The following two-way table of counts summarizes whether respondents smoked or n

Question

The following two-way table of counts summarizes whether respondents smoked or not and whether they have had ever divorced or not for persons who had ever been married Ever Divorced? Smoke?Yes 278 435 No 210 452 Yes No Among those who smoked, what percentage has ever been divorced? [Answer to 2 decimal places. Do not type % symbol in the box.] Submit Answer Tries 0/3 Among those who has ever been divorced, what percentage smoked? [Answer to 2 decimal places. Do not type % symbo/ in the box.] Submit Answer Tries 0/3 Next we intend to test if smoking habits and being divorced are related or not. What is the expected frequency of smoker and ever being divorced? [Answer to 2 decimal places.] Submit Answer Tries 0/3 What is the expected frequency of smoker and never being divorced? [Answer to 2 decimal places.] What is the expected frequency of non-smoker and ever being divorced? [Answer to 2 decimal places.] What is the expected frequency of non-smoker and never being divorced? [Answer to 2 decimal places.] To test independence between smoking habits and being divorced, what is the value of chi-square test statistic? [Answer to 3 decimal places. Suppose we are testing Submit Answer Tries 0/3 Submit Answer Tries 0/3 Submit Answer Tries 0/3 Submit Answer Tries 0/3 Null hypothesis: smoking habit and ever being divorced are not related against Alternative hypothesis: smoking habit and ever being divorced are related If the p-value associated to the ch-square test-statistics is 0.005 and the level of significance is 5%, what will be your conclusion? Do not reject null hypothesis Reject null hypothesis O Not enough information to reach a decision Submit Answer Tries 0/3

Explanation / Answer

here as we know expected value =row total*colun total/grand total

from above; applying chi square test

1) among those who smoked ; % divorced=(278/488)*100=56.97%

2)among those who divorced ; % smoked=(278/713)*100=39.99%

3) expected frequency=713*488/1375=235.05

4)  expected frequency=234.95

5) expected frequency=459.95

6) expected frequency=427.05

7) test statistic chi square =7.921

8)

as p value is less than 0.05 level; reject null hypothesis

Observed OBS Yes No Total Yes 278 210 488 No 435 452 887 Total 713 662 1375 Expected Ei=?row*?column/?total Yes No Total Yes 253.050 234.950 488 No 459.950 427.050 887 Total 713 662 1375 chi square =(Oi-Ei)2/Ei Yes No Total Yes 2.46 2.65 5.11 No 1.35 1.458 2.81 Total 3.81 4.11 7.921

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