Q. 67 - The Practice of Statistics for AP

Question

Bone loss by nursing mothers.: Breastfeeding mothers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone minerals. Researchers measured the per cent 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 per cent 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 minerals? Explain 

Step-by-Step Solution

Answer

(a) The interval to estimate the mean per cent change in BMC in the population is $-4.5754<\mu <-2.5986$

(b) The given data gives that on average nursing mothers lose bone minerals is true.

Step 1: Given information ( Part a)

Randomly selected spines of three months breastfeeding mothers $=47$

The mean change in $BMC =3.587\%$

Standard deviation$=2.506\%$

Find the interval and explain about it.

Step 2: Explanation (Part a)

The given data is

$\overline{x}=-3.587$

$s=2.506$

Find the value of t using the row starting with $d f=n-1=47-1=46>40$

and in row $C=99\%$

$t^{*}=2.704$

Then the margin of error is given as 

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

Thus the confidence interval becomes

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

Thus it is $99\%$ confident that the true population mean is between $4.5754\%$ and $-2.5986$

$-4.5754<\mu <-2.5986$.

Step 1: Given information (Part b)

Randomly selected spines of three months breastfeeding mothers  $=47$

The mean change in $BMC=3.587\%$

Standard deviation $=2.506\%$

Find the interval and explain it. 

Step 2: Explanation (Part B)

From part a

$-4.5754<\mu <-2.5986$

The data shows that on average, breastfeeding mothers lose bone minerals since the confidence interval does not contain $0$ and is fully below $0$.

So the answer is yes.