Question

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 ?=16 n = 16 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. −0.69,−0.19,−0.46,−0.67,−0.41,0.19,−0.37,−0.7,−0.56,−0.35,−1,−0.26,−0.15,−0.46,−0.85,−0.89 −0.69,−0.19,−0.46,−0.67,−0.41,0.19,−0.37,−0.7,−0.56,−0.35,−1,−0.26,−0.15,−0.46,−0.85,−0.89 Data from the sample, are saved in the Download .csv file. Assuming the difference in the LDL-levels is Normally distributed, find a 98% confidence interval for ? μ , 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. equation editorEquation Editor ≤?≤ ≤ μ ≤ equation editorEquation Editor

Answer 1

= (-0.69 + (-0.19) + (-0.46) + (-0.67) + (-0.41) + 0.19 + (-0.37) + (-0.7) + (-0.56) + (-0.35) + (-1) + (-0.26) + (-0.15) + (-0.46) + (-0.85) + (-0.89))/16 = -0.48875

s_d = sqrt(((-0.69 + 0.48875)^2 + (-0.19 + 0.48875)^2 + (-0.46 + 0.48875)^2 + (-0.67 + 0.48875)^2 + (-0.41 + 0.48875)^2 + (0.19 + 0.48875)^2 + (-0.37 + 0.48875)^2 + (-0.7 + 0.48875)^2 + (-0.56 + 0.48875)^2 + (-0.35 + 0.48875)^2 + (-1 + 0.48875)^2 + (-0.26 + 0.48875)^2 + (-0.15 + 0.48875)^2 + (-0.46 + 0.48875)^2 + (-0.85 + 0.48875)^2 + (-0.89 + 0.48875)^2)/15) = 0.30921

At 98% confidence interval the critical value is t^* = 2.602

The 98% confidence interval for is

+/- t^* * s_d/

= -0.48875 +/- 2.602 * 0.30921/sqrt(16)

= -0.48875 +/- 0.20114

= -0.68989, -0.28761