Construct the 95​% confidence interval estimate of the mean wake time for a population with the treatment. enter your response here min<μ<enter your response here min ​

Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment,

16

subjects had a mean wake time of

104.0

min. After​ treatment, the

16

subjects had a mean wake time of

79.4

min and a standard deviation of

24.2

min. Assume that the

16

sample values appear to be from a normally distributed population and construct a

95​%

confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of

104.0

min before the​ treatment? Does the drug appear to be​ effective?

Construct the

95​%

confidence interval estimate of the mean wake time for a population with the treatment.

enter your response here

min<μ<enter your response here

min

​(Round to one decimal place as​ needed.)

Answer 1

I can lay out the steps you should follow to solve the problem, but it does not appear to me that you have all the pieces needed. Have you conveyed all of the pieces of information given to you in the problem?

1.) Describe what is asked for in the problem:

1. Construction of a 95% confidence interval. (I'm assuming that a two-sided interval is being requested)
2. This confidence interval will have the general composition of:
3. mean +- value of test statistic*standard error for that test statistic

2.) Choose a test statistic.

1. You have a repeated measures situation. same people tested more than once
2. You do not know any of the population standard deviations, and you small sample size
3. You will want to perform a repeated measures t test

, where bar-D indicates the mean of the difference scores, s_d the standard deviation of the difference scores, and sqrt(N) the square root of the number of pairs of observations

where

3.) Calculate the test statistic

1. The mean of the difference scores = difference between the means, 104.0 - 79.4 = 24.6
2. The standard deviation of the difference scores does not appear to be given in the problem. is the standard deviation mentioned in the problem (24.2) intended to be this quantity?
3. The square root of the number of pairs of observations = sqrt(16) = 4

4.) Construct your confidence interval.

1. If the interval does NOT cross 0, then you have evidence that pre- and post-treatment means are different. (two-tail tests show differences, not direction, but you can use the direction of the difference in the means to infer improvement.)

Hope this is helpful. Please check back in if you want to provide or ask for any clarifications