Chapter 8.3, Problem 69E

a)

To determine

To construct: The 99% confidence interval to find the mean percent change in breast feeding mothers and interpret it.

Answer to Problem 69E

The confidence interval is $\left(-4.569\text{ ,}-2.605\right)$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Sample mean}\left(\overline{x}\right)=-3.587\\ \text{Sample standard deviation}\left(s\right)=2.506\\ \text{Sample size}\left(n\right)=47\\ \text{Confidence level = 99\%}\end{array}$

Formula used:

The formula to compute the confidence interval for population mean is:

  $CI=\overline{x}\pm t_{\alpha /2}\times \frac{s}{\sqrt{n}}$

Calculation:

The obtained Minitab output is:

  

Interpretation:

There is 99% probability that the mean percent change in BMC for the breast-feeding mothers lies between -4.569 and -2.605.

b)

To determine

To explain: Whether using the above part, it could be concluded that nursing mothers are losing bone mineral

Answer to Problem 69E

No because the confidence interval does not contain 0.

Explanation of Solution

The confidence interval computed in above part lies between -4.569 and -2.605.Since, it does not contain 0. Thus, it cannot be said that the nursing mothers are losing bone mineral.