Question

In a 1985 study of the relationship between contraceptive use and infertility, 89 of 283 infertile women, compared to 640 of 3833 control (fertile) women, had used an intrauterine device (IUD) at some point in their life. Use the contingency table to test for significant differences in contraceptive use patterns between the two groups. Compute a 95% CI for the difference in the proportion of women who have ever used IUDs between the infertile and fertile women. Compute the OR in favor of ever using an IUD for fertile vs. infertile women. Provide a 95% CI for the true OR corresponding to your answer. What is the relationship between your answers to questions 1 and 4? Need help with question 3 and 4

Answer 1

Answer:

In a 1985 study of the relationship between contraceptive use and infertility, 89 of 283 infertile women, compared to 640 of 3833 control (fertile) women, had used an intrauterine device (IUD) at some point in their life. Use the contingency table to test for significant differences in contraceptive use patterns between the two groups. Compute a 95% CI for the difference in the proportion of women who have ever used IUDs between the infertile and fertile women. Compute the OR in favor of ever using an IUD for fertile vs. infertile women. Provide a 95% CI for the true OR corresponding to your answer. What is the relationship between your answers to questions 1 and 4? Need help with question 3 and 4

  

	infertile	fertile
contraceptive use	89	640	729
No use	194	3193	3387
	283	3833	4116

Compute the OR in favor of ever using an IUD for fertile vs. infertile women.

OR = (89*3193)/(194*640) =2.2888

Provide a 95% CI for the true OR corresponding to your answer.

se( log(OR)) = sqrt( 1/89+1/640+1/194+1/3193) =0.1352

95% CI for log(OR) = (log(OR)-1.96* se( log(OR)), log(OR)+1.96* se( log(OR)))

95 % CI for OR:

1.7562 to 2.9830

What is the relationship:

We use a chi-square to test if an association exists. We use an odds ratio to measure or quantify the association between two variables or factors. Therefore, chi-square is a test of association and OR is a measure of association.