fana
Forum Newbie
Topics: 2
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Assuming the use of 2 standard deviations to establish reference interval of a test,what is the approximate percent chance that a normal person could have atleast one false -positive test result if two tests r ordered?
A-5%
B-10%
C-15%
D-20%
E-30%
Plz explain ur answer too.
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| Anil Kumar980
Forum Senior
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Its 5%.2SD means 95 percent fall in the range.2.5 r above and 2.5 r below.So false positive for 1 test is 2.5 n for 2 its 2.5+2.5 or 5.Am I right?
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| docofthebigapple
Forum Senior
Topics: 23
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2 SD is 95% and the range outside it is 5%.
The number of times the test is performed is immaterial, because the chances of the normal person getting the false positive will remain 5%, if the reference interval is 2SD.
This discussion implies to a percentage of tests going wrong in a big number and not a "specific" ratio of tests going wrong.
Let me elaborate, Its like asking, "What are the chances, tests can go wrong"?, and asking "What are the chances "HALF" the tests will go wrong"?. I hope the reader understands the difference in the two.
The chances of half the tests going wrong are half the chances of all the tests going wrong. ( The comparison will always be done to the total number of tests performed. ) The smaller the ratio considered for false positives, the smaller will be the chances. And as we near Zero...0, the chances of tests going wrong will be nearly zero.
Now, the question asks specifically the chances of 1 out of 2 giving false positives. the ratio here is 1/2 or 50% of what would be the answer that comes first in our mind.
I would like to point out that "n" is very very small in this case to reach any significant statistical conclusion. If n were large enough for a significant statistical conclusion, the chances of 1 out of every two tests performed ( only 50% of tests in consideration here ) giving false positives would be 1/2 the chances of the whole lot of tests giving false positives which as question suggests the reference range is 2SD, would be 5%. But as only hals the tests are considered here, i would suggest the correct answer for the question is 2.5%.
So chance of 1 test giving false positive result, if two are performed would be 2.5%.
Thanks for bearing it. 
Please give me messages and your views about my analysis.
Thanks once again.
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| Anil Kumar980
Forum Senior
Topics: 15
Posts: 130
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hiBigapple,I still thinkthe ans is 5.2.5+2.5
you need to add up 2 indipendent variables for the combined probabilities
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| docofthebigapple
Forum Senior
Topics: 23
Posts: 174
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would someone else come with explanations to some answer.
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| fana
Forum Newbie
Topics: 2
Posts: 6
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Act, this is a que from BSS and the answer is B (10%).But i didnt understand the answer even after reading explanation thatsy i posted this question here.Explanation given in the book is "The more test r performed on pt, the greater chance of obtaining a false -positive reslt" as u said docofthebigapple.The formula to calculate the chance of this happening is:100-(0.95n multiplied by 100) where n=no. of test performed on pt.Therefore if two test r ordered (100-0.95 2 multip by 100=10%) there is a 10% chance that one of the test will result in n outlier or false negative"
Thanx a lot both of u.
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| docofthebigapple
Forum Senior
Topics: 23
Posts: 174
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The explanation given by BSS seems inadequate.
Lets say we perform 100 tests. the answer calculated according to the formula then will be:-
100 - (.95n x 100) = formula.
value of n in a hypothetical situation be 100.
it comes to be that
100 - (.95 x 100 x 100) =
100 - 9500 =
- 9400.
Either i am getting it wrong or it has been explained vaguely.
Another thing would be that the chances of getting false positives would not increase more than 5% with increased number of tests performed if eveything else remains unchanged. It will always reach a maximum of 5% with a very large n , because 95% is the reference range for normal results.
However the absolute number of false positives will increase by increasing n.
It will be calculated as .05n.
So number of False positives is directly proportional to n.
"n" represents total number of tests performed.
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| bluedusk
Forum Elite
Topics: 35
Posts: 217
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the BSS question. Really poor wording.
The reference interval = 95% (2SD) means that given you are normal, there is a 2.5% chance your values are higher than this interval and 2.5% your values are lower.
If you define a "positive" result as higher than this interval, as Anil has, (and as I would have been inclined to), then you have a 97.5% chance of not getting a false positive. If you take it again, the chance you would escape a false positive both times is 97.5% x 97.5% = ~95%. This makes the false positive chance over two tries ~5%.
This is actually just another way of saying the BSS explanation posted here.
However, it appears clear that "positive" according to BSS means higher or lower than the reference interval. I think that kind of sucks, but then again, I suppose I will remember it during my exam. In that case, it's 95% x 95% = 90%, or about a 10% chance of at least one false positive.
* One last detail - you can't simply add the %'s, you need to multiply them because you are calculating the chances that they BOTH happen. Think of it this way - if you threw a die six times, would you be guaranteed with coming up with at least one "4"? Of course not - you wouldn't add 1/6 six times. Your chances of getting at least one "4" would be 1 - (missing a "4" all six rolls), or 1 - (5/6 x 5/6 x 5/6 x 5/6 x 5/6 x 5/6). Whatever that is.
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| bluedusk
Forum Elite
Topics: 35
Posts: 217
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Actually, the more I think about it, falling outside of the reference interval SHOULD mean "positive", higher or lower. Damn. OK, maybe it wasn't that poorly worded in BSS.
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| fana
Forum Newbie
Topics: 2
Posts: 6
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The explanation given here is exactly same as the BSS.
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| bluedusk
Forum Elite
Topics: 35
Posts: 217
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Sorry - did it help? Or was it just as confusing as the BSS explanation?
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| Dr. Hussam
Forum Junior
Topics: 5
Posts: 56
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Its B 10%
consider it this way, the chance to get one false positive from a test is 5%=0.05, now to get at least one false positive from two tests you would use the following probability rule
P(A)+P(B)-P(A and B), considering A and B are two independent groups.
so 0.05+0.05 -(0.05*0.05) =9.75% which is almost equal to 10%
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| royalgeorge
Forum Elite
Topics: 10
Posts: 266
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I agree with Dr Hussam , its explained well in the kaplan notes
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