Biostat Que: Ans if you can plz - ebola123

[ArchivalUser]

A new treatment for elevated cholesterol is piloted on a sample of 100 men, ages 45–59 with total serum cholesterol in the range of 260–299 mg/dL at entry. Following 3 months on the medication, the mean cholesterol for the treatment group was 250 mg/dL with a standard deviation of 20 mg/ dL. What is the 95% confidence interval on the mean for this study?
A. 210–290 mg/dL
B. 230–270 mg/dL
C. 246–254 mg/dL
D. 248–252 mg/dL
E. 249–251 mg/dL

I chose A because 95%CI is 2SD of the mean therefore it should be 250+/-(2*20)= 210-290. To my surprise however that option is wrong!!!! Could anyone plz help ans this with explanation.

[ArchivalUser]

CI = mean ± Z X SE


Cal Standard Error*/SE of difference btw means.
The SE equals the SD (σ / square root of the sample size [N]
The SE equals the SD (σ 20 / square root of the sample size [N]100 -> 20/10 -> 2 that's your SE.
"mean cholesterol for the treatment group was 250mg/dl" z score ± mean
CI = mean ± Z X SE at 95% as 2
CI = 250 ± 2 x SE
CI= 250 -/+ 4 ->
Lower CI -> 246
Upper CI-> 254

The End.

[ArchivalUser]

Thanks Cardio!!!!! this makes more sense now. I didn`t think you have to calculate the SE. So does this apply to all CI questions or there are some exceptions???

[ArchivalUser]

The Wechsler Adult Intelligence Scale–Revised (WAIS-R) is a standardized IQ test with a mean of 100 and a standard deviation of 15. A person with an IQ of 115 is at what percentile of IQ?
A. 50th
B. 68th
C. 84th
D. 95th
E. 99th

Ans= B???
My thinking is that 115 is just 1SD of the mean which corresponds to 68th percentile, Right?

[ArchivalUser]

NO, if mean is 100 for those 115 line fall to R in curve -> a mean +1SD. And if mean is 100 like me has 85 is line fall to L in curve -> a mean – 1SD.
The MEAN divided by 50% to the -> R & 50% to the -> L
Rule out the NEG side where I land and take on IQ was her than mean->1/2 OR 50% of 68% (1SD to R -> 34% & 1SD to L get another 34% of population)
http://www.oocities.org/athens/acropolis/7186/normal.gif
Now, you need one more step 68% of population (btw -1SD to +1SD)
The IQ 50%/area below the mean on L + 34 = 84 1SD => 84%
Can't exp simpler than that

[ArchivalUser]

I think I kind of get it but not completely sure. So lets say that the mean is 150 instead of 100 and the SD is 20. the question is asking what the percentile will be for a person with an IQ of 190.

so its gonna be 50% of the mean lies on the right
for an IQ of 190 is 2SD of the mean in both directions but we`re only interested in half of that which is on the right.
therefore 95/2=47.5
final ans is 50+47.5=97.5%

Did i get it right?