Q3) Chi-Square Test for Independence (11 points) This question is an extension f
ID: 3048004 • Letter: Q
Question
Q3) Chi-Square Test for Independence (11 points)
This question is an extension from Q2 above. The researcher was also interested in whether there is a gender difference in terms of students’ choice of one of the four TAs. So she broke the data down by gender in the following table and conducted a Chi-square test for independence with a = .05.
TA
Female
Male
Total
Chang
6
9
15
Peterson
16
16
32
Hernandez
7
6
13
What are the variables in this analysis? What type of variable is each (nominal, ordinal, or continuous)?
TA
Female
Male
Total
Chang
6
9
15
Peterson
16
16
32
Hernandez
7
6
13
Explanation / Answer
The Expected value data are in the table below. Each Cell = Row total * Column Total/N. N = 60
The degrees of freedom = (r – 1) * (c -1) = (3 - 1) * (2 - 1) = 2
The Hypothesis:
H0: There is no relation between gender and student choice of the TA's.
Ha: There is a relation between gender and student choice of the TA's.
The Test Statistic: The table below gives the calculation of 2.
2test = 0.611
The Critical Value: The critical value at = 0.05, df = 2
2critical = 5.99
The Decision Rule: If 2 test is > 2 critical, then Reject H0.
The Decision: Since If 2 test (0.611) is < 2 critical (5.99), We Fail to Reject H0.
The Conclusion: There is insufficient evidence at the 95% significance level to conclude that there is a relation between gender and student choice of the TA's.
The variables here are the students gender(M,F) and the TA's. These are Nominal variables.
Expected Female Male Total Chang 7.25 7.75 15 Peterson 15.47 16.53 32 Hernandez 6.28 6.72 13 Total 29 31 60Related Questions
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