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Experimental medication \(\quad\) As part of a drug research study, individuals suffering from arthritis take an experimental pain relief medication. Suppose that \(25 \%\) of all individuals who take the new drug experience a certain side effect. For a given individual, let \(X\) be either 1 or 0 , depending on whether \(\mathrm{s} / \mathrm{he}\) experienced the side effect or not, respectively. a. If \(n=3\) people take the drug, find the probability distribution of the proportion who will experience the side effect. b. Referring to part a, what are the mean and standard deviation of the sample proportion? c. Repeat part b for a group of \(n=10\) individuals; \(n=100\). What happens to the mean and standard deviation of the sample proportion as \(n\) increases?

Short Answer

Expert verified
Probability distribution for three people: mean = 0.25, std dev ~ 0.236. Larger \(n\) decreases std dev, mean remains 0.25.

Step by step solution

01

Understanding the Probability Distribution

Given that 25% of individuals experience a side effect, we're looking at a binomial distribution. For a sample of size \(n=3\), we define the random variable \(X\) to count the number of individuals experiencing the side effect. Thus, \(X\) follows a Binomial distribution \(X \sim B(n=3, p=0.25)\). We aim to find the probability distribution of the sample proportion \(\hat{P} = \frac{X}{n}\).
02

Calculate Probability Distribution for n=3

For \(n=3\), possible outcomes for \(X\) are 0, 1, 2, and 3. The probability mass function is given by \(P(X=k) = \binom{3}{k} p^k (1-p)^{3-k}\) where \(k = 0,1,2,3\). Calculate: - \(P(X=0) = \binom{3}{0} (0.25)^0 (0.75)^3 = 0.421875\) - \(P(X=1) = \binom{3}{1} (0.25)^1 (0.75)^2 = 0.421875\) - \(P(X=2) = \binom{3}{2} (0.25)^2 (0.75)^1 = 0.140625\) - \(P(X=3) = \binom{3}{3} (0.25)^3 (0.75)^0 = 0.015625\)
03

Calculate Mean and Standard Deviation for n=3

The mean of \(\hat{P}\) for a binomial distribution is given by \(\mu = p = 0.25\). The standard deviation is \(\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.25*0.75}{3}} \approx 0.236\).
04

Repeat Calculation for n=10 and n=100

Calculate mean and standard deviation for \(n=10\) and \(n=100\). For \(n=10\), - Mean: \(\mu = 0.25\)- Standard deviation: \(\sigma = \sqrt{\frac{0.25*0.75}{10}} \approx 0.137\) For \(n=100\),- Mean: \(\mu = 0.25\) - Standard deviation: \(\sigma = \sqrt{\frac{0.25*0.75}{100}} = 0.0433\)
05

Analyze Effect of Increasing n

Observing that the mean remains constant as \(n\) increases, but the standard deviation decreases, it implies that larger samples provide more precise estimations of \(p\) (less variability around the mean). Thus, increasing \(n\) results in more reliable predictions of the proportion of individuals experiencing side effects.

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Key Concepts

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

Probability Distribution
A probability distribution is essentially a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. In the context of our exercise, we're dealing with a binomial distribution. This is because the outcomes for each individual are binary: they either experience the side effect or they do not.

For a binomial distribution characterized by parameters \(n\) and \(p\), \(n\) represents the number of trials or individuals, and \(p\) signifies the success probability. When you calculate the probability distribution of the sample proportion \(\hat{P} = \frac{X}{n}\), where \(X\) is the number of successes (i.e., individuals experiencing the side effect), you create a probability distribution of possible outcomes of the sample proportion.

This type of probability distribution helps us understand how likely different proportions are when a certain number of individuals are affected by the side effect, which is valuable for evaluating the impact of the medication.
Sample Proportion
The sample proportion, denoted as \(\hat{P}\), is a statistic that estimates the true proportion of a population that exhibits a certain characteristic. In our scenario, \(\hat{P}\) is used to estimate the proportion of individuals experiencing side effects after taking a medication.

This statistic is crucial because it gives us a tangible representation of the phenomenon we're observing in samples of varying sizes (e.g., when \(n=3\), \(n=10\), or \(n=100\)). By calculating \(\hat{P}\) for different outcomes and sample sizes, we derive practical insights into the behavior of the medication's side effects across the population.

As sample size increases, the sample proportion becomes a more accurate estimate of the true population proportion. This is captured in our exercise by observing how \(\hat{P}\) changes with different values of \(n\). Larger samples tend to result in sample proportions that are closer, on average, to the population proportion \(p\).
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In binomial distributions, it tells us how much the sample proportion \(\hat{P}\) is expected to deviate from the actual population proportion \(p\).

For a binomial distribution, the standard deviation is computed using the formula \(\sigma = \sqrt{\frac{p(1-p)}{n}}\). This formula reveals that the standard deviation depends on both the probability of success \(p\) (in this case, the probability of experiencing a side effect) and the sample size \(n\).

As we found in the exercise, increasing \(n\) decreases the standard deviation. This indicates that with larger sample sizes, the variability or spread around the mean decreases, suggesting that we obtain more reliable estimates of the true population proportion.
Mean of Binomial Distribution
The mean of a binomial distribution provides a central value or the average anticipated outcome if the experiment were to be repeated many times. For our exercise, the mean of the sample proportion \(\hat{P}\) simplifies to the probability of an individual experiencing the side effect, which is \(p = 0.25\).

This mean \(\mu = p\) remains constant irrespective of the sample size \(n\). What this tells us is that no matter how many individuals are in the sample, the expected proportion that experiences the side effect is always 25%.

Understanding the mean provides critical insight into the expected behavior of the medication, allowing healthcare professionals to predict outcomes and make informed decisions when applying this medication to larger populations. Thus, the mean serves as a foundational benchmark for evaluating the experimental drug's impact.

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Most popular questions from this chapter

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