91Ó°ÊÓ

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) 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) 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

Identify Sample Proportions

First, we need to calculate the sample proportions for both groups. \( \hat{p}_1 \) is the proportion of patients with no visible scars in the treatment group: \( \hat{p}_1 = \frac{259}{316} \approx 0.8196 \). For the control group, \( \hat{p}_2 = \frac{94}{419} \approx 0.2243 \).

02

Check Normal Approximation Conditions

To use a normal distribution to approximate the distribution of \( \hat{p}_1 - \hat{p}_2 \), both \( n_1\hat{p}_1, n_1(1-\hat{p}_1), n_2\hat{p}_2, \text{ and } n_2(1-\hat{p}_2) \) must be greater than 5. Compute these: \( 316 \times 0.8196 = 259.0016, \ 316 \times 0.1804 = 56.9984, \ 419 \times 0.2243 = 93.9707, \) and \( 419 \times 0.7757 = 325.0293. \) All are greater than 5, so we can use a normal approximation.

03

Calculate Standard Error

The standard error for the difference \( \hat{p}_1 - \hat{p}_2 \) is calculated as \( SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}. \) Substituting the values, \( SE \approx \sqrt{\frac{0.8196 \times 0.1804}{316} + \frac{0.2243 \times 0.7757}{419}} \approx 0.0301. \)

04

Find the Confidence Interval

Use the standard normal distribution to find the confidence interval for \( p_1 - p_2 \). At a \( 95\% \) confidence level, the critical value \( z \approx 1.96. \) Thus, the confidence interval is \( 0.8196 - 0.2243 \pm 1.96 \times 0.0301 \approx (0.5554, 0.6258). \)

05

Interpret the Confidence Interval

The confidence interval for \( p_1 - p_2 \) is \((0.5554, 0.6258)\), which is entirely positive. This implies that the proportion of patients with no visible scars is higher for those treated with the plasma compress than for those who were not treated. At a 95% confidence level, treatment with plasma compress does appear to significantly reduce the incidence of visible scars.

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.

Confidence Intervals

Understanding confidence intervals is crucial for interpreting statistical inference results. A confidence interval gives an estimated range of values which is likely to include an unknown population parameter. It is calculated based on the sample data. In our exercise, the confidence interval helps us understand the difference between the treatment and control groups in terms of the proportion of patients with no visible scars.

To calculate a confidence interval, first, you determine the sample proportions and then compute the standard error, which reflects the precision of your sample estimate. Using the standard normal distribution, particularly the critical value - for common confidence levels (like 95%), this critical value is typically around 1.96 - you finally deduce the interval. This interval expresses the range within which the true population difference between treated and untreated groups likely falls.

• A 95% confidence interval means you are 95% confident that the true difference lies within the calculated interval.
• In simpler terms, if you were to take 100 different samples and compute a confidence interval for each sample, about 95 of those intervals would include the true population parameter.
• It helps determine the statistical significance of your findings.

In this example, the confidence interval for the difference, entirely positive, suggests a substantial effect of the treatment.

Normal Distribution Approximation

The normal distribution is a powerful tool when approximating distributions of sample data, especially when dealing with proportions. For two independent groups as in this problem, it allows us to approximate the distribution of the difference between two proportions because it is computationally simpler and more intuitive than working with more complex probability distributions.

A key assumption when using the normal distribution approximation is the sample size condition. In statistics, for a distribution of a sample proportion to approximate normality adequately, certain criteria should be satisfied:

• Both the expected number of successes and failures (in each group) should be at least 5.
• This criterion ensures that the sample is large enough for the binomial distribution of the sample proportion to be close to the normal distribution.

In our specific case, we verified that these conditions were met for both the plasma compress treatment group and the control group. This validation is crucial for justifying the use of a normal approximation when evaluating the difference between the two sample proportions.

Population Proportions

Population proportions reflect the ratio of members in a population that have a particular attribute. In this exercise, we're dealing with the population proportions of patients with and without visible scars after treatment, whether they received plasma compress treatment or not.

The two proportions of interest here are:

• The proportion of treated patients with no visible scars, denoted as \( p_1 \).
• The proportion of untreated patients with no visible scars, denoted as \( p_2 \).

Estimating population proportions from sample data is a common task in statistical inference. A larger proportion of treated patients without visible scars indicates a potential benefit of the treatment, while a lower proportion in the control group reinforces this finding.

The calculation of these sample proportions was the first step in our solution, using straightforward division of the number of favorable outcomes by the total sample size. Understanding these proportions helps interpret how effective the treatment is compared to not treating the patients. Observing a significant difference between these proportions, as evidenced by a positive confidence interval, suggests that the plasma compress treatment is effective.