Question

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

Short answer

The 90% confidence intervalof the mean weight loss for all such subjects in the Atkins weight loss program is.

From the result, we can say that we are 90% confident that the mean weight loss will lie between confidence interval 0.82 lb to 3.38lb, which shows that the program is effective.

The given value of standard deviation is 4.8 lb which is high, therefore, the confidence interval may not be practical.

Step-by-step solution

Given information

40 adults used the Atkins weight loss program, and their mean weight loss is 2.1 lb and the standard deviation is 4.8 lb.

Check the requirements

Assume that the population of weight loss is normally distributed and samples are collected randomly.

As the population standard deviation is unknown in this case, t-distribution would be used.

Describe the formula for confidence interval

The formula for confidence interval is $\overline{x}-E<\mu <\overline{x}+E$.

Here, E is margin of error which is given by $E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}}$,

For critical value, $t_{\frac{\alpha }{2}}$ is the critical value .

Here, $\overline{x}$represents the sample mean of weight loss program of adults and $\mu$ represents the population mean of weight loss of adults in Atkins weight loss program.

Calculate margin of error.

90% confidence level implies 0.1 level of significance.

The degree of freedom is,

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

The critical value ${\text{t}}_{\frac{\alpha }{2}}$with 39 degrees of freedom is 1.685.

Margin of error is given by,

$$\begin{gathered} E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}} \\ =1.685\times \frac{4.8}{\sqrt{40}} \\ =1.279 \end{gathered}$$

Construct the confidence interval.

For 90% confidence interval,

$$\begin{gathered} \overline{x}-E<\mu <\overline{x}+E \\ 2.1-1.279<\mu <2.1+1.279 \\ 0.821<\mu <3.379 \end{gathered}$$

Therefore, 90% confidence interval is $0.82lb<\mu <3.38lb$.

Interpret the result.

The 90% confidence interval of the mean weight loss for all such subjects in the Atkins weight loss program is$0.82lb<\mu <3.38lb$.

From the result, it can be concluded with 90% confidence that the mean weight loss will lie between confidence interval 0.82 lb to 3.38 lb, which is larger than 0. Thus, the program appears to be effective.

Here, the given value of standard deviation is 4.8 lb which is high, therefore, the confidence interval may not be practical.