Transcript AP Statistics Section 14.2 A

AP Statistics Section 14.2 A
The two-sample z procedures of chapter 13
allowed us to compare the proportions of
successes in two groups (either two
populations or two treatment groups in an
experiment). We need a new statistical test
if we want to compare more than two
groups.
A contingency table (or two-way frequency
table) is a table in which frequencies
correspond to two variables. One variable
categorizes rows and the other columns.
Discussed earlier in section 4.2.
Example 14.1: Market researchers know that
background music can influence the mood and
purchasing behavior of customers. One study in
a supermarket in Northern Ireland compared
three treatments: no music, French accordion
music and Italian string music. Under each
condition, the researchers recorded the
numbers of bottles of French, Italian and other
wine purchased. Here is a table that summarizes
the data:
Music
Wine
Chosen
None
French
Italian
Total
French
30
39
30
99
Italian
11
1
19
31
Other
43
35
35
113
Total
84
75
84
243
Example: Calculate the
conditional distribution
(in proportions) of wine
chosen for each music type.
No music playing:
French :
30
 .357
84
11
 .131
84
Other :
Italian :
1
 .013
75
Other :
Italian :
19
 .226
84
Other :
Italian :
43
 .512
84
French music playing:
French :
39
 .52
75
35
 .467
75
Italian music playing:
French :
30
 .357
84
35
 .417
84
The types of wine chosen seems to
differ considerably across the three
music treatments. The key
question of course is this: “Are the
differences due to random
variation or are the differences
statistically significant?”
Section 14.2 presents two types of hypothesis
testing based on contingency tables.
Tests of homogeneity are used to determine
whether different populations have the same
proportion of some characteristic.
Tests of independence are used to determine
whether a contingency table’s row variable is
independent of its column variable.
While both types of tests use the same basic
methods from section 14.1, the questions these
two tests answer are different. A test for
homogeneity tests whether the distribution of a
categorical variable is the same for each of
several populations or treatments. A test for
independence tests whether two categorical
variables are associated in some population of
interest.
Test Statistic:
2 where E =
(row total)(column total)


O

E
2
 
E
grand total
The degrees of freedom equal ___________________________
(# rows - 1)(# of columns - 1)
Conditions:
Data must come from independent SRS’s of the populations of
interest.
All expected cell counts are greater than 1 and no more than
20% are less than 5
Use a  test to compare the
distribution of wines selected for
each type of music.
2
Hypothesis:
The populations of interest are customers at a Northern
Ireland supermarket when no music is playing, when French
music is playing and when Italian music is playing.
H 0 : Distributions of wine selected is the same for each music type :
H a : Distributions of wine selected are not all the same
Conditions:
Not unreasonable to view the data as random samples of
the population.
All expected counts are greater than 5, the smallest
being 9.57.
Seems reasonable to assume sales are independent.
Must also assume N  10n for each population
since sampling w/o replacement.
Calculations:
2
(
30

34
.
22
)
2 
     18.28
34.22
D of F  (3 - 1)(3 - 1)  4
P - value  .001
18.28
TI83/84 : Input data in a matrix : 2nd x -1 EDIT
STATS TESTS C :  2  Test
The expected values are computed for you and stored
in the given matrix.
Conclusions:
Very difficult to word interpretation of p-value,
so
Our p - value of .001 is less than any common significance level,
so we reject the H 0 and conclude that the type of music played
has an effect on wine sales.