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
None French Italian Total
French 30 39 30 99
Wine Italian 11 1 19 31
Chosen Other 43 35 35 113
Total 84 75 84 243
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.
Both types of tests use the same basic methods from section 14.1. Test Statistic: where E = We find one primary test statistic by finding the sum of the test statistics for each cell in the tableThe degrees of freedom equal ___________________________ 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
E
EO 22
totalgrand
al)column tot)( totalrow(
1)- columns of 1)(# - rows (#
Use a test to compare the distribution of wines selected for
each type of music.
2
Hypothesis: The populations of interest are _______________________ and
____________________
H0:__________________________________________
Ha:__________________________________________
sother wine andItalian French,
music no andItalian French,
typemusiceach for same theis selected wineof onsdistributi
same theallnot are selected wineof onsdistributi
Conditions:
SRS.an as data the view toleunreasonabNot
9.57. being
smallest the5,an greater th are counts expected All
t.replacemen w/osampling since10n N assume alsoMust
t.independen are sales assume toreasonable Seems
Calculations:
28.1822.34
)22.3430( 22
28.18
41) - 1)(3-(3 F of D
.001 value-P
Test2:C TESTS STATS
matrix ain dataInput :TI83/84
Conclusions:
sales. on wineeffect an has
played music of type that theconclude and H reject the weso
level, cesignificancommon any than less is .001 of value-pOur
0
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