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	No
Yes	275	213
No	386	556

1. Among those who smoked, what percentage has ever been divorced? [Answer to 2 decimal places. Do not type % symbol in the box.] %

2. Among those who has ever been divorced, what percentage smoked? [Answer to 2 decimal places. Do not type % symbol in the box.] %

Next we intend to test if smoking habits and being divorced are related or not.
3. What is the expected frequency of smoker and ever being divorced? [Answer to 2 decimal places.]

4. What is the expected frequency of smoker and never being divorced? [Answer to 2 decimal places.]

5. What is the expected frequency of non-smoker and ever being divorced? [Answer to 2 decimal places.]

6. What is the expected frequency of non-smoker and never being divorced? [Answer to 2 decimal places.]

7. 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:

Null hypothesis: smoking habit and ever being divorced are not related,
against
Alternative hypothesis: smoking habit and ever being divorced are related.

8. If the p-value associated to the ch-square test-statistics is 0 and the level of significance is 5%, what will be your conclusion?
A. Do not reject null hypothesis
B. Not enough information to reach a decision
C. Reject null hypothesis

Answer 1

(1)
From the given data, the following Table is calculated:

	Ever divorced - Yes	Ever divorced - No	Total
Smoke - Yes	275	213	488
Smoke - No	386	556	942
Total	661	769	1430

Among those who smoked, percentage that has ever been divorced = (275/488) X 100 = 56.25%

So,

answer is:

56.25

(2)

Among those who have been divorced, percentage that smoked = (275/661) X 100 = 41.60

So,

Answer is:

41.60

(3)

Expected Frequency of Smoker AND ever being divorced = (661 X 488)/ 1430 = 225.57

So,

Answer is:

225.57

(4)

Expected Frequency of Smoker AND never being divorced = (769 X 488)/ 1430 = 262.43

So,

Answer is:

262.43

(5)

Expected Frequency of Non-Smoker AND ever being divorced = (661 X 942)/ 1430 = 435.43

So,

Answer is:

435.43

(6)

Expected Frequency of Non-Smoker AND never being divorced = (769 X 942)/ 1430 = 506.57

So,

Answer is:

506.57

(7)

Chi - square test statistic is got as follows:

Observed (O)	Expected (E)	(O - E)2/E
275	225.57	10.8318
213	262.43	9.3104
386	435.43	5.6113
556	506.57	4.8233
Total = =		30.574

Chi - square test statistic = 30.574

(8)

Correct option:

C. Reject null hypothesis