# 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)
```