SOLUTION: Chi-square test of independence Diploma-- No Diploma-- Total Male: 56-- 32-- 88 Female: 62-- 41-- 103 Total:

Question 1048478: Chi-square test of independence
Diploma-- No Diploma-- Total
Male: 56-- 32-- 88
Female: 62-- 41-- 103
Total: 118-- 73-- 191
a.Have the assumptions for this test been met? Why or why not?
b.State the null and alternative hypotheses for this test.
c. Calculate the test statistic for this test. Explain what this test statistic represents.
d. What is the p-value? Explain what this p-value represents
e. What is the conclusion for this test at the 0.05 level of significance. Are these variables dependent/associated? Why or why not?
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
Diploma-- No Diploma-- Total
Male: 56-- 32-- 88
Female: 62-- 41-- 103
Total: 118-- 73-- 191
a. Yes, random sampling, categorical variables and expected frequency counts > 5
|
b.
Ho : There is not a relationship between gender and whether or not someone is a college graduate (they are independent).
Ha: There is a relationship between gender and whether or not someone is a college graduate (they are dependent).
c.Apply Chi-square test for Independence
(using the chi-square test for independence to determine whether gender is related to getting a diploma)
E1,1 = (88 * 118) / 191 = 54.4, E1,2 = (88 * 73) / 191 = 33.6
E2,1 = (103 * 118) / 191 = 63.6, E2,2 = (103 * 73) / 191 = 39.4
X^2 = (56- 54.4)^2/54.4 + (32- 33.6)^2/33.6 + + (62 -63.6)^2/63.6 + (41 - 39.4)^2/39.4
x^2 = .2282 = .23
d. p-value is used to access the probability of X^2 having 1 degree of freedom is more extreme than .23
DF is (r - 1) * (c - 1) = (2 - 1) * (2 - 1) = 1 (two rows, 2 columns)
P(X^2 > .23) = 1 - .37 = .63
e. 0.05 level of significance
.63 > .05, accept Ho
gender is NOT related to whether or not someone is a college graduate(they are independent).

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