91Ó°ÊÓ

According to The American Heart Association in \(2011,\) about 1 in every 3 deaths in the U.S. was from cardiovascular diseases. (Source: Data from www.heart.org, December \(17,2014 .)\) a. For a random sample of 100 deaths in America in 2011 , find the mean and standard deviation of the proportion of those who died from cardiovascular diseases. b. In a sample of 100 deaths observed in a specific hospital in the U.S. in \(2011,\) half of the deaths were from cardiovascular diseases. Would this have been a surprising result if the sample were a random sample of Americans? Answer by finding how many standard deviations that sample result falls from the mean of the sampling distribution of the proportion of 100 deaths from cardiovascular diseases. c. In part b, identify the population distribution, the data distribution, and the sampling distribution of the sample proportion.

Step by step solution

01

Find the Mean Proportion

Given that 1 in 3 deaths are due to cardiovascular diseases, the probability (P) of dying from cardiovascular diseases is \(P = \frac{1}{3}\). For a random sample of \(n = 100\) deaths, the mean of the sample proportion \(\hat{p}\) is given by \(\mu_{\hat{p}} = P = \frac{1}{3}\approx 0.3333\).

02

Calculate the Standard Deviation

The standard deviation of the sample proportion \(\hat{p}\) is calculated using the formula: \(\sigma_{\hat{p}} = \sqrt{\frac{P(1-P)}{n}}\). Substitute \(P = \frac{1}{3}\) and \(n = 100\) to get \(\sigma_{\hat{p}} = \sqrt{\frac{\frac{1}{3}(\frac{2}{3})}{100}} = \sqrt{\frac{2}{900}} = \sqrt{\frac{1}{450}} \approx 0.047\).

03

Evaluate the Sample Result

The sample proportion in part b is 0.5 since half of the 100 observed deaths were from cardiovascular diseases. Determine how many standard deviations this is from the mean using the formula: \(z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}}\). Substituting the values, we get \(z = \frac{0.5 - 0.3333}{0.047} \approx 3.55\).

04

Identify the Distributions

The population distribution is the binomial distribution of all death causes, with a probability of \(\frac{1}{3}\) for cardiovascular deaths. The data distribution in the hospital sample is a sample proportion of 0.5 for cardiovascular deaths out of 100. The sampling distribution of the sample proportion, \(\hat{p}\), is approximately normally distributed because of the Central Limit Theorem, with mean \(0.3333\) and standard deviation \(0.047\).

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Key Concepts

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

Probability

The concept of probability is at the core of statistics and helps us understand the likelihood of various outcomes. In this exercise, the probability of a death being caused by cardiovascular diseases is essential. We define probability as a number between 0 and 1 that represents how likely an event is to occur. For example, the probability of a death due to cardiovascular diseases in this context is \(\frac{1}{3}\), meaning that in every 3 cases, 1 is likely due to these diseases. This probability value can also be interpreted as 33.33%.

• The probability underpins calculations of expected outcomes, such as the mean of a sampling distribution.
• It also determines the standard deviation by illustrating the variation expected around the mean.

Understanding probability allows you to predict outcomes and measure how likely a surprising event, such as 50% of deaths from cardiovascular diseases in a particular sample, would be.

Sampling Distribution

A sampling distribution is the probability distribution of a statistic obtained from a larger number of samples drawn from a specific population. Here, the focus is on the sample proportion of deaths due to cardiovascular diseases. Each sample mean (proportion) represents a sampling distribution even though individual samples have different sample means.

• In this exercise, the sampling distribution refers to the distribution of all possible sample proportions of deaths from cardiovascular diseases when drawing multiple samples of 100 deaths.
• This distribution allows us to compute both the mean, which we found to be approximately 0.3333, and the standard deviation, calculated to be approximately 0.047.

These calculations are critical as they allow us to determine how extreme a particular sample's results are compared to the expected results of the wider population.

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental statistical principle that explains why sampling distributions tend to be normal. Regardless of the shape of the population distribution, the theorem states that the sampling distribution of the sample mean will be normally distributed if the sample size is sufficiently large. In our problem,

• The Central Limit Theorem supports the assumption that the sampling distribution of the proportion is approximately normal since the sample size, n = 100, is sufficiently large.
• The CLT allows us to perform z-score calculations even though the original population distribution may be binomial.
• This principle helps us assess how typical or atypical a particular sample outcome is, such as observing half of the deaths in our sample due to cardiovascular diseases.

Applying CLT simplifies working with sampling distributions, enabling us to use normal distribution properties to compute probabilities and z-scores.

Binomial Distribution

The binomial distribution is a discrete probability distribution that models the number of successful outcomes in a fixed number of trials. It's applicable when each trial in an experiment has two possible outcomes: success or failure. In this context,

• The binomial distribution models the probability of a specific number of deaths being due to cardiovascular diseases in a sample of 100.
• The population distribution of all deaths, mentioned in step 4 of the solution, follows a binomial distribution with each death being considered a "success" (cardiovascular related) with a probability of \(\frac{1}{3}\).
• Understanding this helps to determine the expected number of successes and to compute variance and probabilities for categories of outcomes, reinforcing assumptions about sampling distributions.

This understanding assists in calculating the expected sample proportion as we can relate these probabilities to real-life scenarios, like determining how unusual the hospital's observations are.