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Chapter 11: Problem 31

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 (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 is greater than that during breast feeding by more than 25 g? State and test the appropriate hypotheses using a significance level of .05 \(\begin{array}{lcccccc}\text { Subject } & 1 & 2 & 3 & 4 & 5 & 6 \\\ \mathrm{~B} & 1928 & 2549 & 2825 & 1924 & 1628 & 2175 \\ \mathrm{P} & 2126 & 2885 & 2895 & 1942 & 1750 & 2184 \\ \text { Subject } & 7 & 8 & 9 & 10 & & \\\ \mathrm{~B} & 2114 & 2621 & 1843 & 2541 & & \\ \mathrm{P} & 2164 & 2626 & 2006 & 2627 & & \end{array}\)

Short Answer

Expert verified
Yes, the data suggest that the true average total body bone mineral content during postweaning is more than 25g higher than that during breast feeding. The computed t statistic 2.52 exceeds the critical value 1.833, leading to the rejection of null hypothesis at 0.05 significance level.

Step by step solution

01

Calculate the differences

Subtract the bone mineral content during breastfeeding (B) from the content during postweaning (P) for each subject. Here are the differences (P-B) for the 10 subjects: 198, 336, 70, 18, 122, 9, 50, 5, 163, 86.
02

Compute Sample Mean and Standard Deviation

Calculate the sample mean (\( \bar{d} \)) and standard deviation (s) of the differences. For this case, mean (\( \bar{d} \)) = 105.7 and standard deviation (s) = 100.9.
03

Setup null hypothesis and alternative hypothesis

Setup the null hypothesis \( H_0: \mu_d \leq 25 \) and the alternative hypothesis \( H_a: \mu_d > 25 \). Here \( \mu_d \) is the mean difference.
04

Compute the test statistic

The test statistic is a t-score (t). Here, t = \( \frac{\bar{d} - \mu_0}{s / \sqrt{n}} \), where \( \mu_0 \) is the value in the null hypothesis, n is the number of subjects and s is the standard deviation. For this data, the t score is 2.52.
05

Rejection Region

For a significance level of .05 and degrees of freedom \( df = n - 1 = 9 \), from t-distribution table, the critical value (t*) is approximately 1.833. The rejection region is t > t*.
06

Statistical conclusion

As our computed t score (2.52) falls in the rejection region (t > 1.833), we reject the null hypothesis.
07

Real-world conclusion

There's enough evidence to support the claim that the true average total body bone mineral content during postweaning is greater than that during breast feeding by more than 25 g.

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Key Concepts

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

Statistical Significance
Statistical significance is a way to determine whether a result from a study or experiment is unlikely to have occurred purely by chance. It provides a measure of confidence in the results of a hypothesis test.

When researchers set a significance level (often represented by the Greek letter \(\alpha\)), they're deciding on the threshold for how extreme the data must be to reject the null hypothesis. Commonly, a significance level of 0.05 is used, which means there is a 5% risk of concluding that a difference exists when there is no actual difference.

In the exercise involving bone mass in breastfeeding and postweaning mothers, a significance level of 0.05 is chosen. By calculating a test statistic and comparing it with a critical value from a statistical distribution, we can infer whether the observed data is statistically significant. In this case, if the test statistic falls beyond the critical value (in the rejection region), the result is considered unlikely to be due to chance, hence statistically significant.
Null Hypothesis
The null hypothesis, denoted as \(H_0\), is a fundamental concept in hypothesis testing. It represents the default statement or position that there is no effect or no difference, and any observed effect is due to sampling variability or chance.

In the context of our exercise, the null hypothesis is that the mean difference in total body bone mineral content between postweaning and breastfeeding \(\mu_d\) is 25 grams or less. Symbolically, it's represented as \(H_0:\mu_d \leq 25\). The data is then analyzed to determine if it provides enough evidence to reject this null hypothesis in favor of the alternative hypothesis.
Alternative Hypothesis
Contrary to the null hypothesis, the alternative hypothesis \(H_a\) posits that there is an actual effect or a difference. It's what a researcher wants to prove or validate through the data. Generally speaking, the alternative hypothesis suggests that the observed data is not simply due to random chance.

For the bone mass study, the alternative hypothesis is that the true average total body bone mineral content during postweaning is greater by more than 25 grams compared to during breastfeeding. Mathematically, it's presented as \(H_a: \mu_d > 25\). The research goal is to gather enough evidence to support this alternative hypothesis.
t-test
The t-test is a statistical test that is used to determine if there is a significant difference between the means of two groups, or a group and a specific value, which are subject to random sampling noise. It is particularly useful when dealing with small sample sizes or when the population standard deviation is unknown.

In the presented exercise, researchers used a one-sample t-test to determine whether the mean difference in bone mass between the two states was significantly greater than 25 grams. To do this, the test statistic was computed by comparing the average difference to the hypothesized difference, and dividing by an estimate of the standard error. A rejection region was then established based on the desired significance level. If the computed t-value exceeds the critical t-value from the t-distribution, the null hypothesis is rejected, indicating a statistically significant difference. Using the data from the study, a t-value of 2.52 exceeded the critical value of 1.833, leading to the rejection of \(H_0\) in favor of \(H_a\), and supporting the claim of a significant increase in bone mass postweaning.

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Most popular questions from this chapter

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