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.

1.What is the expected frequency of smoker and ever being divorced?

2.What is the expected frequency of smoker and never being divorced?
3.What is the expected frequency of non-smoker and ever being divorced?

4.What is the expected frequency of non-smoker and never being divorced?

5.To test independence between smoking habits and being divorced, what is the value of chi-square test statistic? [Answer to 3 decimal places.]What is the expected frequency of non-smoker and never being divorced?

Ever Divorced? Smoke? Yes No Yes 268 274 No 407 458

Explanation / Answer

1.What is the expected frequency of smoker and ever being divorced?

2.What is the expected frequency of smoker and never being divorced?282

542*732/1407=282
3.What is the expected frequency of non-smoker and ever being divorced?415

865*675/1407=415

4.What is the expected frequency of non-smoker and never being divorced?450

5.To test independence between smoking habits and being divorced, what is the value of chi-square test statistic?

here chi-square=sum((O-E)2/E)=0.765 with (r-1)(c-1)=(2-1)(2-1)=1 df

since critical chi-square(0.05,1)=3.84 is more than calcuated chi-square=0.765 , so we fail to reject the null hypothesis

here null hypothesis H0: smoking habits and being divorced are independent

alternate hypothesis Ha: smoking habits and being divorced are not independent

Expected frequency Ever divorced Smoke? Yes No Yes 260 282 No 415 450

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