Los paquetes
library("tidyverse")
There were 38 warnings (use warnings() to see them)
Cargo datos para crear mi df
summary(df2)
Edad Sexo c e o ceod
Min. :1.000 Femenino :58 Min. : 0.000 Min. :0.0000 Min. :0.0000 Min. : 0.000
1st Qu.:2.000 Masculino:55 1st Qu.: 0.000 1st Qu.:0.0000 1st Qu.:0.0000 1st Qu.: 0.000
Median :3.000 Median : 0.000 Median :0.0000 Median :0.0000 Median : 1.000
Mean :3.248 Mean : 1.965 Mean :0.0354 Mean :0.3805 Mean : 2.381
3rd Qu.:4.000 3rd Qu.: 4.000 3rd Qu.:0.0000 3rd Qu.:0.0000 3rd Qu.: 5.000
Max. :5.000 Max. :12.000 Max. :1.0000 Max. :6.0000 Max. :13.000
Visualizo mis datos en graficos

grafico para edad y componente c del ceod

Ahora mis analisis estadisticos:Existe diferencia entre sexo y ceod?
t.test(df2$ceod~df2$Sexo)
Welch Two Sample t-test
data: df2$ceod by df2$Sexo
t = -1.2654, df = 102.87, p-value = 0.2086
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.91623 0.42344
sample estimates:
mean in group Femenino mean in group Masculino
2.017241 2.763636
Existe diferencia entre sexo y componente c del copd?
t.test(df2$c~df2$Sexo)
Welch Two Sample t-test
data: df2$c by df2$Sexo
t = -1.5269, df = 100.83, p-value = 0.1299
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.7874456 0.2325867
sample estimates:
mean in group Femenino mean in group Masculino
1.586207 2.363636
Existe diferencia entre ceod y edad??
anova
Call:
aov(formula = df2$ceod ~ df2$Edad)
Terms:
df2$Edad Residuals
Sum of Squares 379.7070 712.9302
Deg. of Freedom 1 111
Residual standard error: 2.534323
Estimated effects may be unbalanced
summary(anova)
Df Sum Sq Mean Sq F value Pr(>F)
df2$Edad 1 379.7 379.7 59.12 6.45e-12 ***
Residuals 111 712.9 6.4
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
ajusto modelo para prueba POST-HOC
pairwise.t.test(df2$ceod,df2$Edad)
Pairwise comparisons using t tests with pooled SD
data: df2$ceod and df2$Edad
1 2 3 4
2 0.87831 - - -
3 0.32321 0.05806 - -
4 0.15874 0.00269 0.87831 -
5 0.00108 1.1e-10 0.00061 0.00317
P value adjustment method: holm
ajusto modelo para prueba POST-HOC bonferroni
pairwise.t.test(df2$ceod,df2$Edad, p.adjust = "bonferroni")
Pairwise comparisons using t tests with pooled SD
data: df2$ceod and df2$Edad
1 2 3 4
2 1.00000 - - -
3 1.00000 0.11612 - -
4 0.39685 0.00385 1.00000 -
5 0.00135 1.1e-10 0.00068 0.00528
P value adjustment method: bonferroni
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