> jackals <- read.table("http://stat.ethz.ch/Teaching/Datasets/jackals.dat", header=T)
> jackals
     M   W
1  120 110
2  107 111
3  110 107
4  116 108
5  114 110
6  111 105
7  113 107
8  117 106
9  114 111
10 112 111
> t.test(jackals$M, jackals$W)

	Welch Two Sample t-test

data:  jackals$M and jackals$W 
t = 3.4843, df = 14.894, p-value = 0.00336
alternative hypothesis: true difference in means is not equal to 0 
95 percent confidence interval:
 1.861895 7.738105 
sample estimates:
mean of x mean of y 
    113.4     108.6 

> t.test(jackals$M, jackals$W, var.equal=T)

	Two Sample t-test

data:  jackals$M and jackals$W 
t = 3.4843, df = 18, p-value = 0.002647
alternative hypothesis: true difference in means is not equal to 0 
95 percent confidence interval:
 1.905773 7.694227 
sample estimates:
mean of x mean of y 
    113.4     108.6 

> # Was soll die Ausgabe
> # 95 percent confidence interval:
> # 1.905773 7.694227 
> # Woher kommen die Werte?
> # Wir hatten 2.101 (bzw. 1.734 einseitig) als Grenze
> wilcox.test(jackals$M, jackals$W)

	Wilcoxon rank sum test with continuity correction

data:  jackals$M and jackals$W 
W = 87.5, p-value = 0.004845
alternative hypothesis: true mu is not equal to 0 

Warnmeldung:
cannot compute exact p-value with ties in: wilcox.test.default(jackals$M, jackals$W) 
> # Was bedeutet die Warnung?
> # e)
> # erübrigt sich da H_A immer true, würden jedoch Wilcoxon-Test, da dies nach Bühlmann
> # der beste Test sei (Beweis durch Autorität;-), Skript Seite: 47).
>