Question

52. Does breast-feeding weaken bones? Breast-feeding mothers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Researchers compared a random sample of 47 breast-feeding women with a random sample of 22 women of similar age who were neither pregnant nor lactating. They measured the percent change in the bone mineral content (BMC) of the women’s spines over three months. Comparative boxplots and summary statistics for the data from Fathom are shown below.

(a) Based on the graph and numerical summaries, write a few sentences comparing the percent changes in BMC for the two groups.
(b) Is the mean change in BMC significantly lower for the mothers who are breast-feeding? Carry out an appropriate test to support your answer.
(c) Can we conclude that breast-feeding causes a mother’s bones to weaken? Why or why not?
(d) Construct and interpret a $95\%$ confidence interval for the difference in mean bone mineral loss. Explain how this interval provides more information than the significance test in part (b)

Short answer

(a) Both distributions appear to be right-skewed.

(b) Yes, there is sufficient evidence to support the hypothesis that breast-feeding mothers experience a much lower mean change in BMC.

(c) No, because will need to do an experiment.

(d) A $95\%$ confident that the mean difference is between $-4.850$ and $-2.943$.

Step-by-step solution

Part (a) Step 1: Given information

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

Part (a) Step 2: Explanation

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.

Part (b) Step 1: Given information

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.

Part (b) Step 2: Explanation

Let,
${\overline{x}}_1=-3.58723$

${\overline{x}}_2=0.309091$

$s_1=2.50561$

$s_2=1.29832$

$n_1=47$

$n_2=22$

Find the hypothesis as follows:

$H_0:{\mu }_1={\mu }_2$

$H_a:{\mu }_1<{\mu }_2$

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

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

$\approx -8.49$

Part (b) Step 3: Explanation

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

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

$=21$

Since, the P-value is the probability of getting the test statistic's value, or a number that is more extreme.The P-value is the number (or interval) in Table B's column title that corresponds to the t-value in row $df=20$:
$P<0.0005$
The null hypothesis is rejected if the P-value is less than or equal to the significance level:
$P<0.05\Rightarrow \text{Reject}{H}_0$
As a result, yes. There is sufficient evidence to support the hypothesis that breast-feeding mothers experience a much lower mean change in BMC.

Part (c) Step 1: Given information

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

Part (c) Step 2: Explanation

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.

Part (d) Step 1: Given information

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).

Part (d) Step 2: Explanation

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

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

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

$=21$

Part (d) Step 3: Explanation

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

$=2.080$

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

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

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

$\approx -4.850$

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

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

$\approx -2.943$

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