Question

For the fish mortality data set, use an appropriate ANOVA design to determine whether age affects proportional mortality while accounting for variation in mortality due to life history strategy. If age has a significant influence on sunfish mortality, see if you can determine which age results in a different mortality rate.

MORTALITY OF A SUNFISH AFFECTED BY LIFE HISTORY STRATEGY AND AGE

% MORTALITY: 38, 42, 14, 41, 41, 16, 36, 39, 18, 32, 36, 15, 28, 33, 17

LIFE HISTORY STRATEGY: 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5

AGE: 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3

Answer 1

Use the following R Code:

mortality=c(38,42,14,41,41,16,36,39,18,32,36,15,28,33,17)
lhs=factor(rep(1:5,rep(3,5)))
age=factor(rep(c(1,2,3),5))
res.aov=aov(mortality~lhs+age)
summary(res.aov)

Output:

Df Sum Sq Mean Sq F value Pr(>F)
lhs 4 92.9 23.2 2.45 0.131
age 2 1440.1 720.1 75.93 6.27e-06 ***
Residuals 8 75.9 9.5

Thus we can see that p-value for age is significant. Now as age is significant we need to conduct a Tukey's HSD Test to check which level of age results in different mortality rate.

Use the following R code:

diffage=TukeyHSD(res.aov,which = "age")
diffage

Output:

Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = mortality ~ lhs + age)

$`age`
diff lwr upr p adj
2-1 3.2 -2.365296 8.765296 0.2836958
3-1 -19.0 -24.565296 -13.434704 0.0000267
3-2 -22.2 -27.765296 -16.634704 0.0000083

Now confidence interval for 2-1 contains 0 and is not significant. Hence level 1 and 2 don't differ among themselves, but 3-1 and 3-2 are significant. So, 3rd level of age results in a different mortality rate.