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#1
05-19-2010
 USMLE Forums Scout Steps History: Not yet Posts: 73 Threads: 7 Thanked 40 Times in 21 Posts Reputation: 60
95% Confidence Interval

A study is conducted to evaluate intelligence quotient (IQ) scores for patients with various types of schizophrenia. Twenty patients were dropped from the sample because they could not complete two or more portions of the test. Four additional patients refused to take the test. Results for the 100 patients who completed the test showed an average IQ of 110 with a standard deviation of 20. Which of the following is the best estimate of the 95% confidence interval for this sample?
A. 70 to 130
B. 70 to 150
C. 85 to 115
D. 90 to 130
E. 105 to 115
F. 106 to 114

#2
05-19-2010
 USMLE Forums Master Steps History: Step 1 Only Posts: 919 Threads: 134 Thanked 1,325 Times in 364 Posts Reputation: 1348

i have NO clue. (really have to revise my biostatistics again!!)

if i take a wild guess its C. whats the answer & explanation?
#3
05-19-2010
 USMLE Forums Guru Steps History: --- Posts: 487 Threads: 69 Thanked 297 Times in 134 Posts Reputation: 317
F??

is it F..??
mean = 110
for 95% CI Z score is 2
24 patients were dropped from the sample,i think it not relavent in answering this Q so...N is 100...
now the calculation is easy..
CI= 110+/-2(20/10)
= 110+/-4
= 106-114

#4
05-20-2010
 USMLE Forums Scout Steps History: Not yet Posts: 73 Threads: 7 Thanked 40 Times in 21 Posts Reputation: 60
Choice F is correct

The formula for the confidence interval of the mean (CI) is: CI = ± Z x SD*s/ n
The mean is given as 110. To achieve a 95% confidence interval use a Z-score of 1.96 (or 2.0. to make the calculation easier). The standard deviation (SD) is given as 20 and the sample size is given as 100 Inserting these values into the formula: CI = 110 ± 4 or a range of 106 to 114.

khushboo you are right.

 Tags Biostatistics-Epidemiology, Step-1-Questions

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