Q. 67 - The Practice of Statistics for AP

Question

Breast-feeding mothers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Researchers measured the percent change in bone mineral content (BMC) of the spines of $47$ randomly selected mothers during three months of breastfeeding. The mean change in BMC was $-3.587\%$ and the standard deviation was $2.506\%$.

(a) Construct and interpret a $99\%$ confidence interval to estimate the mean percent change in BMC in the population.

(b) Based on your interval from (a), do these data give good evidence that on average nursing mothers lose bone mineral? Explain. 

Step-by-Step Solution

Answer

(a) We are $99\%$ confident that the true population mean is between $-4.5754\%$ and $-2.5986\%$.

(b) Yes, the data give good evidence that on average nursing mothers lose bone mineral.

Part (a) Step 1: Given Information

Given

$\overline{x}=-3.587$

$s=2.506$

Part (a) Step 2: Explanation

Determine the t-value by looking in the row starting with degrees of freedom $d f=n-1=47-1=46>40$ and in the row with $c=99\%$ in table B:

$t^{*}=2.704$

The margin of error is then:

$$\begin{gathered} E=t^{*}·\frac{s}{\sqrt{n}} \\ =2.704\times \frac{2.506}{\sqrt{47}} \\ \approx 0.9884 \end{gathered}$$

Then the confidence interval becomes:

$$\begin{gathered} -4.5754=-3.587-0.9884 \\ =\overline{x}-E<\mu <\overline{x}+E \\ =-3.587+0.9884 \\ =-2.5986 \end{gathered}$$

We are $99\%$ confident that the true population mean is between $-4.5754\%$ and $-2.5986\%$.

Part (b) Step 1: Given Information

Given

$\overline{x}=-3.587$

$s=2.506$

Part (b) Step 2: Explanation

Confidence interval found in part (a):

$-4.5754<\mu <-2.5986$

Since the confidence interval does not contain $0$ and lies completely below $0$, the data gives good evidence that on average nursing mothers lose bone mineral.