Question

(1 point) LDL- low density lipoprotien - cholesterol is considered the bad cholesterol. A pharmaceutical company conducts a clinical trial on people with high levels of LDL cholesterol. The trial involves a random sample of n= 15 persons who have high levels of LDL. Each person has their LDL level measured prior to the beginning of the drug trial. Each person is then to take a new drug that is believed to lower ones LDL levels. After a period of 4-weeks, the LDL level of each person is measured again. The difference between the initial LDL level and the LDL-level four weeks after taking the drug is then computed. -0.03,-0.11, -0.82, -0.55,-0.51,-0.39, -0.34,-0.63,-0.7,-1.13,-0.88,-0.26,-0.18,-0.79,-0.74 Data from the sample, are saved in the Download .csv file. Assuming the difference in the LDL-levels is Normally distributed, find a 94% confidence interval for fi, the mean difference in the LDL-level of a person after taking the drug for four weeks. Use at least four decimals for your lower and upper bounds. To avoid Rounding errors you use use R-Studio and not Tables. - Sus

Answer 1

X -0.03 -0.11 -0.82 -0.55 -0.51 -0.39 -0.34 -0.63 -0.7 -1.13 -0.88 -0.26 -0.18 -0.79 -0.74 X - mean (X-mean2 0.5073 0.2574 0.4273 0.1826 -0.2827 0.0799 -0.0127 0.0002 0.0273 0.0007 0.1473 0.0217 0.1973 0.0389 -0.0927 0.0086 -0.1627 0.0265 -0.5927 0.3513 -0.3427 0.1174 0.2773 0.0769 0.3573 0.1277 -0.2527 0.0639 -0.2027 0.0411

Since we know that

Mean() - Σ1 ;

Where n is the number of data points
Now

I;= -8.06 21

and n = 15
This implies that

-8.06 Mean(i) = - 15 Mean(T) = -0.5373

Since we know that

Variance(82) – Eti – 7)? ti - 7) = 1.3948 n = 15 1.3948 Variance(s?) = -2 Variance(s?) = 0.0996 Standard Deviation(s) = V Variance Standard Deviation(s) = 0.3156 14

Mean ($\small \bar{x}$) = -0.5373
Sample size (n) = 15
Standard deviation (s) = 0.3156
Confidence interval(in %) = 94

ta/2n-1 = 2.046

Since we know that

Confidence interval = i ta/2.n-1

Required confidence interval =
10.3156 -0.5373 +2.0465 (-0.5373 -2.046 15 -0.5373 +2.046 0.3156

Required confidence interval = (-0.5373-0.1667, -0.5373+0.1667)
Required confidence interval = (-0.704, -0.3706)
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