R Notebook

Integrantes: Santiago Guerrero Rondon- Valentina Valle Velasco

Datos en vectores

colombia = c(0.45,0.41,0.43,0.46,0.39,0.44,0.48,0.42,0.44,0.48,0.50,0.47,0.44,0.52)
ocarina = c(0.28,0.25,0.32,0.34,0.36,0.40,0.33,0.36,0.39,0.41,0.37,0.42,0.41)  

Media de los datos

desc_col = (psych::describe(colombia))
desc_oca = (psych::describe(ocarina))
desc_col;desc_oca

Prueba de igualdad de varianzas

\[H_o: \sigma_{colombia} = \sigma_{ocarina}\\ H_a: \sigma_{colombia} \neq \sigma_{ocarina}\]

pr_var = var.test(colombia,ocarina); pr_var

    F test to compare two variances

data:  colombia and ocarina
F = 0.46714, num df = 13, denom df = 12, p-value = 0.1879
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
 0.1442133 1.4729897
sample estimates:
ratio of variances 
         0.4671449 

ifelse(pr_var$p.value<0.05, 'varianzas desiguales', 'varianzas iguales')

[1] "varianzas iguales"

\[Ho: med_{col} = med_{oca}\\ Ha: med_{col} \neq med_{oca}\]

med_col = mean(colombia); med_oca = mean(ocarina);med_col;med_oca

[1] 0.4521429
[1] 0.3569231

pru_t = t.test(colombia,ocarina); pru_t

    Welch Two Sample t-test

data:  colombia and ocarina
t = 5.5114, df = 21.018, p-value = 1.805e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 0.05929236 0.13114720
sample estimates:
mean of x mean of y 
0.4521429 0.3569231 

ifelse(pru_t$p.value<0.05, 'Rechazo Ho y medias desiguales', 'NO rechazo Ho y medias iguales')

[1] "Rechazo Ho y medias desiguales"

Conclusion

A partir de los resultados que se pueden apreciar a traves de la prueba t- student se puede concluir que el cultivar colombia muestra una mejor conductancia estomatica frente al deficit de riego, comparandolo con el cultivar ocarina. Esto se evidencia mediante el rechazo de la prueba de hipotesis y de esta forma se toma la hipotesis alternativa: medias desiguales.