QUESTION 8 (10 points) Testing independence of two categorical/nominal variables

Question

QUESTION 8 (10 points) Testing independence of two categorical/nominal variables

Based on the article "California's Likely Voters" published by PPIC.ORG in August 2015 and a hypothetical simple random sample of 1000 Californians, we have some observed frequencies in terms of individuals regarding political ideology and voting behavior as follows:

At the level of significance = 0.10, you will need to test the null hypothesis that voting behavior is independent of political ideology. Follow the steps as follows:
(a)[1] Clearly state the null and alternative hypotheses H0 and H1.
(b)[1] Specify appropriate equations for the 2 statistic and degrees of freedom df. (c)[1] Calculate marginal probabilities given sample data.

(d)[1] Based on marginal probabilities, construct the table of expected frequencies.
(e)[2] Based on expected and observed frequencies, calculate the 2 statistic and its df. (f)[1] Given = 0.10 and df, use CHISQ.INV(*,*) to look up for the righttail 2* critical value. (g)[2] Compare 2 statistic with 2* critical to reject or accept H0. Make a verbal statement. (h)[1] Find the pvalue of the test. Comment on the pvalue.
Hint: Round off to 5 decimal places in calculations. Refer to some Excel lookups:

Registered & likely voter

Registered & infrequent voter

Unregistered adult

Liberal

200

68

59

Moderate

176

64

59

Conservative

236

68

Registered & likely voter

Registered & infrequent voter

Unregistered adult

Liberal

200

68

59

Moderate

176

64

59

Conservative

236

68

70

Explanation / Answer

answer of (1)

null hypothesis Ho:voting behavior is independent of political ideology

alternative hypothesis H1:voting behavior is not independent of political ideology

answer of (2) chi-squre = sum((O-E)2/E), O is observed value, E is expected value

degree of freedom will be (R-1)(C-1) where R is number of row anc C is number of column

answer of remaining part

chi-square (.1,4)=7.779 (tabulated)

since calculated chi-squre is less than the tabulated so we accept null hypothesis at alpha=0.1

p-value=0.7827 since p-value is more than the alpha=.1 so we accept null hypothesis

RL RI UA L 200 68 59 327 M 176 64 59 299 C 236 68 70 374 612 200 188 1000 observed (O) Expected(E) (O-E) (O-E)2 (O-E)2/E E(200) 200 200.124 -0.124 0.015376 7.68324E-05 E(68) 68 65.4 2.6 6.76 0.103363914 E(59) 59 61.476 -2.476 6.130576 0.099723079 E(176) 176 182.988 -6.988 48.832144 0.266859816 E(64) 64 59.8 4.2 17.64 0.294983278 E(59) 59 56.212 2.788 7.772944 0.138279086 E(236) 236 228.888 7.112 50.580544 0.220983817 E(68) 68 74.8 -6.8 46.24 0.618181818 E(70) 70 70.312 -0.312 0.097344 0.001384458 calculated chi-square= 1.743836099 df= 4

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