Deep Singh

I'm sure this is really simple but I'm confusing myself.

Mr X works in a dept where they all do the same job/task. This job has a known complication. This complication is measured.

For the dept as a whole the complication rate is 5% ie 5% of all jobs have this complication.

Mr Xs own results are 2% ie only 2% of his own jobs have this complication.

How much less likely is Mr X to have this complication compared to the avge?

Or whats the best way to express this to make Mr X look as good as possible


There is a certain amount of time allowed to do this job in, but sometimes it does take longer ie over run

For the dept as a whole the avge number of jobs that over run are 27%

The number of jobs that over run in Mr Xs hands are 10%.

Whats the best way to express this to make Mr X look as good as possible.


Bit long winded, but thanks for looking

PeteBrant

Well you look at it the other way round - Ie. -

Percentage of jobs completed on time:

Department average: 72%

Mr X Average: 90%


And for complication


Percentage of Jobs resulting in complication

Departmental avergae 5%

Mr X 2%


I.e. shift the figures round to suit what you are saying - In this example you want a high number for jobs on time - and a low number for jobs with complications.

Deep Singh

I've modified the real scenario for the sake of privacy.

What I really want to be able to say is ' In Mr Xs hands a complication is y% less likely and the job is z% less likely to be delayed'

Chip Sengravy

You don't want to big MrX's abilities up as a percentage, do it as a multiplier.

i.e, Mr X is 2.5 times less likely to encounter complication, and 1.25 times quicker at the task.


I suppose to put that back to percentage, it's 150% and 25% over and above the avg.



Who is this bloke anyway? Sound's like a bit of a smart **** to me. I'd spread some rhumors about him quicksharp. He's making you all look inept

PeteBrant

Well if the departmental average is 5% and MrX's average is 2% then he is 40% less likely to have the complication

(i.e you are finding what 2 is of 5 as a percentage (2/5*100)

If the departmental average for going over limit is 27% and Mr X is 10%, then Mr X is 37% less likely to go over limit

(and here you are finding what 10 is of 27 as a percentage 10/27*100)

alcazar

iTrader: (2)

My thought exactly chip, get the b*gger the sack!

Alcazar

what would scooby do

I was going to say the easy answer to this problem is to outsource

john banks

40% and 37% as likely
60% and 63% less likely
Or you could say 2.5 and 2.7 times less likely.

Second or third one sounds better. Did they teach you epidemiology or were you too busy trying nitrous and halothane Deep?

Deep Singh

Thanks guys. I did think that was the answer but had to be sure or it would have been very embarrasing!