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Chapter 11: Problem 33

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

Expert verified
The short answer will depend on the results obtained from the statistical test conducted. If the null hypothesis is rejected, it means that the data suggest that true average total body bone mineral content during postweaning exceeds that during breast feeding by more than 25g. If not, then there's insufficient evidence to make that claim.

Step by step solution

01

Calculating the Difference

First, calculate the difference \(d_i = P_i - B_i\) between the postweaning and breastfeeding measurements for each subject. Subtract each B value from the corresponding P value.
02

Checking the Significance

Perform a one-sample t-test for the differences \(d_i\). The null hypothesis (\(H_0\)) is that the true mean difference is less than or equal to 25g. The alternative hypothesis (\(H_a\)) is that the true mean difference is more than 25g. If the test statistic falls in the rejection region (or p-value is less than the significance level, 0.05), the null hypothesis should be rejected, indicating that there is a significant increase in bone mineral content during the postweaning period that exceeds 25g.
03

Calculate Test Statistic

The test statistic for a one-sample t-test is \(t = \frac{\bar{d} - \mu_0}{s_d / \sqrt{n}}\), where \( \bar{d}\) is the sample mean difference, \( s_d\) is the standard deviation of the differences, \(n\) is the number of pairs and \( \mu_0\) is the specified value in the null hypothesis (here, 25g).
04

Conduct Hypothesis Testing

Conduct the hypothesis testing. Compare the test statistic value to the critical value, or calculate the p-value. The critical value can be found from the t-distribution table with \(n-1\) degrees of freedom. The rejection rule is: If \(t > t_{critical}\) (for upper-tail test), reject \(H_0\). Likewise, if p-value < 0.05, reject \(H_0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

One-Sample T-Test
A one-sample t-test is a statistical test used to determine if the mean of a single sample is significantly different from a known value or a hypothesized population mean. In this context, we are comparing the mean difference in bone mineral content between two conditions: breastfeeding and postweaning. The main goal is to see if the average increase in bone mineral content (BMC) during the postweaning period exceeds 25 grams. The one-sample t-test is performed on the differences. We calculate the difference for each subject by subtracting the BMC during breastfeeding from the BMC postweaning. This provides us with a new dataset of differences, for which we calculate a mean and standard deviation. To proceed with the test:
  • Calculate the sample mean difference ( ar{d} )
  • Calculate the standard deviation of the differences ( s_d )
  • Use these values to compute the test statistic
The test statistic follows the t-distribution with degrees of freedom equal to the number of sample pairs minus one (n-1). If we find that the test statistic is greater than the critical value from the t-distribution table, or if the p-value is less than 0.05, it suggests that the true mean difference indeed exceeds the specified 25 grams threshold.
Significance Level
The significance level, often denoted as a (alpha), is a critical component in hypothesis testing that helps determine the threshold at which we reject the null hypothesis. In our exercise, we are using a significance level of 0.05. The selection of a significance level depends on how much risk we are willing to take on making a Type I error鈥攔ejecting a true null hypothesis. A significance level of 0.05 implies a 5% risk of concluding that a difference exists when there is none. In hypothesis testing, when the calculated p-value is less than the significance level (0.05 in our case), we reject the null hypothesis. This would suggest that the mean difference in bone mineral content exceeds 25 grams, indicating a significant effect. Conversely, if the p-value is greater than 0.05, we would fail to reject the null hypothesis, suggesting that any difference we observed could likely be due to random chance.
Bone Mineral Content Analysis
Bone Mineral Content (BMC) analysis is a crucial measure in understanding changes in bone health, particularly in scenarios like breastfeeding which can impact maternal calcium stores. In this scientific inquiry, we examine how BMC changes from breastfeeding to the postweaning period. These changes are determined by measuring the total body BMC at two different points: during breastfeeding and after weaning. The main interest is whether the BMC significantly increases after breastfeeding has concluded and by how much, informed by the hypothesis that postweaning BMC should exceed the breastfeeding BMC by at least 25 grams. To analyze BMC properly:
  • Ensure accurate and consistent measurement of BMC at each stage
  • Calculate the differences between the two measurements for each subject
  • Use statistical tests (like the one-sample t-test) to determine significance
This analysis can help establish the effects of breastfeeding on a mother's bone health and can inform dietary and healthcare recommendations. Understanding these changes is vital in ensuring that mothers recover well post-lactation, maintaining optimal bone health.

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