Mydream Forum Elite

Topics: 10 Posts: 331
| | 11/17/07 - 07:51 AM  
 
   
 
|   #1 |
Two experimental drugs, drug X and Y, are being evaluated for the treatment of Congestive heart failure. patients receiving drug X hv a cardiac index of 2.5 L/m2 with a 95% confidence interval of 1.5 to 3.5. patients recvng drug Y hv a cardiac index of 1.7 L/m2 with a CI of 0.7 to 2.7. A test of the significance of diff shows a p-value of 0.1. which of the foll. is the liklihood that the diff in mean cardiac index of patients recvng drug X and drug Y is due to chance ? A. 0% B. 2.5% C. 5% D. 7.5% E. 10% F. 66.7% G. 95%
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| DRFP Forum Elite

Topics: 16 Posts: 269
| | 11/17/07 - 06:12 PM  
 
   
 
|   #2 |
p-value of 0.1 so D 10%
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| SmokyWaters Forum Elite
Topics: 6 Posts: 447
| | 01/22/08 - 05:27 AM  
 
   
 
|   #3 |

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| Yulik08 Forum Newbie

Topics: 1 Posts: 9
| | 03/24/08 - 07:55 AM  
 
   
 
|   #4 |
so hard q and so easy to answer 
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| Ig F Forum Elite

Topics: 3 Posts: 436
| | 05/18/08 - 11:22 AM  
 
   
 
|   #5 |
answer is E...which is 10%....
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| Ig F Forum Elite

Topics: 3 Posts: 436
| | 05/18/08 - 11:23 AM  
 
   
 
|   #6 |
10%.....E
___________________ i m not perfect but i wanna get close to it......
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