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 | 290 | 204 |
No | 436 | 455 |
Among those who smoked, what percentage has ever been divorced? [Answer to 2 decimal places. Do not type % symbol in the box.] %
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Among those who has ever been divorced, what percentage smoked? [Answer to 2 decimal places. Do not type % symbol in the box.] %
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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.]
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What is the expected frequency of smoker and never being divorced? [Answer to 2 decimal places.]
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What is the expected frequency of non-smoker and ever being divorced? [Answer to 2 decimal places.]
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What is the expected frequency of non-smoker and never being divorced? [Answer to 2 decimal places.]
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To test independence between smoking habits and being divorced, what is the value of chi-square test statistic? [Answer to 3 decimal places.]
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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.
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?
Do not reject null hypothesis
Not enough information to reach a decision
Reject null hypothesis
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