91Ó°ÊÓ

We have when comparing the percent changes in BMC for the two groups,

Shape: Because the box in the box plots lies to the left between the whiskers, both the distributions appear to be skewed to the right.

Because the boxplot for the Not-pregnant lies more to the right of the boxplot for breastfeeding, the center for the Not-pregnant appears to be higher than the center for breastfeeding.

Because the width between the whickers of the boxplot for the breastfeeding group is greater than the width between the whickers for the Not-pregnant group, the spread for the breastfeeding group appears to be greater than the spread for the Not-pregnant group.

$$\begin{gathered} {\stackrel{¯}{x}}_1=-3.59 \\ {\stackrel{¯}{x}}_2=0.31 \\ {n}_1=47 \\ {n}_2=22 \\ {s}_1=2.51 \\ {s}_2=1.30 \\ \mathrm{Î\pm }=0.05 \end{gathered}$$

The appropriate hypotheses for this are:

$$\begin{gathered} H_0:{\mathrm{μ}}_{BF}={\mathrm{μ}}_{NP} \\ {H}_0:{\mathrm{μ}}_{BF}<{\mathrm{μ}}_{NP} \end{gathered}$$

Locate the following test statistics:

$$\begin{gathered} t=\frac{\left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)-\left({\mathrm{μ}}_1-{\mathrm{μ}}_2\right)}{\sqrt{\frac{{\frac{2}{2}}_2^2}{n_1}+\frac{\frac{2}{2}}{n_2}}} \\ =\frac{-3.59-0.31-0}{\sqrt{\frac{{251}^2}{47}+\frac{1.{30}^2}{22}}} \\ =-8.493 \end{gathered}$$

The degree of liberty will now be:

$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(47-1,22-1) \\ =21 \end{gathered}$$

So the $P$-value will be:

$P<0.0005$

On the other hand by using the calculator command: $2×$tcdf$(-8.493,1\mathrm{E}99,21)$which results in the -values as $0$

And we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, then,

$P<0.05⇒\text{Reject}{H}_0$

Therefore, we conclude that the true mean percentage change in BMC for breast-feeding women is lower than the true mean percentage change in BMC for non-breast-feeding women.

Part (b) of our conclusion states, "There is convincing evidence that the true mean percentage change in BMC for breast-feeding women is less than the true mean percentage change in BMC for non-pregnant women."

In a completely randomized experiment, all subjects are assigned to a group at random.

The experiment is not completely randomized because the women were not randomly assigned to a treatment group.

As a result, it's possible that members of the same group share characteristics with members of the other group. For example, it's possible that age influences the outcome because older women have a lower BMC and are less likely to breastfeed.

Part (b) of our conclusion states, "There is convincing evidence that the true mean percentage change in BMC for breast-feeding women is less than the true mean percentage change in BMC for non-pregnant women."

When we reject a null hypothesis when the null hypothesis is true, we make a type I error. When we fail to reject the null hypothesis when the null hypothesis is false, we make a Type II error.

As a result, if we reject the null hypothesis, we have made a Type I error.