construct a 90​% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments.

Question

A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with the​ drug, 17 subjects had a mean wake time of 93.6 min and a standard deviation of 40.7 min. Assume that the 17 sample values appear to be from a normally distributed population and construct a

90​% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments. Does the result indicate whether the treatment is​ effective?

Find the confidence interval estimate.



Answer 1

n=17, x-bar=93.6 and s=40.7

Need to consult a t-table for t_.10 with n-1 = 16 degrees of freedom. From the table t=1.746

the 90% CI calculation is x-bar -/+ t * s / sqrt(n) = 93.6 -/+ 1.746*40.7/sqrt(17)

The limits are (93.6 - 1.746*40.7/4.123 , 93.6 + 1.746*40.7/4.123) = (76.365, 110.835)

Does the result indicate whether the treatment is​ effective? That very much depends on what the hypothesis is! If the goal was that the insomniac would get at least 60 minutes of sleep, then the data suggest success. However, if the goal was at least 120 minutes of sleep, then the data are not as promising.

Answer 2

The formula to calculate confidence interval = M ± t∗√s²/n

Given information:

1. t at 90% confidence interval = 1.745.
2. M or mean = 93.6
3. n or sample size = 17
4. s or standard deviation = 40.7

Therefore, Confidence Interval = 93.6 ±1.745∗√(40.7)²/17 = or 93.6 ± 17.23

i.e., You can be 90% confident that the population mean falls between 76.37 and 110.83