1
$\begingroup$

We have been given the following for home work and I have calculated the OR however I am confused with the next 3 questions on percentages. Can someone help and explain and also what is the significance of only women in the study?

You have brought to the attention of the Alcohol and Tobacco control unit, a published case control study that reported an association between smoking and SLE (Systemic Lupus Erythematous). You suggested that given the association the director should publish an informative sheet for women. The results of the study are depicted in the table below:

After the Director read the paper he asked you to answer the following questions so an information sheet can be published:

a) What are the odds of SLE among all women included in the study, women exposed to smoking and women not exposed to smoking? What is the odds ratio of smoking compared to not smoking with regards to SLE?

My answer to (a):

  • The odds of SLE among all women $(a+c)/(a+b+c+d) = 163/636 = 0.256$
  • The odds ratio of cases of SLE who are exposed to smoking: $a/c = 51/112 =0.455$
  • The odds ratio for women without SLE who are exposed to smoking: $b/d= 92/381= 0.241$
    $$OR= (a/c)/(b/d) = ad/bc = (51 \times 381) / (92 \times 112) = 19431 /10304 = 1.886$$
  • The odds of developing SLE in women who smoke is 1.9 times higher than those who do not smoke.

These next question are the ones I am having trouble doing and understanding:

b) What percent of the cases of SLE among all women who smoke is due to smoking?

c) What percent of the total cases of SLE among all women is due to smoking? (Assume that controls are population based and have a 18.7% smoking rate)

d) How does question 5b differ from question c?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

Sorry. I am not converting odds ratios to percentages I was talking about the percentage of the variation among the women that is attributable to smoking. The odds ratio is $(p_1/(1-p_1))/(p_2/(1-p_2))$. I guess you are interested in $p_1/(p_1+p_2)$. First remember that you are computing estimates of the parameters and not the parameters themselve. An odds ratio of 1 means $p_1=p_2$ or $p_1/(p_1+p_2)=1/2$. An odds ratio of $1.9$ means $p_1/(1-p_1)= 1.9 p_2/(1-p_2)$ or $p_1(1-p_2)=1.9 p_2(1-p_1)$ or $$p_1-p_1 p_2 =1.9 p_2-1.9 p_1 p_2$$ This implies $p_1=1.9 p_2-.9 p_1 p_2$ or $p_1= 1.9 p_2(1-.9 p_1)$ or $$p_1/(p_1+p_2)=1.9 p_2(1-.9 p_1)/[p_2(2-.9 p_1)]= 1.9 (1-.9 p_1)/(2-.9 p_1)]$$ This depends on $p_1$. If $p_1=0$ this is $1.9/2=0.95$ and if $p_1=1$ it is $1.9 \cdot .1/1.1=0.19/1.1=0.1727$. for any $p_1>0$ the ratio ranges from $0.95$ to $0.173$. However given the complete contingency table you can calculate $p_1$.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ I've TeXed your equations, verbatim. Hopefully this looks right. $\endgroup$
    – Macro
    Commented Jun 11, 2012 at 2:16
  • 1
    $\begingroup$ Can you put this into simpler term for my old brain please? $\endgroup$
    – Trying Hard
    Commented Jun 11, 2012 at 2:20
  • $\begingroup$ @TryingHard I just did some simple algebra to show that the odds ratio of 1.8 does not imply a specific proportion for p1/(p1+p2). It actually ranges between 0.9 and 0.3. Note that I corrected an initial error. $\endgroup$
    – Michael R. Chernick
    Commented Jun 11, 2012 at 2:35
  • 1
    $\begingroup$ Can you also help with part b) and c) please? $\endgroup$
    – Trying Hard
    Commented Jun 11, 2012 at 2:43
  • $\begingroup$ My answer above does answer part b. There is no unique value for this proportion based on the odds ratio alone. The estimate ranges from 0.95 to 0.173 depending on the value for p1. $\endgroup$
    – Michael R. Chernick
    Commented Jun 11, 2012 at 14:12
0
$\begingroup$

If you have a linear model that includes smoking yes or no along with other factors then the partial $R^2$ for smoking tells the percentage of variance due to smoking when other factors are held fixed.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Would the person that downvoted this response explain their reasoning? $\endgroup$
    – Michael R. Chernick
    Commented Jun 11, 2012 at 1:17
  • $\begingroup$ So then would the following be the correct anser for question b)??An odds ratio of 1.0 means there is no increase of decrease in risk of developing SLE. Therefore for an odds ratio of 1.80, there is an increase in risk for developing SLE due to smoking of 0.90 or 90/100 over the general population risk. $\endgroup$
    – Trying Hard
    Commented Jun 11, 2012 at 1:22
  • 2
    $\begingroup$ I did not downvote, Michael, but I am inclined to convert this reply into a somewhat mysterious comment to the original question. I am puzzled why you are invoking a "linear model" and "a partial $R^2$" when the question appears to concern a 2 x 2 contingency table and, by virtue of its homework tag, it begs us for elementary and clearly explained responses. $\endgroup$
    – whuber ♦
    Commented Jun 11, 2012 at 13:36
  • $\begingroup$ I recognize that the question asked for a conversion from odds ratios to proportions and addressed that in my second answer. My feeling here was that this conversion idea is not a very good one and I think that address the question in terms of variance explained is a better way to look at things. The contingency table expresses counts for SLE as a function of smoking or not and therefore can be looked at as a simple ANOVA model. I don't see what is unclear about my response. R square provides an an estimated of percentage of variance explained by the factor smoking (yes or no). $\endgroup$
    – Michael R. Chernick
    Commented Jun 11, 2012 at 14:00
  • $\begingroup$ I am really sorry Michael, but I agree with "whuber" comments as I dont understand your answer however it is probably that this subject is really new to me and I need "baby" talk to help me understand. So sorry. $\endgroup$
    – Trying Hard
    Commented Jun 12, 2012 at 9:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.