A study was undertaken relating political partyidentification to residential are
ID: 3172096 • Letter: A
Question
A study was undertaken relating political partyidentification to residential area. The following sample data were observed. Determine if this is a goodness-of-fit test or a test for independence, and conduct the chi-square analysis. (30 points) Residential ar Party affiliation Suburbs City Country 40 60 10 Democrat Republican 40 20 30 Independent 20 20 10 a. State the null and alternate hypotheses in sentences, referring tothe specificvariables being evaluated b. Calculate X2 c Write up these results in anAPA style research report. B sure toinclude all appropriate information in your written statement.Explanation / Answer
Solution:
Here, we have to use the chi square test for independence.
Part a
The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: The party affiliation and residential area are independent from each other.
Alternative hypothesis: Ha: The party affiliation and residential area are not independent from each other.
Part b
Here, we have to calculate the test statistic Chi square. The formula for test statistic is given as below:
Chi square = [(O – E)^2/E]
Where, O is the observed frequencies and E is the expected frequencies
E = (Row total * Column total) / Grand total
The calculation tables are given as below:
Chi-Square Test
Observed Frequencies
Residential area
Party affiliation
City
Suburbs
Country
Total
Democrat
40
60
10
110
Republican
40
20
30
90
Independent
20
20
10
50
Total
100
100
50
250
Expected Frequencies
Residential area
Party affiliation
City
Suburbs
Country
Total
Democrat
44
44
22
110
Republican
36
36
18
90
Independent
20
20
10
50
Total
100
100
50
250
Calculations
(O – E)
-4
16
-12
4
-16
12
0
0
0
(O – E)^2/E
0.363636
5.818182
6.545455
0.444444
7.111111
8
0
0
0
Sum = 28.28283
Test statistic = Chi square = [(O – E)^2/E] = 28.28283
Part c
For the above test, we have
Number of rows = r = 3
Number of columns = c = 3
Degrees of freedom = (r – 1)(c – 1) = (3 – 1)(3 – 1) = 2*2 = 4
Critical value = 9.487729
Test statistic value = Chi square = 28.28283
P-value = 0.00001
We assume level of significance or alpha value as 1% or 0.01.
P-value < Alpha value 0.01
So, we reject the null hypothesis at 1% level of significance.
This means we reject the null hypothesis that the party affiliation and residential area are independent from each other. This means we conclude that there is sufficient evidence that the party affiliation and residential area are not independent from each other.
Chi-Square Test
Observed Frequencies
Residential area
Party affiliation
City
Suburbs
Country
Total
Democrat
40
60
10
110
Republican
40
20
30
90
Independent
20
20
10
50
Total
100
100
50
250
Expected Frequencies
Residential area
Party affiliation
City
Suburbs
Country
Total
Democrat
44
44
22
110
Republican
36
36
18
90
Independent
20
20
10
50
Total
100
100
50
250
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