**Problem 4: (18 points)** A treatment trial is proposed to test the efficacy of vitamin E as a preventive agent for cancer. One problem with such a study is how to assess compliance among participants. A small pilot study is undertaken to establish criteria for compliance with the proposed study agents. In this study, 100 patients were given 400 IU/day of vitamin E, and 100 patients were given similar-sized tablets of placebo over a 3-month period. Their serum vitamin-E levels are measured before and after the 3-month period, and the change (3-month – baseline) is distributed with a Normal distribution in both treatment and control groups with the following parameters:

Mean (µ)

Standard deviation (σ)

Treatment

0.80

0.48

Control

0.05

0.16

Suppose we consider the serum vitamin-E assay a screening test for compliance with vitamin-E supplementation. This study aims to identify a threshold Δ, such that participants whose change in serum vitamin E is ≥ Δ mg/dL will be considered vitamin-E takers, and participants whose change is < Δ mg/dL will be considered placebo takers.

Suppose a change of 0.30 mg/dL in serum levels is proposed as a test criterion for compliance. What percentage of the vitamin-E group would be expected to show a change of at least 0.30 mg/dL? **(3 points)**

What percentage of the placebo group would be expected to show a change of not more than 0.30 mg/dL? **(3 points)**

To solve this problem, we need to calculate the percentage of participants in both the treatment (vitamin-E) and control (placebo) groups that meet the given threshold change in serum vitamin-E levels. This can be done using the properties of the normal distribution.

### 1. Percentage of the vitamin-E group showing a change of at least 0.30 mg/dL:

The change in serum vitamin-E levels in the treatment group is normally distributed with a mean of \( \mu = 0.80 \) mg/dL and a standard deviation of \( \sigma = 0.48 \) mg/dL.

We need to find the probability that the change in serum levels is at least \( 0.30 \) mg/dL. This means we want to calculate:

\[
P(X \geq 0.30)
\]

To do this, we first compute the z-score corresponding to a change of 0.30 mg/dL using the formula:

\[
z = \frac{X - \mu}{\sigma}
\]

For \( X = 0.30 \), \( \mu = 0.80 \), and \( \sigma = 0.48 \), the z-score is:

\[
z = \frac{0.30 - 0.80}{0.48} = \frac{-0.50}{0.48} \approx -1.04
\]

Using the z-table, the probability of getting a z-score less than -1.04 is approximately 0.1492, which represents the proportion of the vitamin-E group with a change less than 0.30 mg/dL.

Thus, the percentage of participants showing a change of at least 0.30 mg/dL is:

\[
P(X \geq 0.30) = 1 - P(X < 0.30) = 1 - 0.1492 = 0.8508
\]

So, about **85.08%** of the vitamin-E group would be expected to show a change of at least 0.30 mg/dL.

### 2. Percentage of the placebo group showing a change of not more than 0.30 mg/dL:

The change in serum vitamin-E levels in the placebo group is normally distributed with a mean of \( \mu = 0.05 \) mg/dL and a standard deviation of \( \sigma = 0.16 \) mg/dL.

We need to find the probability that the change in serum levels is no more than \( 0.30 \) mg/dL. This means we want to calculate:

\[
P(X \leq 0.30)
\]

First, we compute the z-score for \( X = 0.30 \) mg/dL:

\[
z = \frac{0.30 - 0.05}{0.16} = \frac{0.25}{0.16} \approx 1.56
\]

Using the z-table, the probability corresponding to a z-score of 1.56 is approximately 0.9406.

Thus, the percentage of the placebo group showing a change of not more than 0.30 mg/dL is:

\[
P(X \leq 0.30) = 0.9406
\]

So, about **94.06%** of the placebo group would be expected to show a change of not more than 0.30 mg/dL.

### Summary:

- **85.08%** of the vitamin-E group would be expected to show a change of at least 0.30 mg/dL.
- **94.06%** of the placebo group would be expected to show a change of not more than 0.30 mg/dL.

Would you like more details on any part of this process, or have further questions?

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Here are 5 related questions to deepen understanding:

1. How would the percentages change if the test criterion was raised to 0.50 mg/dL?
2. How is the z-score related to standard deviations in the normal distribution?
3. What would happen to the placebo group percentages if their mean serum vitamin-E level was higher?
4. How could we compute the percentage of false positives or false negatives in this scenario?
5. What assumptions do we make about the normality of the distributions?

**Tip:** When dealing with normal distributions, calculating z-scores helps you quickly find probabilities using standard tables or software.

A treatment trial is proposed to test the efficacy of vitamin E as a preventive agent for cancer. One problem with such a study is how to assess compliance among participants. A small pilot study is undertaken to establish criteria for compliance with the proposed study agents. In this study, 100 patients were given 400 IU/day of vitamin E, and 100 patients were given similar-sized tablets of placebo over a 3-month period. Their serum vitamin-E levels are measured before and after the 3-month period, and the change (3-month – baseline) is distributed with a Normal distribution in both treatment and control groups with the following parameters: Mean (µ) and Standard deviation (σ) for Treatment (0.80, 0.48) and Control (0.05, 0.16). What percentage of the vitamin-E group would be expected to show a change of at least 0.30 mg/dL? What percentage of the placebo group would be expected to show a change of not more than 0.30 mg/dL?

In this problem, we calculate the percentage of participants showing a change in serum vitamin-E levels using a threshold of 0.30 mg/dL. We use the normal distribution and z-scores to determine the expected outcomes for both treatment and placebo groups. The solution provides step-by-step calculations and is suitable for university-level students studying statistics.

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