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Medical: Plasma Compress At Community Hospital, the burn center is experimenting with a new plasma compress treatment. A random sample of \(n_{1}=316\) patients with minor burns received the plasma compress treatment. Of these patients, it was found that 259 had no visible scars after treatment. Another random sample of \(n_{2}=419\) patients with minor burns received no plasma compress treatment. For this group, it was found that 94 had no visible scars after treatment. Let \(p_{1}\) be the population proportion of all patients with minor burns receiving the plasma compress treatment who have no visible scars. Let \(p_{2}\) be the population proportion of all patients with minor burns not receiving the plasma compress treatment who have no visible scars. (a) Check Requirements Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation Explain the meaning of the confidence interval found in part (b) in the context of the problem. Does the interval contain numbers that are all positive? all negative? both positive and negative? At the \(95 \%\) level of confidence, does treatment with plasma compresses seem to make a difference in the proportion of patients with visible scars from minor burns?

Step by step solution

01

Define Sample Proportions

First, we define the sample proportions for both groups. For the plasma compress treatment group, the sample proportion is \( \hat{p}_1 = \frac{259}{316} \approx 0.8196 \). For the non-treatment group, the sample proportion is \( \hat{p}_2 = \frac{94}{419} \approx 0.2243 \).

02

Check Normal Approximation Requirements

To use the normal approximation for the difference in sample proportions, we check if \( n_1\hat{p}_1(1-\hat{p}_1) \geq 5 \) and \( n_2\hat{p}_2(1-\hat{p}_2) \geq 5 \). This evaluates to \( 316 \times 0.8196 \times 0.1804 \approx 46.42 \) and \( 419 \times 0.2243 \times 0.7757 \approx 72.34 \). Both conditions are satisfied, so a normal approximation is appropriate.

03

Calculate the Standard Error

The standard error of the difference in sample proportions is given by \[ SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \].Substitute the values to get: \[ SE = \sqrt{\frac{0.8196 \times 0.1804}{316} + \frac{0.2243 \times 0.7757}{419}} \approx 0.0332 \].

04

Calculate the 95% Confidence Interval

The confidence interval for \( \hat{p}_1 - \hat{p}_2 \) at a 95% confidence level is given by \[ (\hat{p}_1 - \hat{p}_2) \pm Z \times SE \],where \( Z \approx 1.96 \) for a 95% confidence level. Thus, the interval is:\[ (0.8196 - 0.2243) \pm 1.96 \times 0.0332 = 0.5953 \pm 0.065 \].This results in the interval \( (0.5303, 0.6603) \).

05

Interpret the Confidence Interval

The 95% confidence interval ranges from 0.5303 to 0.6603. This indicates that patients who receive the plasma compress treatment have a significantly higher proportion of no visible scars by about 53.03% to 66.03% compared to those who do not receive the treatment. Since the interval contains only positive values, it's very likely that the treatment makes a positive difference.

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Key Concepts

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

Normal Approximation

Normal approximation is a statistical technique that uses the normal distribution to estimate the probabilities of a given event, usually for large sample sizes. In this exercise, we are interested in the difference between two sample proportions: those who received the plasma compress treatment and those who did not. To check if the normal distribution can be apt for approximating the distribution of the difference in sample proportions, we need to meet specific requirements.

A normal approximation can be used if the sample size is large enough so that both the number of successes and failures in each group are at least 5. This ensures that the distribution is approximately normal. Mathematically, we check:\[ n_1\hat{p}_1(1-\hat{p}_1) \geq 5 \] and \[ n_2\hat{p}_2(1-\hat{p}_2) \geq 5 \].

In this exercise, these values evaluate to approximately 46.42 and 72.34, respectively, for the treatment and non-treatment groups. Since both values are greater than 5, the condition for normal approximation is satisfied.

This step is crucial because it validates our use of the normal distribution to approximate the distribution of the difference in sample proportions. Without satisfying this condition, the conclusions drawn from the statistical analysis might not be reliable.

Sample Proportions

Sample proportions represent the ratio of a specific outcome within a sample group. They provide insightful information about a population when collecting and analyzing data from a limited subset of the wider population. For this exercise, we calculated sample proportions of patients with no visible scars in two groups.

For the plasma compress treatment group, the sample proportion \( \hat{p}_1 \) is calculated as:

• \( \hat{p}_1 = \frac{259}{316} \approx 0.8196 \)

For the non-treatment group, the sample proportion \( \hat{p}_2 \) is:

• \( \hat{p}_2 = \frac{94}{419} \approx 0.2243 \)

These numbers give us a percentage of patients in each group who had no visible scars after treatment or non-treatment, respectively. By looking at these proportions, we can compare the effectiveness of the plasma compress treatment.

The difference in these proportions helps us understand if and how the treatment is affecting the number of patients with no visible scars. These numbers are the building blocks for further statistical analysis, leading to insights into the effectiveness of the treatment.

Standard Error

Standard error measures the variability or standard deviation of the sampling distribution of a statistic, often the sample mean. It acts as an indicator of how much the sample proportion is expected to fluctuate between different samples. The smaller the standard error, the closer the sample mean is likely to be to the population mean.

For the difference in sample proportions, the standard error is given by:\[ SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \]
Substituting the values from this exercise, we get:\[ SE = \sqrt{\frac{0.8196 \times 0.1804}{316} + \frac{0.2243 \times 0.7757}{419}} \approx 0.0332 \]

The smaller standard error (around 0.0332) tells us that our sample proportions are reliable estimates of the true population proportions. A larger standard error would suggest more fluctuation and hence less reliable estimates. Understanding standard error is crucial because it directly relates to how confident we can be in our statistical inference from the sample to the population.

Difference in Proportions

The difference in proportions is a key statistical concept used to compare the probability or success rate of two different groups. Here, it's used to understand the effect of the plasma compress treatment on the scarring chances.

The difference in the sample proportions is calculated as:\[ \hat{p}_1 - \hat{p}_2 = 0.8196 - 0.2243 = 0.5953 \]

This difference indicates that there is a noticeable improvement, approximately 59.53%, in the outcome (no visible scars) for the treatment group compared to the non-treatment group.

Next, we calculate the 95% confidence interval for this difference using the standard error and a Z-value of approximately 1.96, which is typical for 95% confidence intervals:\[ (0.8196 - 0.2243) \pm 1.96 \times 0.0332 = 0.5953 \pm 0.065 \]

The confidence interval results in (0.5303, 0.6603), meaning we are 95% confident that the true difference in population proportions lies within this range. In the context of this problem, since this interval does not include zero and is entirely positive, we can be quite confident that the plasma compress treatment indeed has a positive effect on reducing visible scars.