A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with the drug, 12 subjects had a mean wake time of 95.6 min and a standard deviation of 42.1 min. Assume that the 12 sample values appear to be from a normally distributed population and construct a 90% confidence interval estimate to the standard deviation of the wake times for a population with the drug treatments. Does the result indicate whether the treatment is effective?
Solution
Given that the sample size is $n=12$, sample standard deviation is $s=42.1$. We wish to construct a $90$ % confidence interval for population standard deviation $\sigma$.
Step 1 Specify the confidence level $(1-\alpha)$
Confidence level is $1-\alpha = 0.9$. Thus, the level of significance is $\alpha = 0.1$.
Step 2 Given information
Given that sample size $n=12$ and sample standard deviation $s =42.1$.
Step 3 Specify the formula
$100(1-\alpha)$% confidence interval estimate of population standard deviation $\sigma$ is
$$ \begin{aligned} \bigg(\sqrt{\frac{(n-1)s^2}{\chi^2_{(\alpha/2,n-1)}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{(1-\alpha/2,n-1)}}}\bigg) \end{aligned} $$
where $\chi^2_{(\alpha/2,n-1)}$ and $\chi^2_{(1-\alpha/2,n-1)}$ are the critical values from $chi^2$ distribution with $\alpha$ level of significance and $n-1$ degrees of freedom.
Step 4 Determine the critical value
The critical values of $\chi^2$ for $\alpha$ level of significance and $n-1$ degrees of freedom are $\chi^2_{(\alpha/2,n-1)}=\chi^2_{(0.05,11)}=19.675$ and $\chi^2_{(1-\alpha/2,n-1)}=\chi^2_{(0.95,11)}=4.575$.
Step 5 Determine the confidence interval
$90$% confidence interval estimate for population standard deviation is
$$ \begin{aligned} \sqrt{\frac{(n-1)s^2}{\chi^2_{(\alpha/2,n-1)}}} &\leq \sigma \leq \sqrt{\frac{(n-1)s^2}{\chi^2_{(1-\alpha/2,n-1)}}}\\ \sqrt{\frac{11*1772.41}{19.675}} &\leq \sigma \leq \sqrt{\frac{11*1772.41}{4.575}}\\ 31.479 &\leq \sigma \leq 65.28. \end{aligned} $$
Thus $90$% confidence interval for population standard deviation is $(31.479,65.28)$.
The result indicate that the treatment is effective.
