Question

How did they get that percentage?

Calculate the relative risk of diabetes for those who fractured their hips compared with those who didn't

\frac{ Diabetes.fractured.hip }{ Diabetes.no.hip.fracture }=\frac{ \frac{ 12 }{ 67 } }{ \frac{ 1488 }{ 9933 } }=\frac{ 0.1791 }{ 0.1498 }=1.19


A person who has had a hip fracture is 1.2 times more likely to have diabetes than a person who does not have a fractured hip.

The answer in the book wrote 20% .. and I dont know where they got 20% from.

Answer 1

\[\frac{ Diabetes.fractured.hip }{ Diabetes.no.hip.fracture }=\frac{ \frac{ 12 }{ 67 } }{ \frac{ 1488 }{ 9933 } }=\frac{ 0.1791 }{ 0.1498 }=1.19\]

Answer 2

i'm not sure...
but saying a person with a hip fracture getting diabetes is 1.2 times the other possibility is the same as saying chance of getting diabetes is 20% more if you nave a hip fracture

Answer 3

*have

Answer 4

how did u work out that 20%

Answer 5

if A= 1.2B
then A/B = 1.2
(A/B) in percentage is 120%

Answer 6

what it means is: if B was 100, then A is 120
so the difference is 20%

Answer 7

i know...its not very convincing. i'm not too sure either