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Chapter 4: Problem 21

Body fat in women and men. The third National Health and Nutrition Examination Survey collected body fat percentage (BF) data from 13,601 subjects whose ages are 20 to \(80 .\) A summary table for these data is given below. Note that BF is given as mean \(\pm\) standard error. Construct a \(95 \%\) confidence interval for the difference in average body fat percentages between men and women, and explain the meaning of this interval. Tip: the standard error can be calculated as \(S E=\sqrt{S E_{M}^{2}+S E_{W}^{2}}\)

Short Answer

Expert verified
Calculate SE, difference in means, and use them to find the 95% confidence interval for the difference in body fat percentages between genders.

Step by step solution

01

Extract Given Information

We have two groups: men and women. The data given includes the mean and standard error (SE) for body fat percentage for both groups. For men, let's denote the mean as \( \bar{x}_M \) and the SE as \( SE_M \). Similarly, for women, denote the mean as \( \bar{x}_W \) and the SE as \( SE_W \).
02

Compute Combined Standard Error

To find the standard error of the difference between means of the two groups, use the formula: \[ SE = \sqrt{SE_{M}^{2} + SE_{W}^{2}} \] Plug in the values for \( SE_M \) and \( SE_W \) from the data to compute \( SE \).
03

Determine the Mean Difference

Calculate the difference in mean body fat percentages between men and women: \[ \Delta \bar{x} = \bar{x}_W - \bar{x}_M \]
04

Calculate Confidence Interval

The 95% confidence interval for the difference in means is given by: \[ (\Delta \bar{x} - 1.96 \times SE, \Delta \bar{x} + 1.96 \times SE) \] Use the computed \( \Delta \bar{x} \) and \( SE \) to find this confidence interval.
05

Interpret the Confidence Interval

The 95% confidence interval suggests that we are 95% confident that the true difference in average body fat percentages between men and women falls within this interval. If the interval does not include zero, it indicates a statistically significant difference between the two groups.

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Key Concepts

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

Body Fat Percentage
Body fat percentage is an essential measurement in understanding human health. It refers to the proportion of your body that is composed of fat. In this context, body fat percentage was measured for both men and women in a sample population. Knowing the average body fat percentage helps us assess and compare the health status across different groups. For the National Health and Nutrition Examination Survey, researchers collected data on individuals aged 20 to 80.
The mean body fat percentage provides an average level across all surveyed individuals within the group. It is crucial because it gives a snapshot of a population's health related to body weight. High or low averages might suggest trends or health concerns across a population.
When looking at differences between groups, such as between men and women, it highlights potential physiological differences. Men and women naturally have different distributions of body fat. Often, women have a higher body fat percentage due to biological differences. However, these statistics allow us to quantify and rigorously analyze those differences.
Standard Error
The standard error (SE) is a statistical term that reflects the variability or precision of the sample mean. In the survey, both men's and women鈥檚 body fat percentages are given as a mean \(\pm\) SE. It indicates how much the sample mean differs from the true population mean.
A smaller SE suggests that the sample mean is a more accurate estimate of the population mean. It is calculated from the sample standard deviation divided by the square root of the sample size. For the body fat data, the combined SE for the difference in means is crucial in constructing a confidence interval.
To compute it for two groups, such as men and women, apply the formula: \[ SE = \sqrt{SE_{M}^{2} + SE_{W}^{2}} \]This formula combines the SEs of both groups, allowing us to understand the error associated with the difference in means. It helps us determine the reliability of comparing these two averages.
Difference of Means
The difference of means refers to the disparity between two groups' average values. In this survey, we are interested in the difference between men's and women's average body fat percentages. To find the difference, subtract the mean body fat percentage of men from that of women:\[ \Delta \bar{x} = \bar{x}_W - \bar{x}_M \]This calculation yields the absolute difference in average body fat percentage. Such a difference helps in understanding biological variations or the impact of different lifestyles, diets, and habits between the groups.
To assess the significance of this difference, a confidence interval is constructed. The 95% confidence interval gives us a range within which the true difference likely lies:\[ (\Delta \bar{x} - 1.96 \times SE, \Delta \bar{x} + 1.96 \times SE) \]If this interval does not include zero, it indicates a statistically significant difference, meaning the body fat percentages are truly different between the groups, and not just due to sampling variability.

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Most popular questions from this chapter

Identify the critical t. An independent random sample is selected from an approximately normal population with unknown standard deviation. Find the degrees of freedom and the critical \(t\) value \(\left(\mathrm{t}^{*}\right)\) for the given sample size and confidence level. (a) \(n=6, \mathrm{CL}=90 \%\) (c) \(n=29, \mathrm{CL}=95 \%\) (b) \(n=21, \mathrm{CL}=98 \%\) (d) \(n=12, \mathrm{CL}=99 \%\)

Work hours and education, Part II. The General Social Survey described in Exercise 4.16 included random samples from two groups: US residents with a college degree and US residents without a college degree. For the 505 sampled US residents with a college degree, the average number of hours worked each week was 41.8 hours with a standard deviation of 15.1 hours. For those 667 without a degree, the mean was 39.4 hours with a standard deviation of 15.1 hours. Conduct a hypothesis test to check for a difference in the average number of hours worked for the two groups.

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Weight gain during pregnancy. In \(2004,\) the state of North Carolina released to the public a large data set containing information on births recorded in this state. This data set has been of interest to medical researchers who are studying the relationship between habits and practices of expectant mothers and the birth of their children. The following histograms show the distributions of weight gain during pregnancy by 867 younger moms (less than 35 years old) and 133 mature moms ( 35 years old and over) who have been randomly sampled from this large data set. The average weight gain of younger moms is 30.56 pounds, with a standard deviation of 14.35 pounds, and the average weight gain of mature moms is 28.79 pounds, with a standard deviation of 13.48 pounds. Calculate a \(95 \%\) confidence interval for the difference between the average weight gain of younger and mature moms. Also comment on whether or not this interval provides strong evidence that there is a significant difference between the two population means.

Does the Paleo diet work? The Paleo diet allows only for foods that humans typically consumed over the last 2.5 million years, excluding those agriculture-type foods that arose during the last 10,000 years or so. Researchers randomly divided 500 volunteers into two equal-sized groups. One group spent 6 months on the Paleo diet. The other group received a pamphlet about controlling portion sizes. Randomized treatment assignment was performed, and at the beginning of the study, the average difference in weights between the two groups was about 0. After the study, the Paleo group had lost on average 7 pounds with a standard deviation of 20 pounds while the control group had lost on average 5 pounds with a standard deviation of 12 pounds. (a) The \(95 \%\) confidence interval for the difference between the two population parameters (Paleo \- control) is given as (-0.891,4.891) . Interpret this interval in the context of the data. (b) Based on this confidence interval, do the data provide convincing evidence that the Paleo diet is more effective for weight loss than the pamphlet (control)? Explain your reasoning. (c) Without explicitly performing the hypothesis test, do you think that if the Paleo group had lost 8 instead of 7 pounds on average, and everything else was the same, the results would then indicate a significant difference between the treatment and control groups? Explain your reasoning.
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