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The sample number of values is $n=16$.

The mean wake time is 98.9 min.

The sample standard deviation is $s=42.3$min.

The level of confidence is 98%.

The degrees of freedom is computed as,

$$\begin{gathered} df=n-1 \\ =16-1 \\ =15 \end{gathered}$$

The level of confidence is 98%, which implies that the level of significance is 0.02.

Using the Chi-square table, the critical values at 0.02 level of significance and at 15 degrees of freedom arerole="math" localid="1648109418028" ${蠂}_L^2=5.2293$ and ${蠂}_R^2=30.5779$.

The 98% confidence interval estimate of the standard deviation of the wake times for a population with zopiclone treatments is computed as,

$$\begin{gathered} \sqrt{\frac{\left(n-1\right)s^2}{{蠂}_R^2}}<蟽<\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_L^2}} \\ \sqrt{\frac{\left(16-1\right){42.3}^2}{30.5779}}<蟽<\sqrt{\frac{\left(16-1\right){42.3}^2}{5.2293}} \\ 29.6<蟽<71.6 \end{gathered}$$

Therefore, the 90% confidence interval estimate of the standard deviation of the wake times for a population with zopiclone treatments is role="math" localid="1648109480702" $29.6min<蟽<71.6min$.

And, the estimated confidence interval of the standard deviation of the wake times for a population with zopiclone treatments does not indicate whether the treatment is effective.