Question

In Exercises 9–16, assume that each sample is a simple

random sample obtained from a population with a normal distribution.

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

about the effectiveness of the diet?

Short answer

The 90% confidence interval estimate of the standard deviation of the weight loss for all such subjects is\(4.06\;{\rm{lb}} < \sigma < 5.91\;{\rm{lb}}\).

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

Step-by-step solution

Given information

The sample number of adults is\(n = 40\).

The mean weight loss is 2.1 lb.

The sample standard deviation is\(s = 4.8\;{\rm{lb}}\).

The level of confidence is 90%.

Compute the confidence intervalestimate of \({\bf{\sigma }}\)

The degrees of freedom is computed as,

\(\begin{array}{c}df = n - 1\\ = 40 - 1\\ = 39\end{array}\)

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,

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

Therefore, the 90% confidence interval estimate of the standard deviation of the weight loss for all such subjects is\(4.06\;lb < \sigma < 5.91\;lb\).

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