# Difference between revisions of "TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Bura)/Übungen 2019W/8.2"

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load('waitingtimes2.Rdata') | load('waitingtimes2.Rdata') | ||

par(mfrow=c(2,1)) | par(mfrow=c(2,1)) | ||

− | + | x = wt[[1]] | |

− | + | y = wt[[2]] | |

+ | hist(x) | ||

+ | hist(y) | ||

</syntaxhighlight> | </syntaxhighlight> | ||

− | b) | + | b) |

− | c) | + | <syntaxhighlight lang=r> |

+ | semx = sd(x)^2/length(x) | ||

+ | semy = sd(y)^2/length(y) | ||

+ | |||

+ | t = (mean(x) - mean(y))/sqrt(semx + semy) | ||

+ | t | ||

+ | |||

+ | pnorm(t)*2 | ||

+ | </syntaxhighlight> | ||

+ | |||

+ | c) | ||

+ | <syntaxhighlight lang=r> | ||

+ | t.test(x,y) | ||

+ | </syntaxhighlight> |

## Revision as of 12:17, 3 December 2019

- Two-sample t-test using normal approximation

Messages are frequently sent from a sender to either receiver 1 or receiver 2. For both receivers, several times for the transfer were measured (in seconds) and stored in the file `waitingtimes2.Rdata`

.

- (a) Plot both data sets. Is their distribution approximately bell-shaped?
- (b) Test the null-hypothesis of equal mean transfer times for both receivers on the 1%-level with a two sample t-test (using the normal approximation).
- (c) Compare your result to the output of
`t.test()`

## Lösungsvorschlag

a)

```
load('waitingtimes2.Rdata')
par(mfrow=c(2,1))
x = wt[[1]]
y = wt[[2]]
hist(x)
hist(y)
```

b)

```
semx = sd(x)^2/length(x)
semy = sd(y)^2/length(y)
t = (mean(x) - mean(y))/sqrt(semx + semy)
t
pnorm(t)*2
```

c)

```
t.test(x,y)
```