Suppose two hospitals are willing to participate in an exper

Suppose two hospitals are willing to participate in an experiment to test a new treatment, and both hospitals agree to include 1100 patients in the study. Because the researchers who are conducting the experiment are on the staff of hospital A, they decide to perform the majority of cases with the new procedure. They randomly assign 1000 patients to the new treatment, with the remaining 100 receiving the standard treatment. Hospital B, which is a bit reluctant to try something new on too many patients, agrees to randomly assign 100 patients to the new treatment, leaving 1000 to receive the standard treatment. The following table displays the results: Which treatment was more successful in hospital A? Justify your answer with relevant percents or measures of risk. Which treatment was more successful in hospital B? Justify your answer with relevant percents or measures of risk. Combine the data from the two hospitals into a single contingency table that shows the relationship between treatment and outcome. Which treatment has the higher survival rate in this combined table? Justify your answer with relevant percents or measures of risk. Explain how this exercise is an example of Simpson\'s paradox.

Solution

6.a) for hospital A we have

let p1=proportion of petients in hospital A survived on taking the standard treatment.

so p1=5/100

let p2=proportion of petients in hospital A survived on taking the new treatment.

so p2=100/1000

so the measure of relative risk is r=p1/p2=(5/100)/(100/1000)=0.5<1

so p1<p2

so the proportion of petients in hospital A survived on taking the standard treatment is less than proportion of petients in hospital A survived on taking the new treatment..

hence in hospital A new treatment was more successful.   [answer]

b) for hospital B we have

let p1=proportion of petients in hospital B survived on taking the standard treatment.

so p1=500/1000

let p2=proportion of petients in hospital B survived on taking the new treatment.

so p2=95/100

so the measure of relative risk is r=p1/p2=(500/1000)/(95/100)=0.52<1

so p1<p2

so the proportion of petients in hospital B survived on taking the standard treatment is less than proportion of petients in hospital B survived on taking the new treatment..

hence in hospital B new treatment was more successful.   [answer]

c) on comining the two hospitals A and B we have

let p1=proportion of petients survived on taking the standard treatment.

so p1=505/1100

let p2=proportion of petients survived on taking the new treatment.

so p2=195/1100

so the measure of relative risk is r=p1/p2=(505/1100)/(195/1100)=2.589>1

so p1>p2

so the proportion of petients survived on taking the standard treatment is greater than proportion of petients survived on taking the new treatment..

hence in the combined table standard treatment has higher survival rate. [answer]

d) this is indeed an example of simpson\'s paradox.

because when the factor hospital was ignored that is when we had the data as a whole we find that the proportion of petients survived on taking the standard treatment is greater than proportion of petients survived on taking the new treatment..hence the conclusion was that standard treatment has higher survival rate.

but when we looked into each hospital individually we find that in both the hospitals new treatment has higher survival rate.

so when the groups are combined the trend reverses.

hence it is an example of simpson\'s paradox [answer]

treatment	survive	die	total
standard	5	95	100
new	100	900	1000
total	105	995	1100