303

And would get as a result:

Two Sample t-test

data: GroupA and GroupB

t = 1.0377, df = 14, p-value = 0.317

alternative hypothesis: true difference in means is not equal to

095 percent confidence interval:

-1.066768 3.066768

sample estimates:

mean of x mean of y

10 9

A p-value of 0.317 was found, which means that we cannot reject the null hypothesis.

Thus, the new therapy does not show any significant improvement with regard to the aver­

age duration of illness.

We can now use a chi-square test (test for independence) to determine whether the two

variables are independent, that is, whether the number of airbags is independent of the

type of car.

Analogous to the two previous examples, we must also formulate the test hypothesis

here (p-value < 0.05). The null hypothesis (H0) would be: Both variables are independent.

The alternative hypothesis H1: The number of airbags depends on the type of car.

In R we would use the following script (clipboard loads data from cache, just copy the

table to do this):

> table<-read.table(clipboard)

> chisq.test(table)

And would get as a result:

Pearson's Chi-squared test

data: table

Example 19.3

19.3

Table 19.3  Number of airbags in different car types

Compact

Large

Medium

Small

Sports

Van

Driver/passenger

2

4

7

0

3

0

Driver

9

7

11

5

8

3

None

5

0

4

16

3

6

19.6  Introduction to Programming (Meta Tutorial)