To test the efficacy of a new cholesterollowering medication

To test the efficacy of a new cholesterol-lowering medication, 10 people are selected at random. Each has their LDL levels measured (shown below as Before), then take the medicine for 10 weeks, and then has their LDL levels measured again (After).

Give a 96.1% confidence interval for BA , the difference between LDL levels before and after taking the medication.

Confidence Interval =
at 96.1% confidence.

Give your answer as an open interval, in the form (A,B) where A is the lower bound and B is the upper bound.

Solution

Consider:

Getting the mean and standard deviation of the last column,

X = 3.9
s = 18.25407108

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.0195          
X = sample mean =    3.9          
t(alpha/2) = critical t for the confidence interval =    2.413881427          
s = sample standard deviation =    18.25407108          
n = sample size =    10          
df = n - 1 =    9          
Thus,              
Margin of Error E =    13.93399565          
Lower bound =    -10.03399565          
Upper bound =    17.83399565          
              
Thus, the confidence interval is              
              
(   -10.03399565   ,   17.83399565   ) [ANSWER]