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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 accompanying data on total body bone mineral content (in grams) for a representative sample of mothers both during breast feeding (B) and in the post-weaning period (P). Use a \(95 \%\) confidence interval to estimate the difference in mean total body bone mineral content during post- weaning and during breast feeding.

Short Answer

Expert verified
The 95% confidence interval for the difference in mean total body bone mineral content during post-weaning and during breastfeeding can be calculated using the following steps: 1. Calculate the sample mean and sample standard deviation for both breastfeeding and post-weaning data. 2. Compute the difference in sample means (\(\bar{X}_{D} = \bar{X}_{P} - \bar{X}_{B}\)). 3. Estimate the standard error of the difference: \(SE = \sqrt{\frac{s_{B}^2}{n_{B}} + \frac{s_{P}^2}{n_{P}}}\). 4. Compute the margin of error using a 95% confidence interval: \(ME = t^* \times SE\). 5. Calculate the 95% confidence interval for the difference in means: Lower limit = \(\bar{X}_{D} - ME\), Upper limit = \(\bar{X}_{D} + ME\). This interval estimates the difference in mean total body bone mineral content during post-weaning and during breastfeeding with 95% confidence.

Step by step solution

01

Calculate the sample mean and sample standard deviation of both breastfeeding and post-weaning data

Let's denote the breastfeeding data as sample B and the post-weaning data as sample P. First, calculate the sample mean (\(\bar{X}_{B}\) and \(\bar{X}_{P}\)) and sample standard deviation (s_B and s_P) for both samples.
02

Compute the difference in sample means

Now, calculate the difference in sample means by subtracting the breastfeeding mean (\(\bar{X}_{B}\)) from the post-weaning mean (\(\bar{X}_{P}\)): \(\bar{X}_{D} = \bar{X}_{P} - \bar{X}_{B}\)
03

Estimate the standard error of the difference

We need to estimate the standard error of the difference. Assuming that the two samples are independent, we can calculate the standard error (SE) of the difference in means as: \(SE = \sqrt{\frac{s_{B}^2}{n_{B}} + \frac{s_{P}^2}{n_{P}}}\) where \(n_{B}\) and \(n_{P}\) are the sample sizes of breastfeeding and post-weaning data, respectively.
04

Compute the margin of error using a 95% confidence interval

The margin of error (ME) can be calculated using the t-distribution critical value (t*) and standard error (SE) as follows: \(ME = t^* \times SE\) Since we want a 95% confidence interval, and we know the degrees of freedom (which should be the smaller of \(n_{B}-1\) and \(n_{P}-1\)), we can find the t-distribution critical value (t*) from a two-tailed t-distribution table.
05

Calculate the confidence interval for the difference in means

Finally, we can calculate the 95% confidence interval for the difference in mean total body bone mineral content during post-weaning and during breastfeeding: Lower limit of the confidence interval = \(\bar{X}_{D} - ME\) Upper limit of the confidence interval = \(\bar{X}_{D} + ME\) This interval provides an estimate of the difference in mean total body bone mineral content during post-weaning and during breastfeeding with 95% confidence.

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Key Concepts

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

Sample Mean
The sample mean, often denoted as \( \bar{X} \) for a given sample, is a fundamental statistical calculation representing the average value of a set of numbers. It is computed by adding together all the values in a sample (in this case, the total body bone mineral content of mothers during breastfeeding and post-weaning periods) and dividing by the number of observations in the sample.

For educational clarity, consider if a classroom of students took a test. The sample mean of their scores would be found by adding up all the individual scores and dividing by the total number of students. In the context of our exercise, this concept is used to assess the central tendency in bone mineral content among the sample of mothers at different stages: breastfeeding and post-weaning.
Sample Standard Deviation
The sample standard deviation is a measurement of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values are close to the mean, while a high standard deviation indicates a wider dispersion around the mean. To calculate this, we subtract the sample mean from each observation, square these differences, add them up, divide by the sample size minus one, and finally take the square root of that result.

An easily relatable example could be measuring the height variation in a group of plants grown under similar conditions. The sample standard deviation would give us insight into how much the plant heights vary from the average plant height. Applying this to our mother's bone mineral content, it helps us understand how much individual measurements deviate from the average bone mineral content in each group.
Standard Error
The standard error of the mean (SEM) is a statistical term that measures how far the sample mean of the data is likely to be from the true population mean. It's calculated by dividing the sample standard deviation by the square root of the sample size. The smaller the standard error, the more representative the sample mean is likely to be of the overall population.

Imagine throwing a dart at a target several times. Your aim (the sample mean) might improve with practice, but there's always some variation in where the darts land. The standard error helps us understand how much this variation affects our estimation of the true center of the target. In the exercise context, calculating the standard error gives us an understanding of the precision of our estimate for the difference in bone mineral content between the two states.
T-distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, like the standard normal distribution, but has heavier tails, meaning it accounts for more variability. It's especially useful when dealing with small sample sizes or when the population standard deviation is unknown.

Let's say you're tasting different batches of a chef's soup to determine which has the best flavor. However, you only have a few samples to taste from each batch. The t-distribution would give you a more accurate assessment of the likely flavor (the true mean) across all batches, considering there are fewer taste tests (samples) for evaluation. When we are working with the sample data of bone mineral content in mothers, the t-distribution helps us understand the probabilities of different means we can expect to find and thereby helps to calculate the margin of error and confidence interval for the actual mean difference between breastfeeding and post-weaning bone mineral content.
Bone Mineral Content
Bone mineral content (BMC) is a measure of the amount of minerals, such as calcium, contained in bone tissue, which gives bones their strength and rigidity. BMC can vary due to several factors including dietary intake, physical activity, and physiological conditions like pregnancy and lactation.

As an analogy, imagine the bones as the framework of a building. Just like the building's strength depends on the quality of materials used, bone strength relies on its mineral content. In our exercise, researchers are interested in the fluctuation of BMC in adolescent mothers during and after breastfeeding, a period when their bodies might have different mineral demands. Understanding the change in BMC can provide insights into the health implications of breastfeeding on bone density.

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Most popular questions from this chapter

The paper "The Effect of Multitasking on the Grade Performance of Business Students" (Research in Higher Education Journal [2010]: 1-10) describes an experiment in which 62 undergraduate business students were randomly assigned to one of two experimental groups. Students in one group were asked to listen to a lecture but were told that they were permitted to use cell phones to send text messages during the lecture. Students in the second group listened to the same lecture but were not permitted to send text messages during the lecture. Afterwards, students in both groups took a quiz on material covered in the lecture. The researchers reported that the mean quiz score for students in the texting group was significantly lower than the mean quiz score for students in the no-texting group. In the context of this experiment, explain what it means to say that the texting group mean was significantly lower than the no-text group mean. (Hint: See discussion on page \(662 .\) )

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