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The sample number of adults is $n=40$.

The mean weight loss is 2.1 lb.

The sample standard deviation is $s=4.8\text{lb}$.

The level of confidence is 90%.

The degrees of freedom is computed as,

$$\begin{gathered} df=n-1 \\ =40-1 \\ =39 \end{gathered}$$

The level of confidence is 90%, which implies that the level of significance is 0.10.

Using the Chi-square table, the critical values at 0.10 level of significance and at 39 degrees of freedom are${蠂}_L^2=25.6954$ and${蠂}_R^2=54.5722$.

The 90% confidence interval estimate of the standard deviation of the weight loss for all such subjects is computed as,

$$\begin{gathered} \sqrt{\frac{\left(n-1\right)s^2}{{蠂}_R^2}}<蟽<\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_L^2}} \\ \sqrt{\frac{\left(40-1\right){4.8}^2}{54.5722}}<蟽<\sqrt{\frac{\left(40-1\right){4.8}^2}{25.6954}} \\ 4.06<蟽<5.91 \end{gathered}$$

Therefore, the 90% confidence interval estimate of the standard deviation of the weight loss for all such subjects is role="math" localid="1648108991166" $4.06lb<蟽<5.91lb$.

And, the confidence interval does not provide any information about the effectiveness of the diet.