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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.

Short Answer

Expert verified
The mean of the sampling distribution is 0.409 and the standard deviation is approximately 0.0078.

Step by step solution

01

Define Binary Random Variable

Here, each voter in the sample either votes for Whitman or not. We define a binary random variable \(X\) such that \(X = 1\) if a voter votes for Whitman, and \(X = 0\) if the voter votes for another candidate. The probability distribution is given by \(P(X = 1) = 0.409\) and \(P(X = 0) = 1 - 0.409 = 0.591\). This represents the population distribution for \(X\).
02

Determine the Mean of Sampling Distribution

The sample size \(n\) is 3889. The mean of the sampling distribution of the proportion \(\hat{p}\) of voters who voted for Whitman is the population proportion \(p\). Thus, \(\mu_{\hat{p}} = p = 0.409\).
03

Calculate the Standard Deviation of the Sampling Distribution

The standard deviation of the sampling distribution of \(\hat{p}\) is given by \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\). Substituting \(p = 0.409\) and \(n = 3889\), we have: \[ \sigma_{\hat{p}} = \sqrt{\frac{0.409 \cdot 0.591}{3889}} \approx 0.0078 \]

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Key Concepts

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

Probability Distribution
Probability distributions are essential in statistics as they provide a comprehensive description of how probabilities are distributed over the different possible outcomes of a random variable. In this case, we explore the probability distribution of voters choosing a particular candidate. Here, we focus on a discrete probability distribution since we deal with a binary random variable, which we explain later.

The probability distribution in our case tells us the likelihood of one voter casting their vote for Whitman (\(X = 1\)) or not (\(X = 0\)). For Whitman, the probability is \(P(X = 1) = 0.409\). The probability for the alternative, which is not voting for Whitman, is \(P(X = 0) = 0.591\). These probabilities should always sum up to 1, as they represent the complete set of outcomes for this scenario.

In general, understanding how each possible outcome is weighted is crucial for analyzing behavior patterns in any statistical study. This foundational knowledge supports calculations such as finding expected values or analyzing the variability in the data.
Sampling Distribution
Sampling distributions provide us with a picture of how sample statistics, such as proportions or means, would behave if we repeatedly drew samples from a population and computed the statistic for each one. They are pivotal in inferential statistics, enabling us to estimate population parameters based on sample data.

For this election's exit poll, the sampling distribution concerns the proportion of voters among a sample of 3889 people who cast their vote for Whitman. The mean of this sampling distribution, traditionally indicated as \(\mu_{\hat{p}}\), is equal to the population proportion \(p = 0.409\). This means that if we took many samples of 3889 voters, the average of these sample proportions would hover around 0.409.

The standard deviation of the sampling distribution, known as the standard error, offers an understanding of how much variation exists among sample proportions. It is computed using the formula \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\). Substituting in our values, we calculated it to be approximately 0.0078. This tiny standard deviation indicates a high level of precision in the sample proportion estimate, given the large sample size. A good grasp of sampling distributions is essential for predicting how sample results deviate from the true population parameter.
Binary Random Variable
Binary random variables are a specific type of discrete random variable that can take only two outcomes. In the context of our voter scenario, the outcomes are straightforward: a voter can either vote for Whitman (\(X = 1\)) or for another candidate (\(X = 0\)). These are mutually exclusive outcomes, meaning if one occurs, the other cannot, which makes the binary categorization quite clear.

Recognizing a binary random variable is important because it simplifies the analysis and calculations. The probability distribution for a binary variable is simple: it can have two values - think of this like a yes or no situation with corresponding probabilities that add up to one.
  • The probability of choosing Whitman: \(P(X = 1) = 0.409\)
  • The probability of not choosing Whitman: \(P(X = 0) = 0.591\)

Understanding how to define and work with binary random variables helps statisticians convert real-world situations into manageable numerical formats. They are the building blocks for many statistical models, including logistic regression and various machine learning algorithms.

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Most popular questions from this chapter

Multiple choice: Sampling distribution The sampling distribution of a sample mean for a random sample size of 100 describes a. How sample means tend to vary from random sample to random sample of size 100 . b. How observations tend to vary from person to person in a random sample of size 100 . c. How the data distribution looks like the population distribution when the sample size is larger than 30 . d. How the standard deviation varies among samples of size 100 .

Bank machine withdrawals An executive in an Australian savings bank decides to estimate the mean amount of money withdrawn in bank machine transactions. From past experience, she believes that \(\$ 50\) (Australian) is a reasonable guess for the standard deviation of the distribution of withdrawals. She would like her sample mean to be within \(\$ 10\) of the population mean. Estimate the probability that this happens if she randomly samples 100 withdrawals. (Hint: Find the standard deviation of the sample mean. How many standard deviations does \(\$ 10\) equal?)

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?

Finite populations The formula \(\sigma / \sqrt{n}\) for the standard deviation of \(\bar{x}\) actually is an approximation that treats the population size as infinitely large relative to the sample size \(n\). The exact formula for a finite population size \(N\) is Standard deviation \(=\sqrt{\frac{N-n}{N-1}} \frac{\sigma}{\sqrt{n}}\) The term \(\sqrt{(N-n) /(N-1)}\) is called the finite population correction. a. When \(n=300\) students are selected from a college student body of size \(\mathrm{N}=30,000\), show that the standard deviation equals \(0.995 \sigma / \sqrt{n}\). (When \(n\) is small compared to the population size \(\mathrm{N},\) the approximate formula works very well.) b. If \(n=\mathrm{N}\) (that is, we sample the entire population), show that the standard deviation equals \(0 .\) In other words, no sampling error occurs, since \(\bar{x}=\mu\) in that case.

Canada lottery In one lottery option in Canada (Source: Lottery Canada), you bet on a six-digit number between 000000 and \(999999 .\) For a \(\$ 1\) bet, you win \(\$ 100,000\) if you are correct. The mean and standard deviation of the probability distribution for the lottery winnings are \(\mu=0.10\) (that is, 10 cents) and \(\sigma=100.00\). Joe figures that if he plays enough times every day, eventually he will strike it rich, by the law of large numbers. Over the course of several years, he plays 1 million times. Let \(\bar{x}\) denote his average winnings. a. Find the mean and standard deviation of the sampling distribution of \(\bar{x}\). b. About how likely is it that Joe's average winnings exceed \(\$ 1,\) the amount he paid to play each time? Use the central limit theorem to find an approximate answer.

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