A clinical trial was conducted to test the

Answered question

2022-05-02

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 25 subjects had a mean wake time of 104.0 min. After treatment, the 25 subjects had a mean wake time of 94.7 min and a standard deviation of 22.4 min. Assume that the 25 sample values appear to be from a normally distributed population and construct a 99% confidence interval estimate of the mean wake time for a population with drug treatments

 

(a) Construct the 99% confidence interval estimate of the mean wake time for a population with the treatment.

(b) What does the result suggest about the mean wake time of 104.0 min before the treatment? 

(c) Does the drug appear to be effective?

Answer & Explanation

karton

karton

Expert2022-07-07Added 613 answers

Given :

Sample size = n = 25

Sample mean = x= 94.7

Standard deviation = s = 22.4

Significance level = α = 1-0.99 = 0.01

a) The 99% confidence interval estimate of the mean wake time for a population with the treatment :

The number of degrees of freedom are df = 25 — 1 = 24, and the significance level is α=0.01.

Based on the provided information, the critical t-value for α=0.01 and df = 24 degrees of freedom is te= 2.797.

The 99% confidence for the population mean ju is computed using the following expression

CI=(X¯-te×sn,X¯+te×sn)

Therefore, based on the information provided, the 99 % confidence for the population mean μ is

CI=(94.7¯-2.797×22.425,94.7¯+2.797×22.425)

= (94.7 — 12.53, 94.7 + 12.53)

= (82.17, 107.23)
Therefore the 99% confidence interval estimate of the mean wake time for a population with the treatment is 82.2 < μ < 107.2

b) The confidence interval contain the mean wake time of 104.0 min before the treatment, so the means before and after the treatment is not different.

c) This result suggests that the drug treatment have not significant effect.

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