A clinical trial was conducted to test the effectiveness of a drug for treating

Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before? treatment, 17 subjects had a mean wake time of 105.0 min. After? treatment, the 17 subjects had a mean wake time of 99.8 min and a standard deviation of 43.5 min. Assume that the 17 sample values appear to be from a normally distributed population and construct a 90?% 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 105.0 min before the? treatment? Does the drug appear to be? effective?

Explanation / Answer

TRADITIONAL METHOD
given that,
sample mean, x =99.8
standard deviation, s =43.5
sample size, n =17
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 43.5/ sqrt ( 17) )
= 10.55
II.
margin of error = t ?/2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, ? = 0.1
from standard normal table, two tailed value of |t ?/2| with n-1 = 16 d.f is 1.746
margin of error = 1.746 * 10.55
= 18.421
III.
CI = x ± margin of error
confidence interval = [ 99.8 ± 18.421 ]
= [ 81.379 , 118.221 ]
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DIRECT METHOD
given that,
sample mean, x =99.8
standard deviation, s =43.5
sample size, n =17
level of significance, ? = 0.1
from standard normal table, two tailed value of |t ?/2| with n-1 = 16 d.f is 1.746
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 99.8 ± t a/2 ( 43.5/ Sqrt ( 17) ]
= [ 99.8-(1.746 * 10.55) , 99.8+(1.746 * 10.55) ]
= [ 81.379 , 118.221 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 81.379 , 118.221 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

the drug does not appear to be? effective since before value is lies in the interval

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