A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects.

Question

After treatment with the​ drug, 20 subjects had a mean wake time of 90.7 min and a standard deviation of

43.7 min. Assume that the 20 sample values appear to be from a normally distributed population and construct a

98​% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments. Find the confidence interval estimate.

____min < σ < ___min

Answer 1

Hi Trinity,

This problem involves using the normal-distribution confidence interval formula that deals with samples, namely:

X - z_α/2(σ/√n) < μ < X + z_α/2(σ/√n)

After treatment with the​ drug, 20 subjects had a mean wake time of 90.7 min and a standard deviation of

43.7 min. Assume that the 20 sample values appear to be from a normally distributed population and construct a

98​% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments. Find the confidence interval estimate.

____min < σ < ___min

Given information in the problem: Confidence Interval (CI) = .98, X = 90.7, n = 20, σ = 43.7

Step 1): Find alpha (α) = 1-0.98 = 0.02, and divide this answer by two to get (_α/2) = 0.01

Step 2): Find ± z_α/2 subtracting 1.0000 - 0.0100 = .9900 and locate z-score in normal distribution table, yielding ± 2.33

Step 3): Evaluate confidence interval formula and arithmetic as follows:

X ± z_α/2(σ/√n) = 90.7 ± 2.33 (43.7/√20) = 90.7 ± 22.76786775

90.7 - 22.76786775 < μ < 90.7 + 22.76786775

67.9 min < μ < 113.5 min