
20-29 Female 14216   
20-29 Female 15402   
20-29 Female 15326
20-29 Female 12047
30-39 Female 15357
30-39 Female 16771   
30-39 Female 15036   
30-39 Female 16297
30-39 Female 17636
40+ Female 17260
40+ Female 25399
40+ Female 18647   
40+ Female 15077
40+ Female 25898
")
if(!require(psych)){install.packages("car")}
library(car)
Data = read.table(textConnection(Input),header=TRUE)
model = lm(Time ~ AgeClass + Gender + AgeClass:Gender,
data=Data)
Anova(model, type="II")

Output:

Anova Table (Type II tests)

Response: Time
Sum Sq Df F value Pr(>F)
AgeClass 92086979 2 5.0998 0.01427 *
Gender 15225413 1 1.6864 0.20642
AgeClass:Gender 21042069 2 1.1653 0.32885
Residuals 216683456 24
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Since p-value for interaction=0.32885>0.05 so interaction effect is not significantly present. Moreover p-value for gender=0.20642>0.05 so Gender is not significantly present. But Age Classes are not all same since P-value=0.01427<0.05.