Estou com uma dúvida sobre path analysis no pacote agricolae.
No exemplo do manual do pacote se usa a correlação de Pearson para depois computar os coeficientes da análise.
Minha dúvida é se posso utliizar correlação de Spearman ao invés, já que meus dados não são normais, gostaria de saber se isso bastaria.
y <- c(77, 92, 94, 95, 79, 87, 91, 70, 65, 69, 68, 56, 71, 67, 62, 95, 91, 78,
90, 85, 83, 78, 65, 77, 87, 43, 55, 63, 66, 87, 67, 70, 69, 30, 28, 44,
67, 89, 82, 36, 60, 52, 14, 19, 51, 22, 37, 15, 86, 79, 31, 35, 15, 96,
89, 88, 92, 90, 18, 15, 13, 93, 95, 83, 95, 100, 100, 84, 90, 93, 74, 78,
73, 69, 75, 67, 68, 100, 95, 94, 94, 88, 93, 83, 73, 88, 95, 71, 69, 71,
74, 100, 71, 77, 73, 56, 52, 62, 75, 93, 88, 44, 72, 60, 27, 33, 69, 35,
44, 27, 93, 87, 56, 52, 27, 98, 94, 90, 96, 95, 26, 20, 16, 94, 97, 100)
temp <- c(26.26, 26.26, 26.26, 26.26, 26.26, 26.26, 26.26, 25.83, 25.83, 25.83, 25.83, 25.83,
25.83, 25.83, 25.83, 26.40, 26.40, 26.40, 26.40, 26.40, 26.40, 26.40, 26.40, 26.13,
26.13, 26.13, 26.13, 26.13, 26.13, 26.13, 26.13, 25.60, 25.60, 25.60, 25.60, 25.60,
25.60, 25.60, 25.60, 24.96, 24.96, 24.96, 24.96, 24.96, 24.96, 24.96, 24.96, 25.30,
25.30, 25.30, 25.30, 25.30, 25.30, 25.30, 25.30, 26.76, 26.76, 26.76, 26.76, 26.76,
26.76, 26.76, 26.76, 24.36, 24.36, 24.36, 24.36, 24.36, 24.36, 24.36, 23.80, 23.80,
23.80, 23.80, 23.80, 23.80, 23.80, 24.26, 24.26, 24.26, 24.26, 24.26, 24.26, 24.26,
24.26, 23.70, 23.70, 23.70, 23.70, 23.70, 23.70, 23.70, 23.70, 23.06, 23.06, 23.06,
23.06, 23.06, 23.06, 23.06, 23.06, 21.93, 21.93, 21.93, 21.93, 21.93, 21.93, 21.93,
21.93, 22.30, 22.30, 22.30, 22.30, 22.30, 22.30, 22.30, 22.30, 24.00, 24.00, 24.00,
24.00, 24.00, 24.00, 24.00, 24.00, 24.00)
pluv <- c(136.6, 136.6, 136.6, 136.6, 136.6, 136.6, 136.6, 136.0, 136.0, 136.0, 136.0, 136.0,
136.0, 136.0, 136.0, 112.3, 112.3, 112.3, 112.3, 112.3, 112.3, 112.3, 112.3, 65.3,
65.3, 65.3, 65.3, 65.3, 65.3, 65.3, 65.3, 66.0, 66.0, 66.0, 66.0, 66.0,
66.0, 66.0, 66.0, 65.0, 65.0, 65.0, 65.0, 65.0, 65.0, 65.0, 65.0, 53.0,
53.0, 53.0, 53.0, 53.0, 53.0, 53.0, 53.0, 53.0, 53.0, 53.0, 53.0, 53.0,
53.0, 53.0, 53.0, 254.6, 254.6, 254.6, 254.6, 254.6, 254.6, 254.6, 256.6, 256.6,
256.6, 256.6, 256.6, 256.6, 256.6, 231.3, 231.3, 231.3, 231.3, 231.3, 231.3, 231.3,
231.3, 142.6, 142.6, 142.6, 142.6, 142.6, 142.6, 142.6, 142.6, 126.6, 126.6, 126.6,
126.6, 126.6, 126.6, 126.6, 126.6, 118.0, 118.0, 118.0, 118.0, 118.0, 118.0, 118.0,
118.0, 110.6, 110.6, 110.6, 110.6, 110.6, 110.6, 110.6, 110.6, 57.6, 57.6, 57.6,
57.6, 57.6, 57.6, 57.6, 57.6, 57.6)
# shapiro-wilk
shapiro.test(y)
shapiro.test(x$temperatura) # não é normal!
shapiro.test(x$pluviosidade)
x <- data.frame(temp,pluv)
# path analysis
require(agricolae)
cor.y <- correlation(y,x)$correlation
cor.x <- correlation(x)$correlation
path.analysis(cor.x,cor.y)