Chi-Square Test of Independence

H0: The two variables are independentHa: The two variables are dependent

Test Statistic: =

Rejection Region: Reject Ho if >

NOTE: This test is valid as long as every expected cell count is at least 5.

Example Opinion polls often provide information on how different groups’ opinions vary on controversial issues. A random sample of 102 registered voters was taken from the Supervisor of Election’s roll. Each of the registered voters was asked the following two questions:

1. What is your political party affiliation?2. Are you in favor of increased arms spending?

The results are summarized in the table below.Party

Opinion Democrat Republican NoneFavor 16 21 11No favor 24 17 13

Conduct test to determine if opinion and party affiliation are related in the population. Use =.05.

IN MINITAB:ENTERING THE DATA:

Party Opinion CountDemocrat Favor 16Democrat Not In Favor 24Republican Favor 21Republican Not In Favor 17None Favor 11None Not In Favor 13

COMMANDS: STAT> TABLES> CROSS-TABULATION > CLASSIFICATION VARIABLES PARTY OPINION

FREQUENCIES ARE IN COUNTCLICK ON CHI-SQUARE ANALYSIS – ABOVE AND EXPECTED COUNT

OUTPUT:Favor Not In F All

Democrat 16 24 40 18.82 21.18 40.00 None 11 13 24 11.29 12.71 24.00 Republic 21 17 38 17.88 20.12 38.00 All 48 54 102 48.00 54.00 102.00 Chi-Square = 1.841, DF = 2, P-Value = 0.398

H0: Party affiliation and Opinion regarding increased arms spending are independentHa: Party affiliation and Opinion regarding increased arms spending are dependent (related)Calculate test statistic: p-value

Since p-value > , we fail to reject H0. The sample data do not provide sufficient evidence to conclude that party affiliation and opinion regarding increased arms spending are related in the population.

Chi-Square Goodness-of-Fit Test

H0: 1 = hypothesized population proportion for category 1

.

.

.

k = hypothesized population proportion for category k

Ha: H0 is not true, so at least one of the category proportions differs from the corresponding hypothesized value.

Test Statistic: =

Rejection Region: Reject H0 if >

NOTE: This test is valid as long as every expected cell count is at least 5.

Example In previous presidential elections in a given locality, 50% of the registered voters were Republicans, 40% were Democrats, and 10% were registered as independents. Prior to the upcoming election, a random sample of 200 registered voters showed that 90 were registered as Republicans, 80 as Democrats, and 30 as independents. Is there sufficient evidence to conclude that the distribution of registered voters is different from that in previous election years. Use =.01.

In Minitab Spreadsheet:

C1 C290 0.580 0.430 0.1

COMMANDS: STAT> TABLES> Chi-Square Goodness-of-Fit Test > Observed counts C1 > specific proportions C2

Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: C1

Test ContributionCategory Observed Proportion Expected to Chi-Sq1 90 0.5 100 12 80 0.4 80 03 30 0.1 20 5

N DF Chi-Sq P-Value200 2 6 0.050

R = the proportion of all of this year’s registered voters that are Republicans D = the proportion of all of this year’s registered voters that are DemocratsI = the proportion of all of this year’s registered voters that are Independents

H0: R = .5D = .4I = .1

Ha: H0 is not true, so at least one of the category proportions differs from the corresponding hypothesized value.

=.01

Calculate test statistic: p-value

Since p-value > , we fail to reject H0. The sample data do not provide sufficient evidence to conclude that the distribution of registered voters in the given locality is different from that in previous election years.