Chapter 10.2, Problem 64E

(a)

To determine

To write a few sentences comparing the percent changes in BMC for the two groups.

Explanation of Solution

When comparing the percent changes in BMC for the two groups, we have,

Shape: Both the distribution appears to be skewed to the right because the box in the box plots lies to the left between the whiskers.

Center: The center for the Not-pregnant seems to be higher than the center for Breast feed because the boxplot for the Not-pregnant lies more to the right of the boxplot for Breast feed.

Spread: The spread for the Breast feed group seems to be more than the spread for the Not-pregnant group because the width between the whickers of the boxplot for the Breast feed group is greater than the width between the whickers for the Not-pregnant group.

(b)

To determine

To find out what conclusion should the researcher make at the $\alpha =0.05$ significance level.

Answer to Problem 64E

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 women who are not pregnant.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=-3.59\\ {\overline{x}}_2=0.31\\ n_1=47\\ n_2=22\\ s_1=2.51\\ s_2=1.30\end{array}$

  $\alpha =0.05$

Therefore, the appropriate hypotheses for this is:

  $\begin{array}{l}{H}_0:{\mu }_{BF}={\mu }_{NP}\\ H_a:{\mu }_{BF}<{\mu }_{NP}\end{array}$

Now, find the test statistics:

  $\begin{array}{l}t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{-3.59-0.31-0}{\sqrt{\frac{{2.51}^2}{47}+\frac{{1.30}^2}{22}}}\\ =-8.493\end{array}$

Now, the degree of freedom will be:

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

So the P -value will be:

  $P<0.0005$

On the other hand by using the calculator command: $2\times \text{tcdf(-8}\text{.493,1E99,21)}$ which results in the P -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,

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

Thus, we conclude that 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 women who are not pregnant.

(c)

To determine

To explain can we conclude that breast feeding causes a mother’s bones to weaken.

Answer to Problem 64E

No, we cannot conclude that.

Explanation of Solution

We conclude in part (b) that,

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 women who are not pregnant.

A completely randomized experiment randomly assigns all subjects to a group.

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

This then implies that it is possible that the individual in the same group have common characteristics in the other group. For example, it is possible that the age confounds the result since the women tend to have a lower BMC when the women is older and the women is less likely to be breast feeding when the women is older.

This then implies we cannot conclude that the breast feeding causes a mother’s bones to weaken because the age could be the possible cause.

(d)

To determine

To explain what type of error could you have made- Type I error or Type II error, based on the conclusion in part (b).

Answer to Problem 64E

Type I error.

Explanation of Solution

We conclude in part (b) that,

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 women who are not pregnant.

A type I error occurs if we reject a null hypothesis when the null hypothesis is true. And the Type II error occurs if we fails to reject the null hypothesis when the null hypothesis is false.

Thus, in this case we reject the null hypothesis then it is a Type I error.