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Chapter 5: Problem 68

A study compared different treatments for preventing bone loss among postmenopausal women younger than 60 years of age \([6] .\) The mean change in bone-mineral density of the lumbar spine over a 2 -year period for women in the placebo group was \(-1.8 \%\) (a mean decrease), with a standard deviation of \(4.3 \% .\) Assume the change in bone-mineral density is normally distributed. 5.68 If a decline of \(2 \%\) in bone-mineral density is considered clinically significant, then what percentage of women in the placebo group can be expected to show a decline of at least this much?

Short Answer

Expert verified
Approximately 48.14% of the women in the study may show a decline of at least 2%.

Step by step solution

01

Define the problem

We need to find the percentage of women whose bone-mineral density declined by at least 2% over two years. Since decline is negative from a starting point of 0, we are interested in finding where the change is less than or equal to -2%.
02

Standardize the value

For normally distributed data, we can use the z-score formula to standardize the target value: \( Z = \frac{X - \mu}{\sigma} \), where \( X = -2\% \), \( \mu = -1.8\% \), and \( \sigma = 4.3\% \). Substituting these values gives: \[ Z = \frac{-2 - (-1.8)}{4.3} = \frac{-0.2}{4.3} \approx -0.0465 \]
03

Use the standard normal distribution

Next, we will use the standard normal distribution to find the probability that corresponds to the z-score we calculated. Since we are looking for a decline of at least 2%, we find the probability of getting a z-score less than or equal to -0.0465.
04

Calculate the percentage

Using the standard normal (z) table or a calculator, find the probability corresponding to \( Z = -0.0465 \). This gives a cumulative probability of about 0.4814. This value represents the percentage of women whose bone-mineral density decreased by at least 2%.
05

Interpret the result

About 48.14% of women in the placebo group are expected to show a decline in bone-mineral density of at least 2%.

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Key Concepts

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

Bone-mineral density
Bone-mineral density (BMD) is a measurement that reflects the strength and density of bones. It is a crucial indicator of bone health, especially in postmenopausal women, who are at a higher risk of osteoporosis due to hormonal changes.
  • BMD is typically measured using a dual-energy X-ray absorptiometry (DEXA) scan. This scan produces a quantitative reading of bone mineral content per unit area, expressed in terms of percentage change.
  • In studies, changes in BMD are often used to assess the effectiveness of treatments aimed at preventing bone loss.
Understanding BMD helps healthcare providers make informed decisions regarding osteoporosis treatment and prevention.
A decline in BMD suggests weakened bones and a higher likelihood of fractures, so tracking changes over time is essential for maintaining bone health.
Standard deviation
Standard deviation is a statistical measure of how much variation exists from the average (mean) in a dataset. It's a crucial concept in biostatistics and helps us understand the dispersion of data points.
  • A high standard deviation means the data points are spread out over a wide range of values.
  • A low standard deviation indicates the data points tend to be very close to the mean.
In the context of the BMD study, the standard deviation of 4.3% indicates the typical amount by which the percentage changes in bone-mineral density vary from the mean change of -1.8% over two years.
This variability can influence how healthcare providers and researchers interpret the effectiveness of different treatments.
Normal distribution
The normal distribution is a pivotal concept in statistics, often referred to as the "bell curve" due to its characteristic shape. It describes how values of a dataset are expected to distribute themselves around the mean.
  • In a normal distribution, most data points are clustered around the mean, with fewer data points appearing as you move towards the tails.
  • The properties of the normal distribution allow us to make predictions about data behavior.
In the BMD study, the assumption that changes are normally distributed enables researchers to use tools like the z-score to predict outcomes, like the percentage of women experiencing a specific level of bone density change.
Recognizing when data follows a normal distribution is key to applying various statistical methods accurately.
Z-score
A z-score is a statistical measure that quantifies the number of standard deviations a data point is from the mean. It is a useful tool for standardizing different datasets so they can be compared fairly.
  • A z-score tells us how far and in what direction the data point lies from the mean, in terms of standard deviations.
  • The z-score formula is: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
In the BMD study, a z-score of approximately -0.0465 was found for a decline of 2%. This z-score tells us how this specific value compares to the mean change of -1.8% in bone-mineral density.
Once calculated, the z-score can be used with standard normal distribution tables or calculators to find the probability of a specific outcome, making it a powerful tool in biostatistics.

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