 # Construct the 95% confidence interval estimate of the mean wake time for a population with the treatment. Sinead Mcgee 2021-08-09 Answered

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 13 subjects had a mean wake time of 101.0 min. After​ treatment, the 13 subjects had a mean wake time of 94.6 min and a standard deviation of 24.9 min. Assume that the 13 sample values appear to be from a normally distributed population and construct a $95\mathrm{%}$ 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 101.0 min before the​ treatment? Does the drug appear to be​ effective?
Construct the $95\mathrm{%}$ confidence interval estimate of the mean wake time for a population with the treatment.
$min<\mu ​(Round to one decimal place as​ needed.)
What does the result suggest about the mean wake time of 101.0 min before the​ treatment? Does the drug appear to be​ effective?
The confidence interval ▼ does not include| includes the mean wake time of 101.0 min before the​ treatment, so the means before and after the treatment ▼ could be the same |are different. This result suggests that the drug treatment ▼ does not have | has a significant effect.

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Step 1
Let ${\mu }_{a}$ denotes the true population mean for after treatment. Let $\stackrel{―}{{x}_{b}}$ and $\stackrel{―}{{x}_{a}}$ denote the sample mean for before and after treatment data. Then, provided that $\stackrel{―}{{x}_{b}}=101.0min$ and $\stackrel{―}{{x}_{a}}=94.6min$. The sample size, $n=13$ and the sample standard deviation for after treatment data is ${s}_{a}=24.9min$.
A $95\mathrm{%}$ confidence interval for the true population mean of after treatment data is given by $\stackrel{―}{{x}_{a}}±{t}_{c}\cdot \frac{{s}_{a}}{\sqrt{n}}$. Where, ${t}_{c}$ is the critical t-score.
Step 2
The critical t-score for $95\mathrm{%}$ confidence level with 12 degree of freedom is ${t}_{c}=2.18$.
Now, substituting, $\stackrel{―}{{x}_{a}}=94.6min,{s}_{a}=24.9min$, and $n=13$ in the expression,
$\stackrel{―}{{x}_{a}}±{t}_{c}\cdot \frac{{s}_{a}}{\sqrt{n}}$
$\stackrel{―}{{x}_{a}}±{t}_{c}\cdot \frac{{s}_{a}}{\sqrt{n}}=94.6±2.18\cdot \frac{24.9}{\sqrt{13}}$
$\approx 94.6±15.1$
$=\left(79.5,109.7\right)$
As, it can be observe that the confidence interval, $79.5min<{\mu }_{a}<109.7min$ includes $\stackrel{―}{{x}_{b}}=101.0min$. Therefore, the true mean for before and after treatment could be the same. This tells that the treatment does not have the significance effect.