Chapter 10.2, Problem 52E

(a)

To determine

To Write: a few lines on the basis of provided information comparing the percent variations in BMC for the two classes.

Explanation of Solution

Given:

  

The Not Pregnant Centre tends to be greater than the Breastfeeding Centre, since There is a higher mean for nonpregnant, and the boxplot much more to the right. The distribution appears to be more than the distribution for the Breastfeed community Non-pregnant party, since it has a higher standard deviation and there is more distance between the boxplot whiskers. Both distributions seem to be right-skewed, as the median to be the box of the boxplot further to the left.

(b)

To determine

To explain: for mothers who are breast-feeding, the mean change in BMC is significantly lower and an adequate test is conducted to confirm the response.

Answer to Problem 52E

Yes

Explanation of Solution

Given:

  $\begin{array}{l}{\overline{x}}_1=-3.58723\\ {\overline{x}}_2=0.309091\\ s_1=2.50561\\ s_2=1.29832\\ n_1=47\\ n_2=22\end{array}$

Formula used:

  $t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Calculation:

The hypothesis:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_1:{\mu }_1<{\mu }_2\end{array}$

The test statistic:

  $\begin{array}{l}t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{-3.58723-0.309091}{\sqrt{\frac{{2.50561}^2}{47}+\frac{{1.29832}^2}{22}}}\\ =-8.499\end{array}$

The test statistic:

  $\begin{array}{c}df=\min \left(n_1-1,n_2-1\right)\\ =\min \left(47-1,22-1\right)\\ =21\end{array}$

P-value in the row $df=20$ is

  $P<0.0005$

  $P<0.05\Rightarrow \text{ Reject }{H}_0$

There is ample evidence to support the argument that, for mothers who are breast-feeding, the mean change in BMC is substantially lower.

(c)

To determine

To explain: it may mean that breast-feeding causes weakening of the bones of a woman, why or why not.

Answer to Problem 52E

No

Explanation of Solution

In order to observe their reactions, an experiment purposefully imposes certain care on individuals. Without disrupting the scene, they are studying, an observational study tries to gather details. Observation Study Because the study is not an experiment, correlation (namely that breast-feeding causes mother bones to weaken) cannot be inferred because the study may be affected by lurking variables. A lurking variable is a variable which has a major impact in an analysis on the relationship between the variables, but is not one of the studied explanatory variables. If want to prove causation, then it needs an experiment.

(d)

To determine

To Construct: and interpreting a 95 percent confidence interval for the difference in mean bone mineral loss and explaining how more information is given by this interval than in part (b) the significance test.

Answer to Problem 52E

(-4.850, -2.943)

Explanation of Solution

Given:

  $\begin{array}{l}{\overline{x}}_1=-3.58723\\ {\overline{x}}_2=0.309091\\ s_1=2.50561\\ s_2=1.29832\\ n_1=47\\ n_2=22\end{array}$

Formula used:

For the confidence interval

  $\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Calculation:

Degrees of freedom is

  $df=\min \left(n_1-1,n_2-1\right)=\min \left(47-1,22-1\right)=21$

Determine $t^{\ast }$ with $df=21$ and c=95%

  $t^{\ast }=t_{\alpha /2}=2.080$

Confidence interval are for ${\mu }_1-{\mu }_2$ are:

  $\left({\overline{x}}_1-{\overline{x}}_2\right)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\left(-3.58723-0.309091\right)-2.080\times \sqrt{\frac{{2.50561}^2}{47}+\frac{{1.29832}^2}{22}}=-4.850$

  $\left({\overline{x}}_1-{\overline{x}}_2\right)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\left(-3.58723-0.309091\right)+2.080\times \sqrt{\frac{{2.50561}^2}{47}+\frac{{1.29832}^2}{22}}=-2.943$

95 percent are confident that the mean distance is between -4.850 and -2.943.