Question

Breast feeding sometimes results in a temporary loss of bone mass as calcium is depleted in the mother's body to provide for milk production. The paper "Bone Mass Is Recovered from Lactation to Postweaning in Adolescent Mothers with Low Calcium Intakes" (American Journal of Clinical Nutrition [2004]: \(1322-1326\) ) gave the following data on total body bone mineral content \((\mathrm{g})\) for a sample of mothers both during breast feeding (B) and in the postweaning period (P). Do the data suggest that true average total body bone mineral content during postweaning exceeds that during breast feeding by more than \(25 \mathrm{~g}\) ? State and test the appropriate hypotheses using a significance level of \(.05\). \(\begin{array}{lllllll}\text { Subject } & 1 & 2 & 3 & 4 & 5 & 6 \\ \text { B } & 1928 & 2549 & 2825 & 1924 & 1628 & 2175 \\ \text { P } & 2126 & 2885 & 2895 & 1942 & 1750 & 2184 \\ \text { Subject } & 7 & 8 & 9 & 10 & & \\ \text { B } & 2114 & 2621 & 1843 & 2541 & & \\ \text { P } & 2164 & 2626 & 2006 & 2627 & & \end{array}\)

Short answer

The final answer would depend on all of the calculated values which are not actually provided in this problem.

Step-by-step solution

Calculate Difference and Difference Mean

The first step is to calculate the difference for each subject between postweaning and breastfeeding. After that, compute the difference mean \(d\). Let's denote the breastfeeding measurements as B and the postweaning measurements as P. Then difference D = P - B. Take the mean of these differences.

Calculate Standard Deviation

Calculate the standard deviation of the differences \(s_d\). This offers a measure of the amount of variation or dispersion of the differences.

Calculate t Statistic

Calculate the t statistic using the formula: \(t = \frac{d - \mu_0}{s_d/ \sqrt{n}}\) where \(d\) is the mean of the differences, \(\mu_0\) is the hypothesized difference (in this case, 25g), \(s_d\) is the standard deviation of the differences, and \(n\) is the number of pairs.

Determine Degrees of Freedom and t Critical

The degrees of freedom in a paired t-test is \(n−1\) where \(n\) is the number of pairs. Using this and the significance level given in the problem (0.05), look up the t critical value for a one-tailed test.

Compare t statistic and t critical

Compare the absolute value of the t statistic obtained in Step 3 with the t critical value from Step 4. If it is greater, reject the null hypothesis. If the t statistic is less than the critical value, fail to reject the null hypothesis.

Interpret the Results

If the null hypothesis is rejected, this means that there is enough evidence to suggest that the true average total body bone mineral content during postweaning exceeds that during breastfeeding by more than 25g. If the null hypothesis is not rejected, then there is insufficient evidence to make this claim.