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Chapter 5: Problem 123

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 a woman is measured at two visits with two readings per visit. If her true long-term SBP \(=130 \mathrm{mm} \mathrm{Hg}\) then what is the probability that her observed mean SBP is \(\geq 140 \mathrm{mm}\) Hg? (lgnore any continuity correction.) ( Note : By true mean SBP we mean the average SBP over a large number of visits for that subject.)

Short Answer

Expert verified
The probability that her observed mean SBP is greater than or equal to 140 mmHg is approximately 0.0212.

Step by step solution

01

Identify Given Values

The problem provides several key values: \(\sigma_A^2 = 42.9\), \(\sigma^2 = 12.8\), \(n = 2\) (number of visits), \(k = 2\) (readings per visit), and the mean SBP value \(\mu = 130\) mmHg.
02

Calculate Variance Sample Mean

We use the formula for the variance of the sample mean: \(\sigma_w^2 = \frac{\sigma_A^2}{n} + \frac{\sigma^2}{n k}\). Plug in the given values: \[\sigma_w^2 = \frac{42.9}{2} + \frac{12.8}{4} = 21.45 + 3.2 = 24.65.\]
03

Determine Standard Deviation

Calculate the standard deviation of the sample mean by taking the square root of the variance: \[\sigma_w = \sqrt{24.65} \approx 4.96.\]
04

Calculate Z-score

Next, find the Z-score for the observed mean SBP of 140 mmHg using the formula: \[Z = \frac{X - \mu}{\sigma_w},\] where \(X = 140\). Substitute the values: \[Z = \frac{140 - 130}{4.96} \approx 2.016.\]
05

Find Probability from Z-table

Using the standard normal distribution table, find the probability corresponding to a Z-score of 2.016. This is the probability that \(\bar{X}_{n, k} \leq 140\), which is approximately 0.9788.
06

Calculate the Complement Probability

Since we need the probability that the mean SBP \(\geq 140\), take the complement of the Z-table result from 1: \[P(\bar{X}_{n, k} \geq 140) = 1 - 0.9788 = 0.0212.\]

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Key Concepts

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

Blood Pressure Variability
Understanding blood pressure variability is crucial in biostatistics as it provides insights into how blood pressure measurements can change over time and under different conditions. Blood pressure readings can fluctuate for a variety of reasons, often due to factors like stress, activity level, time of day, or even measurement error. To effectively analyze these changes, it's crucial to factor in variability both within a single visit and across different visits.
- **Between-visit variability** ( \(\sigma_A^2\) ) reflects the fluctuations in blood pressure readings between different times the patient comes in for measurements.
- **Within-visit variability** ( \(\sigma^2\) ) accounts for differences in readings that occur during a single visit, perhaps due to changes in the patient鈥檚 condition or the precision of the measurement method.
In the context of the exercise, we calculate the overall variability of repeated blood pressure measurements using a combined variance formula that considers both of these factors. This provides a comprehensive picture of the blood pressure variability for an individual over a determined period.
Normal Distribution
The normal distribution is a fundamental concept in statistics, particularly in biostatistics, as it models natural phenomena efficiently. When an individual鈥檚 blood pressure is analyzed, the assumption often is that these measurements form a normal distribution.
This distribution, sometimes known as the bell curve due to its shape, is characterized by:
  • A symmetric curve where the mean, median, and mode are all located at the peak.
  • The spread of the distribution is defined by its standard deviation.
  • The tails of the distribution approach the horizontal axis asymptotically without ever touching it.
In this exercise, the individual's mean systolic blood pressure (SBP) is thought to be normally distributed around their long-term average. This allows for the application of statistical methods, such as the calculation of z-scores, to determine probabilities related to the blood pressure readings. Knowing that the distribution is normal provides a powerful tool for making inferences about the data and the likelihood of different outcomes.
Z-score Calculation
Z-score calculation is a statistical technique used to determine how many standard deviations a data point is from the mean. This method is particularly useful in biostatistics for comparing individual data points to the overall distribution.
In the context of our blood pressure exercise, after calculating the variance and obtaining the standard deviation ( \(\sigma_w\) = 4.96 for this example), we use these in the Z-score formula: \[Z = \frac{X - \mu}{\sigma_w}\]Where \(X\) is the observed mean SBP of interest (140 mmHg in this case) and \(\mu\) is the mean SBP (130 mmHg).
After substituting the values into the formula, we identify a Z-score of approximately 2.016.
This Z-score tells us how many standard deviations away the observed mean is from the average mean. Using standard normal distribution tables, we can convert this Z-score into a probability, aiding in decision-making and predictions based on the SBP measurements.

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