In Exercises 9–16, assume that each sample is a simplerandom sample obtained from a population with a normal distribution.

Atkins Weight Loss Program In a test of weight loss programs, 40 adults used the Atkins weight loss program. After 12 months, their mean weightlosswas found to be 2.1 lb, with a standard deviation of 4.8 lb. Construct a 90% confidence interval estimate of the standard deviation of the weight loss for all such subjects. Does the confidence interval give us informationabout the effectiveness of the diet?

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${\chi }_L^2=25.6954$ and${\chi }_R^2=54.5722$.

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

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

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

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