91Ó°ÊÓ

A previous study found that people consuming large quantities of vegetables containing lutein (mainly spinach) were less likely to develop macular degeneration, a common eye disease among older people (age \(65+)\) that causes a substantial loss in visual acuity and in some cases can lead to total blindness. To follow up on this observation, a clinical trial is planned in which participants 65+ years of age without macular degeneration will be assigned to either a high-dose lutein supplement tablet or a placebo tablet taken once per day. To estimate the possible therapeutic effect, a pilot study was conducted in which 9 people \(65+\) years of age were randomized to placebo and 9 people \(65+\) years of age were randomized to lutein tablets (active treatment). Their serum lutein level was measured at baseline and again after 4 months of follow-up. From previous studies, people with serum lutein \(\geq 10 \mathrm{mg} / \mathrm{dL}\) are expected to get some protection from macular degeneration. However, the level of serum lutein may vary depending on genetic factors, dietary factors, and study supplements. Suppose that among people randomized to placebo, at a 4-month follow-up mean serum lutein level \(=6.4 \mathrm{mg} / \mathrm{dL}\) with standard deviation \(=3 \mathrm{mg} / \mathrm{dL}\). If we presume a normal distribution for serum lutein, then what percentage of placebo subjects will have serum lutein in the therapeutic range \((\geq 10 \mathrm{mg} / \mathrm{dL}) ?\) (For the following problems, assume that lutein can be measured exactly, so that no continuity correction is necessary.)

Step by step solution

01

Understand the Problem

We need to determine what percentage of placebo group participants have serum lutein levels greater or equal to 10 mg/dL, given that these levels are normally distributed with a mean of 6.4 mg/dL and a standard deviation of 3 mg/dL.

02

Calculate the Z-score

The Z-score is given by the formula \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value of interest, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For \(X = 10 \mathrm{mg/dL}\), \(\mu = 6.4 \mathrm{mg/dL}\), and \(\sigma = 3 \mathrm{mg/dL}\), the Z-score is calculated as follows: \[Z = \frac{10 - 6.4}{3} = \frac{3.6}{3} = 1.2\].

03

Find the Probability from Z-score

Use the standard normal distribution table to find the probability that corresponds to a Z-score of 1.2. The table gives the probability that a value is less than 1.2, which is approximately 0.8849.

04

Calculate the Desired Percentage

To find the percentage of subjects with serum lutein \(\geq 10 \mathrm{mg/dL}\), calculate the complement of the probability found in Step 3. So, the percentage is \(1 - 0.8849 = 0.1151\), or 11.51%.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Clinical Trial

A clinical trial is a research study designed to evaluate the effectiveness of medical, surgical, or behavioral interventions. In this context, clinical trials are crucial to understand whether a new treatment or supplement, such as lutein, can prevent diseases like macular degeneration. Trials involve randomly assigning participants to different groups, like a placebo group and an active treatment group, to control for other variables that could influence the outcome.

In our scenario, participants aged 65 and older are either given a high-dose lutein supplement or a placebo. This setup helps ensure any observed differences in health outcomes can be attributed to the supplement rather than other factors. It's an essential step toward confirming whether increased lutein levels can effectively reduce the risk of macular degeneration.

Normal Distribution

Normal distribution, also known as the bell curve, is a statistical concept that describes how values (data points) are distributed around a mean. In a normal distribution, most values cluster around a central peak, and probabilities for values further from the mean taper off symmetrically.

This distribution is critical when analyzing serum lutein levels. Knowing that these levels are normally distributed means we can apply statistical methods to estimate the likelihood of obtaining extreme values, like high lutein levels. In the placebo group, we have a mean lutein level of 6.4 mg/dL with a standard deviation of 3 mg/dL, which means most participants' lutein levels will be close to 6.4 mg/dL, but we can also predict how many might have significantly higher or lower levels.

Macular Degeneration

Macular degeneration is a leading cause of vision loss in older adults. It affects the macula, the part of the retina responsible for clear central vision. Over time, this condition can severely impact one's quality of life, making everyday activities like reading and driving challenging or even impossible.

Age-related macular degeneration (AMD) doesn't have a cure, which is why prevention is vital. Studies suggest that nutritional factors, such as high lutein intake, might offer some protective benefits against AMD. That's why this clinical trial is significant; it explores whether boosting serum lutein through supplements could reduce the incidence of this debilitating disease.

Z-score Calculation

The Z-score is a statistical measure that expresses how many standard deviations a value, like serum lutein, is from the mean of a normal distribution. Calculating a Z-score helps us understand the position of a particular data point relative to the average, allowing for precise probability estimations.

In our case, we calculated the Z-score for a serum lutein level of 10 mg/dL in the placebo group, which has a mean of 6.4 mg/dL and a standard deviation of 3 mg/dL. The formula is \[Z = \frac{X - \mu}{\sigma}\]where \(X = 10\), \(\mu = 6.4\), and \(\sigma = 3\). This calculates to a Z-score of 1.2.

Using a standard normal distribution table, we find that a Z-score of 1.2 corresponds to a probability of 0.8849 for having levels below 10 mg/dL. To find those above it, we take the complement, resulting in approximately 11.51% of placebo subjects reaching the therapeutic range.