1 2 3 4

10

15

20

trat

resp

Dilan Vergara. Actividad 21-11-2014

BANANAS

> #tratamiento empaques

> pok<-c(14.1,9.4,9.3,10.9,6.3);pok

[1] 14.1 9.4 9.3 10.9 6.3

> pol<-c(10.0,7.1,14.2,14.8,18.9);pol

[1] 10.0 7.1 14.2 14.8 18.9

> saco<-c(18.3,17.5,14.4,16.2,23.1);saco

[1] 18.3 17.5 14.4 16.2 23.1

> nada<-c(21,18,17.3,16.8,21.9)nada

nada<-c(21,18,17.3,16.8,21.9)nada

> spaq<-c(21.0,18.0,17.3,16.8,21.9);spaq

[1] 21.0 18.0 17.3 16.8 21.9

> resp<-c(pok,pol,saco,spaq)

> mean(resp)

[1] 14.975

> var(resp)

[1] 23.48092

> sd(resp)

[1] 4.845712

> boxplot(resp)

> trat<-c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4);trat

[1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4

> datos<-data.frame(trat,resp);datos

trat resp

1 1 14.1

2 1 9.4

3 1 9.3

4 1 10.9

5 1 6.3

6 2 10.0

7 2 7.1

8 2 14.2

9 2 14.8

10 2 18.9

11 3 18.3

12 3 17.5

13 3 14.4

14 3 16.2

15 3 23.1

16 4 21.0

17 4 18.0

18 4 17.3

19 4 16.8

20 4 21.9

> tapply(resp,trat,summary)

$`1`

Min. 1st Qu. Median Mean 3rd Qu. Max.

6.3 9.3 9.4 10.0 10.9 14.1

$`2`

Min. 1st Qu. Median Mean 3rd Qu. Max.

7.1 10.0 14.2 13.0 14.8 18.9

$`3`

Min. 1st Qu. Median Mean 3rd Qu. Max.

14.4 16.2 17.5 17.9 18.3 23.1

$`4`

Min. 1st Qu. Median Mean 3rd Qu. Max.

16.8 17.3 18.0 19.0 21.0 21.9

> tapply(resp,trat,var)

1 2 3 4

8.040 20.825 10.625 5.285

> tapply(resp,trat,mean)

1 2 3 4

10.0 13.0 17.9 19.0

> t.test(pok,mu=15)

One Sample t-test

data: pok

t = -3.943, df = 4, p-value = 0.01692

alternative hypothesis: true mean is not equal to 15

95 percent confidence interval:

6.479275 13.520725

sample estimates:

mean of x

10

> t.test(pol,mu=15)

One Sample t-test

data: pol

t = -0.98, df = 4, p-value = 0.3826

alternative hypothesis: true mean is not equal to 15

95 percent confidence interval:

7.333739 18.666261

sample estimates:

mean of x

13

> t.test(saco,mu=15)

One Sample t-test

data: saco

t = 1.9894, df = 4, p-value = 0.1175

alternative hypothesis: true mean is not equal to 15

95 percent confidence interval:

13.85267 21.94733

sample estimates:

mean of x

17.9

> t.test(spaq,mu=15)

One Sample t-test

data: spaq

t = 3.8907, df = 4, p-value = 0.01768

alternative hypothesis: true mean is not equal to 15

95 percent confidence interval:

16.14552 21.85448

sample estimates:

mean of x

19

> #analisis de varianza

> trat<-factor(trat)

> mod<-aov(resp~trat,data=datos);mod

Call:

aov(formula = resp ~ trat, data = datos)

Terms:

trat Residuals

Sum of Squares 254.4025 191.7350

Deg. of Freedom 1 18

Residual standard error: 3.263732

Estimated effects may be unbalanced

> anova(mod)

Analysis of Variance Table

Response: resp

Df Sum Sq Mean Sq F value Pr(>F)

trat 1 254.40 254.403 23.883 0.0001186 ***

Residuals 18 191.74 10.652

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

> #boxplot de los tratamientos

> plot(resp~trat)

>

>

> #Kruskal Wallis

> comparacion<-kruskal(resp,trat,group=TRUE, main="Empaques")

Study: Empaques

Kruskal-Wallis test's

Ties or no Ties

Value: 11.48

degrees of freedom: 3

Pvalue chisq : 0.00939431

trat, means of the ranks

resp replication

1 4.2 5

2 8.4 5

3 14.0 5

4 15.4 5

t-Student: 2.119905

Alpha : 0.05

LSD : 5.437877

Means with the same letter are not significantly different

Groups, Treatments and mean of the ranks

a 4 15.4

a 3 14

b 2 8.4

b 1 4.2