Chi-Square Test of Independence

Name________________________________ Score________________
Due 05/07/2013 Each additional day late will cost you 4 points.
SHOW ALL WORK AND DO BY HAND. YOU CAN CHECK ON CALCULATOR.
12. (20 points) A public opinion poll surveyed a simple random sample of voters. Respondents were classified by gender (male or female) and by voting preference (Republican, Democrat, or Independent). Results are shown below. | Voting Preferences | | Republican | Democrat | Independent | Row total | Male | 200 | 150 | 50 | 400 | Female | 250 | 300 | 50 | 600 | Column total | 450 | 450 | 100 | 1000 |

a) If you conduct a chi-square test of independence, what is the expected frequency count of male Independents?

b) If you conduct a chi-square test of independence, what is the expected frequency count of female Democrats?

c) If you conduct a chi-square test of independence, what is the observed count of female Independents?

d) If you conduct a chi-square test of independence, what is the expected frequency count of male Republicans?

e) If you conduct a chi-square test of independence, what is the observed count of male Independents? f) If you conduct a chi-square test of independence, what is the expected frequency count of female Republicans?
g) The table represents a

(A) 2 by 3 table

(B) 3 by 3 table

(C) 4 by 4 table

(D) 4 by 3 table

h) The null hypothesis is

(A) Gender and voting preferences are independent

(B) Gender and voting preferences are not independent

i) The degrees of freedom is ___________________

j) The chi-square statistic by hand is ________________ (show all work below for part (j))

k) The p-value from the table is ___________________

l) Therefore using a 0.05 level of significance, we (A) fail to reject the null hypothesis, insufficient evidence (B) reject the null hypothesis, sufficient