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
1 132 114
2 174 175
3 147 140
4 179 152
5 122 100
6 192 158
7 125 98
8 119 93
9 200 191
10 173 139
Give a 93.1% confidence interval for μB−μA, the difference between LDL levels before and after taking the medication.

Confidence Interval =
at 93.1% 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.

Answer 1

Answer on Question #44740 – Math - Statistics and Probability

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

Give a 93.1% confidence interval for $\mu B - \mu A$, the difference between LDL levels before and after taking the medication.

Confidence Interval = at 93.1% 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

This is a "paired t-test", so we need to take the difference between the columns Before and After:

Subject Difference

Next, we find the mean and standard deviation of the "differences" column:

$$\bar{x} = \frac{18 - 1 + 7 + 27 + 22 + 34 + 27 + 26 + 9 + 34}{10} = 20.3$$ $$s = 11.85$$ $$n = 10$$

93.1% with 9 d.f. has $t = 2.064$.

$$CI = 20.3 \pm \frac{(2.064)(11.85)}{\sqrt{10}}$$ $$CI = (12.566; 28.034).$$

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