91Ó°ÊÓ

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.

The given data is

$\stackrel{¯}{x}=-3.587$

$s=2.506$

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

and in row $C=99\%$

$t^{*}=2.704$

Then the margin of error is given as

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

Thus the confidence interval becomes

$-4.5754=-3.587-0.9884=\stackrel{¯}{x}-E<\mathrm{μ}<\stackrel{¯}{x}+E=-3.587+0.9884=-2.5986$

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

$-4.5754<\mathrm{μ}<-2.5986$.

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.

From part a

$-4.5754<\mathrm{μ}<-2.5986$

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.