91Ó°ÊÓ

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.

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.

The formula for confidence interval is $\stackrel{¯}{x}-E<\mathrm{μ}<\stackrel{¯}{x}+E$.

Here, E is margin of error which is given by $E=t_{\frac{\mathrm{Î\pm }}{2}}×\frac{s}{\sqrt{n}}$,

For critical value, $t_{\frac{\mathrm{Î\pm }}{2}}$ is the critical value .

Here, $\stackrel{¯}{x}$represents the sample mean of weight loss program of adults and $\mathrm{μ}$ represents the population mean of weight loss of adults in Atkins weight loss program.

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 ${\text{t}}_{\frac{\mathrm{Î\pm }}{2}}$with 39 degrees of freedom is 1.685.

Margin of error is given by,

$$\begin{gathered} E=t_{\frac{\mathrm{Î\pm }}{2}}×\frac{s}{\sqrt{n}} \\ =1.685×\frac{4.8}{\sqrt{40}} \\ =1.279 \end{gathered}$$

For 90% confidence interval,

$$\begin{gathered} \stackrel{¯}{x}-E<\mathrm{μ}<\stackrel{¯}{x}+E \\ 2.1-1.279<\mathrm{μ}<2.1+1.279 \\ 0.821<\mathrm{μ}<3.379 \end{gathered}$$

Therefore, 90% confidence interval is $0.82lb<\mathrm{μ}<3.38lb$.

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

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.