To test the efficacy of a new cholesterol-lowering medication, 10 people are sel

Question

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.

Subject Before After 1 127 138 2 122 85 3 121 132 4 110 119 5 159 174 6 183 176 7 144 152 8 133 112 9 152 126 10 146 144

Explanation / Answer

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]

Subject Before After xB - xA 1 127 138 -11 2 122 85 37 3 121 132 -11 4 110 119 -9 5 159 174 -15 6 183 176 7 7 144 152 -8 8 133 112 21 9 152 126 26 10 146 144 2

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