confidence interval

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).

Subject. before. after

1. 113. 123
2. 169. 157
3. 163. 163
4. 142. 106
5. 167. 172
6. 145. 136
7. 154. 114
8. 196. 179
9. 133. 143
10. 147. 135

Give a 90% confidence interval for the difference between LDL levels before and after taking the medication.

any help would be appreciated!!!

Answer 1

To find the 90% confidence interval for the difference between before and after taking the new medication, we first need to calculate the mean difference between before and after.

Therefore, the mean difference is calculated as follows:

Sum of (After - Before)= (123-113) + (157-169) + (163-163) + (106-142) + (172-167) + (136-145) + (114-154) + (179-196) + (143-133) + (135-147) = -5

Mean Difference = -5/10 = -0.5

Next, we need to calculate the standard deviation of the differences.

The standard deviation of the differences can be found by calculating the variance and then taking its square root.

The variance of the differences can be found by calculating the sum of the squared differences and dividing by the sample size (10).

Therefore (After - Before)2= (123-113)2 + (157-169)2 + (163-163)2 + (106-142)2 + (172-167)2 + (136-145)2 + (114-154)2 + (179-196)2 + (143-133)2 + (135-147)2 = 364

Variance = 364/10 = 36.4

Standard Deviation = √36.4 = 6.030

The 90% confidence interval is calculated using the following formula:

CI = (Mean Difference) ± (1.645 X Standard Deviation)

CI = (-0.5) ± (1.645 X 6.030)

CI = -0.5 ± 9.923

Therefore, the 90% confidence interval for the difference between LDL levels before and after taking the medication is (-10.423, 9.423).

Answer 2

This is a paired test, so you look at the t-distribution

the confidence interval is (x-bar - t .95,9 s/sqrt(10), x-bar + t .95,9 s/sqrt(10))

t.95,9 = in excel =T.INV(0.95,9) = 1.83311293265624

x-bar is average(sample) = -10.1

s is =STDEV.S(sample) = 17.5084106772843

Thus, (x-bar - t .95,9 s/sqrt(10), x-bar + t .95,9 s/sqrt(10)) =

(-20.2492966436975, 0.0492966436974882)

You might want to round this to (-20.2,0)