Explore BrainMass

# Explanation of odds ratio in a study of smoking.

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

Those who have quit smoking often return to the habit. The authors of a paper concluded that forbidding smoking in the smoker's residence was a significant predictor of the ability to abstain from smoking. They collected data, calculated statistics, and presented the following as support for their conclusion:

OR=1.95
C.I. : (1.05, 3.6)
P < 0.01

Based on these statistics, describe how the authors reached their decision. Is the conclusion they reached statistically significant? If the data had shown p<0.05 rather than p<0.01, would this have been stronger or weaker support for their conclusion?

https://brainmass.com/statistics/probability/explanation-odds-ratio-study-smoking-12892

## SOLUTION This solution is FREE courtesy of BrainMass!

Odds can be defined as the probability of an event not happening divided by the probability of the event happening. If the probability of an event happening is p then the odds associated with the event are:

Odds = (1-p) / p

In this study, samples were taken from the two groups and estimates of the odds were calculated on each of them.

Odds _1 = (proportion not returning to smoking) / (proportion returning to smoking) within the group where smoking is prohibited in the residence.

Odds _2 = (proportion not returning to smoking)/(proportion returning to smoking) within the group where smoking is allowed in the residence.

The test statistic in this study, sometimes called the "odds ratio" or "OR", is the ratio of the odds from the two groups

OR = (odds_1)/(odds_2)

If there is no difference between the two groups, the odds of the populations from which the samples were selected are the same, making the odds ratio of the populations equal to 1. In this case, we would expect odds ratio we calculated from the samples (our test statistic) to differ from 1 only by what is reasonably due to sampling error.

The odds ratio from the samples in this test is 1.95. The sample odds of staying off cigarettes if smoking is forbidden are almost twice the sample odds of staying off cigarettes if smoking is allowed.

Is this seemingly large value of the sample odds ratio statistically significant? If so, we can conclude that the odds of the populations from which the samples were selected are different. We make this decision by using the CI or "confidence interval" and its corresponding p or "p-value".

The confidence interval for the population odds ratio is (1.05, 3.6). Since this test was done at the .01 level of significance, this can be assumed to be a 99% confidence interval. We can be 99% confident that the population odds ratio lies between the endpoints of the interval. These are the "believable" values of the population odds ratio given the information in our samples. Notice that the value of 1 is not one of the "believable" values since the entire interval is greater than 1. Therefore under the assumption that returning to smoking is independent of environment, we have seen an event that happens less than 1% of the time (p-value<0.01). Since such events are rare, the researchers can conclude that house rules affect the decision to return to smoking (the assumption of independence is rejected and we can conclude dependence exists between returning to smoking and environment). We can conclude this with 99% confidence.

If the p-value had been (p < 0.05) rather than (p<0.01), the results of the test would have been less powerful evidence that the 2 variables are related. Under the assumption of independence, the observed event could only be said to occur less than 5% of the time rather than 1% of the time. The researchers would reach the same conclusion but with only 95% confidence.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!