Subjects with preexisting cardiovascular symptoms who were receiving subitramine
ID: 3223581 • Letter: S
Question
Subjects with preexisting cardiovascular symptoms who were receiving subitramine, an appetite suppressant, were found to be at increased risk of cardiovascular events while taking the drug. The study included 9899 overweight or obese subjects with preexisting cardiovascular disease and/ or type 2 diabetes. The subjects were randomly assigned to subitramine (4952 subjects) or a placebo (4947 subjects) in a double-blind fashion. The primary outcome measured was the occurrence of any of the following events: nonfatal myocardial infarction or stroke, resuscitation after cardiac arrest, or cardiovascular death. The primary outcome was observed in 579 subjects in the subitramine group and 478 subjects in the placebo group.
Find the proportion of subjects (±±0.0001) experiencing the primary outcome for both the subitramine and placebo groups. For the subitramine group pˆ1=p^1= 0.1170______________ For the control (placebo) group pˆ2=p^2= 0.1221 is ________________
Use the PLUS four method to find a 96% confidence interval for the difference between the proportions of subitramine and placebo subjects who experienced the primary outcome. The confidence interval (±±0.0001) is from____ to_____
Explanation / Answer
Find the proportion of subjects (±±0.0001) experiencing the primary outcome for both the subitramine and placebo groups.
For the subitramine group pˆ1=p^1= 579/4952 = 0.1169
For the control (placebo) group pˆ2=p^2= 478/4947 = 0.0966
Use the PLUS four method to find a 96% confidence interval for the difference between the proportions of subitramine and placebo subjects who experienced the primary outcome.
For the subitramine group p~1=p~1= (x1 +1)/ (n1 +2) = (579+1)/(4952+2) = 0.1171
For the control (placebo) group p~2=p~2= (x2 +1)/ (n2 +2) = (478+1)/(4947+2) = 0.0968
confidence interval = (p~1 - p~2) +- z SEp˜1p˜2
where SEp˜1p˜2 = sqrt [ p~1 ( 1- p~1)/(n1 + 2) + p~2 ( 1- p~2)/(n2 + 2) ]
= sqrt [ 0.1171 * 0.8829/ 4954 + 0.0968 * 0.9032/ 4949] = 0.0062
confidence interval = (p~1 - p~2) +- z SEp˜1p˜2
= (0.1171 - 0.0968) +- 2.0537 * 0.0062
= 0.0203 - 0.012733
= ( 0.0076, 0.0330)
The confidence interval (±±0.0001) is from 0.0076 to 0.0330.