91影视

To write a few sentences comparing the percent changes in BMC for the two groups, based on the graph and numerical summaries.

Because there is a higher mean for non pregnant and the boxplot is considerably more to the right, the Not Pregnant Center likely to be greater than the Breastfeeding Center.
Because it has a higher standard deviation and more distance between the boxplot whiskers, the distribution seems to be larger than the distribution for the Breastfeed community Non-pregnant party.
As the median of the boxplot farther to the left, both distributions appear to be right-skewed.

To determine the mean change in BMC significantly lower for the mothers who are breast-feeding and to Carry out an appropriate test to support the answer.

Let,
${\stackrel{炉}{x}}_1=-3.58723$

${\stackrel{炉}{x}}_2=0.309091$

$s_1=2.50561$

$s_2=1.29832$

$n_1=47$

$n_2=22$

Find the hypothesis as follows:

$H_0:{渭}_1={渭}_2$

$H_a:{渭}_1<{渭}_2$

Find the test statistic as follows:
$t=\frac{{\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

$=\frac{-3.58723-0.309091}{\sqrt{\frac{2.{50561}^2}{47}+\frac{1.{29832}^2}{22}}}$

$鈮�-8.49$

Find the degrees of freedom as follows:
$df=\min \left(n_1-1,n_2-1\right)$

$=\min (47-1,22-1)$

$=21$

To conclude that breast-feeding causes a mother鈥檚 bones to weaken, why or why not.

An experiment purposely puts specific care on persons in order to examine their reactions. An observational research attempts to obtain details without upsetting the scene they are investigating. Because the study is not an experiment, a correlation (that breastfeeding weakens mother's bones) cannot be inferred because the study could be influenced by unknown circumstances.
A lurking variable is one that has a significant impact on the relationship between variables in an analysis but is not one of the examined explanatory factors. If need to prove causation, will need to do an experiment.

As a result, no because will need to do an experiment.

To construct and interpret a $95\%$confidence interval for the difference in mean bone mineral loss. Then to explain how this interval provides more information than the significance test in part (b).

Let,
${\stackrel{炉}{x}}_1=-3.58723$
${\stackrel{炉}{x}}_2=0.309091$
$s_1=2.50561$
$s_2=1.29832$
$n_1=47$
$n_2=22$
Find the degrees of freedom as follows:

$df=\min \left(n_1-1,n_2-1\right)$

$=\min (47-1,22-1)$

$=21$

Using table B, find $t^{*}$with $df=21$and $c=95\%$ :
$t^{*}=t_{伪/2}$

$=2.080$

For ${渭}_1-{渭}_2$, the confidence interval are:

$\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)-t_{伪/2}脳\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

$=(-3.58723-0.309091)-2.080脳\sqrt{\frac{2.{50561}^2}{47}+\frac{1.{29832}^2}{22}}$

$鈮�-4.850$

$\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)+t_{伪/2}脳\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

$=(-3.58723-0.309091)+2.080脳\sqrt{\frac{2.{50561}^2}{47}+\frac{1.{29832}^2}{22}}$

$鈮�-2.943$

As a result, $95\%$confident that the mean difference is between $-4.850$and $-2.943$.