| Author | 5 Posts |
vnty
Forum Senior
Topics: 24
Posts: 86
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Two experimental drugs, drug X and drug Y, are being evaluated for treatment of congestive heart failure. patients receiving drug X have cardiac index of 2.5 L/m2 with a 95% confidence interval of 1.5 to 3.5. Patients receiving drug Y have cardiac index of 1.7 L/m2 with a confidence interval of 0.7 to 2.7. A test of the significance of diferrence shows a p-value of 0.1. Which of the following is the likelyhood that the difference of the cardiac index of the patients receiving drug X and drug Y due to chance?
A. 0%
B. 2.5%
C. 5%
D. 7.5%
E. 10%
F. 66.7%
G. 95%
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| sfk
Forum Elite
Topics: 43
Posts: 294
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E. 10%
___________________
Forum Elite, But Step 2 Newbie....
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| vnty
Forum Senior
Topics: 24
Posts: 86
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Could you explain why???
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| test123
Forum Newbie
Topics: 5
Posts: 16
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If E is the answer, this can be the possible explanation,
The p (Probability) value is the chance of a type 1 error occurring. If a p value is less than 0.05, then it means that there is less than 5% chance that the data will show something that is not really there. I.e. if p = 0.05 that means that 5 in 100 or 1 in 20 are by chance (A p value equal to or less than 0.05 is generally considered statistically significant.).
Here p-value = 0.1 that means 10 in 100 patients or 1 in 10 patients (10%) shows something by chance.
so, what is the answer?.
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| test123
Forum Newbie
Topics: 5
Posts: 16
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I was going through the previous MCQ's and found a similar question where simple explanations were given.
look into this link
http://www.prep4usmle.com/forum/thread/42230
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