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Standard deviation of a proportion Suppose \(x=1\) with probability \(p,\) and \(x=0\) with probability \((1-p) .\) Then, \(x\) is the special case of a binomial random variable with \(n=1,\) so that \(\sigma=\sqrt{n p(1-p)}=\sqrt{p(1-p)} .\) With \(n\) trials, using the formula \(\sigma / \sqrt{n}\) for a standard deviation of a sample mean, explain why the standard deviation of a sample proportion equals \(\sqrt{p(1-p) / n}\)

Short Answer

Expert verified
The standard deviation of a sample proportion is \(\sqrt{\frac{p(1-p)}{n}}\) because it's derived from the binomial setup and applies the Central Limit Theorem to \(n\) trials.

Step by step solution

01

Understanding the binomial setup

In this setup, we have a binomial random variable where the outcome can be either 1 with probability \(p\) or 0 with probability \(1-p\). Since this is a special case of a binomial distribution where \(n=1\), the standard deviation, \(\sigma\), for a single trial is \(\sqrt{p(1-p)}\).
02

Applying the Central Limit Theorem

When considering multiple trials, specifically \(n\) independent trials, the distribution of the average of these outcomes becomes important. By the Central Limit Theorem, the distribution of the sample mean approaches a normal distribution with mean \(p\) and variance \(\frac{\sigma^2}{n}\).
03

Calculating the sample standard deviation

In the context of multiple trials, we are interested in the standard deviation of the proportion of successes. The standard deviation of the sample mean (or average proportion of successes) is given by \(\frac{\sigma}{\sqrt{n}}\). Since we earlier established \(\sigma = \sqrt{p(1-p)}\), substituting this gives us the sample standard deviation as \(\sqrt{\frac{p(1-p)}{n}}\).

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Key Concepts

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

Binomial Random Variable
A binomial random variable arises from a situation where there are only two possible outcomes: success or failure. In our example, success is represented by the variable \(x = 1\) and failure by \(x = 0\). When the experiment is conducted once—meaning only a single trial occurs—we have \(n=1\). This special case results in the random variable simplifying to a simple situation labeled as a Bernoulli trial. This leads to the simplification of the standard deviation formula to: \(\sigma = \sqrt{p(1-p)}\). Here, \(p\) is the probability of success and \(1-p\) represents the probability of failure.

Understanding these basic components lays the groundwork for tackling other concepts like the Central Limit Theorem and sample proportions, which rely on grasping how these probabilities of success and failure behave across numerous trials.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that the distribution of sample means of a sufficiently large number of independent, identically distributed random variables, tends toward a normal distribution, regardless of the original distribution of the variables. This is especially powerful because it allows us to make inferences about population parameters using sample data.

In the context of a binomial random variable, when we consider \(n\) independent trials, CLT helps predict that the distribution of the sample mean will approximate a normal distribution. This approximation improves as the number of trials \(n\) increases. The mean remains \(p\), while the variance becomes \(\frac{\sigma^2}{n}\). By using the Central Limit Theorem, we can then conclude that the standard deviation of this sample mean is \(\frac{\sigma}{\sqrt{n}}\), allowing us to estimate population parameters with greater precision.
Sample Proportion
The concept of sample proportion is closely tied to the idea of estimating the probability of success in a population based on a subset, or sample, of that population. In our exercise, we examine the proportion of successes in \(n\) trials. The key here is how the individual probability estimates translate when we seek to predict the larger population behavior.

For sample proportions, the standard deviation derived through the Central Limit Theorem turns out to be \(\sqrt{\frac{p(1-p)}{n}}\). This is the adjusted calculation for understanding deviations in a sample proportion rather than a single trial. It accounts for the structural change when moving from a single observation to multiple observations (in our case, multiple trials), ensuring that our statistical inferences remain reliable as \(n\) grows. This insight allows statisticians to use sample data to estimate population parameters, making predictions not only feasible but also statistically sound.

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Most popular questions from this chapter

Other scenario for exit poll Refer to Examples 1 and 2 about the exit poll, for which the sample size was \(3889 . \mathrm{In}\) that election, \(40.9 \%\) voted for Whitman. a. Define a binary random variable \(X\) taking values 0 and 1 that represents the vote for a particular voter \((1=\) vote for Whitman and \(0=\) another candidate \()\) State its probability distribution, which is the same as the population distribution for \(X\). b. Find the mean and standard deviation of the sampling distribution of the proportion of the 3889 people in the sample who voted for Whitman.

What is a sampling distribution? How would you explain to someone who has never studied statistics what a sampling distribution is? Explain by using the example of polls of 1000 Canadians for estimating the proportion who think the prime minister is doing a good job.

Multiple choice: Standard deviation Which of the following is not correct? The standard deviation of a statistic describes a. The standard deviation of the sampling distribution of that statistic. b. The standard deviation of the sample data measurements. c. How close that statistic falls to the parameter that it estimates. d. The variability in the values of the statistic for repeated random samples of size \(n\).

Multiple choice: CLT The central limit theorem implies a. All variables have approximately bell-shaped data distributions if a random sample contains at least about 30 observations. b. Population distributions are normal whenever the population size is large. c. For sufficiently large random samples, the sampling distribution of \(\bar{x}\) is approximately normal, regardless of the shape of the population distribution. d. The sampling distribution of the sample mean looks more like the population distribution as the sample size increases.

Purpose of sampling distribution You'd like to estimate the proportion of all students in your school who are fluent in more than one language. You poll a random sample of 50 students and get a sample proportion of 0.12. Explain why the standard deviation of the sampling distribution of the sample proportion gives you useful information to help gauge how close this sample proportion is to the unknown population proportion.

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