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Chapter 5: Problem 70

The change in bone-mineral density of the lumbar spine over a 2-year period among women in the alendronate \(5-\mathrm{mg}\) group was \(+3.5 \%\) (a mean increase), with a standard deviation of \(4.2 \%\). Suppose \(10 \%\) of the women assigned to the alendronate 5 -mg group are actually not taking their pills (noncompliers). If noncompliers are assumed to have a similar response as women in the placebo group, what percentage of women complying with the alendronate 5 -mg treatment would be expected to have a clinically significant decline? (Hint: Use the total-probability rule.)

Short Answer

Expert verified
32.64% of women complying with alendronate 5-mg treatment are expected to have a clinically significant decline in bone density.

Step by step solution

01

Understand the Problem

We need to determine the percentage of women who are actually taking their alendronate pills (compliers) that have a clinically significant decline in bone density, while the data shows a mean increase of 3.5% for the entire group. We will use the information about noncompliers and the rule of total probability to find this percentage.
02

Identify the Components

Assume that among the total population (99) in the alendronate group, 10% are noncompliers. Since noncompliers are assumed to have a similar response as a placebo, we can reason that for noncompliers, the mean change is 0% (no change). Let c_1 = 75% representing the mean change for compliers, while c_2 = 0% for noncompliers. The known mean change for the group is 3.5%.
03

Apply the Total Probability Rule

Use the rule of total probability to calculate the overall mean change: \[\bar{x} = p_1 c_1 + p_2 c_2 = 0.9 * \overline{x}_1 + 0.1 * \overline{x}_2\]Where \(p_1 = 0.9\) and \(p_2 = 0.1\). Given that \(\bar{x} = 3.5\). Solving for \(\overline{x}_1\), the mean change for compliers.
04

Solve for the Compliers Mean Change

Solving the equation: \[3.5 = 0.9\overline{x}_1 + 0.1 0\]Then \[3.5 = 0.9\overline{x}_1\]Continuing from \[\overline{x}_1 = \frac{3.5}{0.9} = 3.89\]Thus, the mean increase in bone density for compliers is 3.89%.
05

Calculate Expected Clinically Significant Decline

A clinically significant decline could be defined as any decline greater than or equal to 2% from the mean for compliers. Calculate the number of standard deviations below the mean needed for a 2% decline given that the mean is 3.89% and the standard deviation is 4.2%.Calculate z-score: \[z = \frac{2 - 3.89}{4.2} \approx \frac{-1.89}{4.2} \approx -0.45\]Find the probability associated with this z-score.
06

Convert Z-score into Probability

Using a standard z-table, a z-score of -0.45 corresponds approximately to 0.3264. This is the probability that a single individual is below a 2% decrease, meaning 32.64% of compliers would have a clinically significant decline.

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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 crucial indicator of bone health. It measures the amount of minerals, primarily calcium and phosphorous, contained in a specific volume of bone. Here鈥檚 why it鈥檚 important: BMD tests can detect osteoporosis, evaluate how well osteoporosis medicine is working, and predict your risk of fractures. Higher bone mineral density means stronger bones, while lower BMD can indicate weaker bones prone to fractures.
In the context of our exercise, BMD is used to assess the effectiveness of the alendronate treatment, a medication used to treat osteoporosis. A significant change in BMD indicates whether the treatment is successfully enhancing bone density, thereby reducing the risk of fractures. The example provided demonstrates a mean increase in BMD, indicating a positive response among women complying with the treatment.
Total Probability Rule
The total probability rule is instrumental in probabilistic analysis, especially when outcomes have different groups or scenarios affecting the probability. This rule helps when outcomes depend on multiple distinct events, and we need to assess the overall likelihood of a particular outcome.
In the problem, this rule assists in determining the mean change in bone-mineral density across compliers and noncompliers. By accounting for different behaviors within the groups (compliers versus noncompliers), the total probability rule allows us to calculate the combined expected mean change for the group.
  • This approach assumes each subgroup has a distinct probability or contribution to the total.
  • Unifies these probabilities to project the overall mean change reflecting both compliant and noncompliant participants.
Using this logical framework gives insight into the true effect of the medication by separating the influence of compliers and noncompliers.
Z-score Analysis
Z-score analysis is a statistical method used to quantify the deviation of a data point from the mean of a dataset. It is a measure that describes a value's relation to the mean of a group of values. The z-score is calculated by subtracting the mean from the data point and dividing the result by the standard deviation.
This method is especially helpful here as it allows us to determine how far the bone-mineral density changes of the compliers deviate from the average. In our exercise, the mean increase was 3.89%, but we needed to understand the likelihood of a decline, which was defined as a reduction beyond a certain threshold.
  • A negative z-score indicates that the bone density change is below the mean, suggesting a decrease rather than an increase.
  • The probability of a significant decline is assessed using this z-score.
The z-score of -0.45 suggests 32.64% of compliers could experience a clinically significant decline. This type of analysis is critical for identifying potential negative outcomes in medical treatments.

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Most popular questions from this chapter

Serum cholesterol is an important risk factor for coronary disease. We can show that serum cholesterol is approximately normally distributed, with mean \(=219 \mathrm{mg} / \mathrm{dL}\) and standard deviation \(=50 \mathrm{mg} / \mathrm{dL}\). If the clinically desirable range for cholesterol is \(<200 \mathrm{mg} / \mathrm{dL},\) what proportion of people have clinically desirable levels of cholesterol?

The duration of cigarette smoking has been linked to many diseases, including lung cancer and various forms of heart disease. Suppose we know that among men ages \(30-34\) who have ever smoked, the mean number of years they smoked is 12.8 with a standard deviation of 5.1 years. For women in this age group, the mean number of years they smoked is 9.3 with a standard deviation of \(3.2 .\) Assuming that the duration of smoking is normally distributed, what proportion of men in this age group have smoked for more than 20 years?

The study is expanded to include 50 medical interns, of whom 11 report having had an automobile accident over the past year. One issue in the above study is that not all people report automobile accidents to the police. The NHTSA estimates that only half of all auto accidents are actually reported. Assume this rate applies to interns. What is the sum of the 40th and 60th percentiles of a normal distribution with a mean \(=8.2\) and variance \(=9.5 ?\)

Blood pressure readings are known to be highly variable. Suppose we have mean SBP for one individual over \(n\) visits with \(k\) readings per visit \(\left(\bar{X}_{n, k}\right) .\) The variability of \(\left(\bar{X}_{n, k}\right)\) depends on \(n\) and \(k\) and is given by the formula \(\sigma_{w}^{2}=\sigma_{A}^{2} / n+\) \(\sigma^{2} /(n k),\) where \(\sigma_{A}^{2}=\) between visit variability and \(\sigma^{2}=\) within visit variability. For \(30-\) to 49 -year-old Caucasian females, \(\sigma_{A}^{2}=42.9\) and \(\sigma^{2}=12.8 .\) For one individual, we also assume that \(\bar{X}_{n, k}\) is normally distributed about their true long- term mean \(=\mu\) with variance \(=\sigma_{w}^{2}\). Suppose we want to observe the woman over \(n\) visits, where \(n\) is sufficiently large so that there is less than a \(5 \%\) chance that her observed mean SBP will not differ from her true mean SBP by more than \(5 \mathrm{mm}\) Hg. What is the smallest value of \(n\) to achieve this goal? ( Note : Assume two readings per visit.)

It is also known that over a large number of 30 - to 49 -year-old Caucasian women, their true mean SBP is normally distributed with mean \(=120 \mathrm{mm}\) Hg and standard deviation \(=14\) mm Hg. Also, over a large number of African American 30- to 49-year-old women, their true mean SBP is normal with mean \(=130 \mathrm{mm} \mathrm{Hg}\) and standard deviation \(=20 \mathrm{mm} \mathrm{Hg}\). Suppose we select a random \(30-\) to 49 -year-old Caucasian woman and a random 30 - to 49 -year-old African American woman. What is the probability that the African American woman has a higher true SBP?
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