91Ó°ÊÓ

A sample survey contacted an SRS of 663 registered voters in Oregon shortly after an clection and asked respondents whether they had voted. Voter records show that 56\(\%\) of registered voters had actually voted. We will see later that in repeated random samples of size 663 , the proportion in the sample who voted (call this proportion With vary according to the Normal distribution with mean \(\mu=0.56\) and standard deviation \(\sigma=0.019\) (a) If the respondents answer truthfully, what is \(\mathrm{P}(0.52 \leq V \leq 0.60)\) ? This is the probability that the sample proportion \(V\) estimates the population proportion 0.56 within \(\pm 0.04\) (b) In fact, 72\(\%\) of the respondents said they had voted \((V=0.72) .\) If respondents answer truthfully, what is \(P(V \geq 0.72) ?\) This probability is so small that it is good evidence that some people who did not vote claimed that they did vote.

Step by step solution

01

Understanding the Given Data

The problem provides a mean \( \mu = 0.56 \) and a standard deviation \( \sigma = 0.019 \) for the sampling distribution of the sample proportion \( V \). These parameters are derived under the assumption that the respondents all answered truthfully.

02

Set up Part (a) Problem

For part (a), we need to calculate the probability that the sample proportion \( V \) lies between 0.52 and 0.60, or \( P(0.52 \leq V \leq 0.60) \). We will use the standard normal distribution to approximate this probability.

03

Standardize the Boundaries for Part (a)

To find \( P(0.52 \leq V \leq 0.60) \), convert 0.52 and 0.60 to z-scores using the formula: \( z = \frac{x - \mu}{\sigma} \). So, for \( x = 0.52 \), \( z = \frac{0.52 - 0.56}{0.019} = -2.11 \). For \( x = 0.60 \), \( z = \frac{0.60 - 0.56}{0.019} = 2.11 \).

04

Calculate Probability for Part (a)

Use standard normal distribution tables or a calculator to find \( P(-2.11 \leq Z \leq 2.11) \). This gives us approximately \( P(Z \leq 2.11) - P(Z \leq -2.11) = 0.9830 - 0.0170 = 0.9660 \). So, \( P(0.52 \leq V \leq 0.60) \approx 0.9660 \).

05

Set up Part (b) Problem

For part (b), we need to calculate the probability that the sample proportion \( V \) is at least 0.72, \( P(V \geq 0.72) \). Again, we'll standardize \( 0.72 \) into a z-score.

06

Standardize the Value for Part (b)

Convert 0.72 to a z-score: \( z = \frac{0.72 - 0.56}{0.019} = 8.42 \).

07

Calculate Probability for Part (b)

The probability that \( Z \) exceeds 8.42 is extremely small and is effectively 0 (since \( Z \approx 8.42 \) is far out in the tail of the normal distribution), thus \( P(V \geq 0.72) \approx 0 \).

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Key Concepts

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

Standard Deviation

Standard deviation is a measure that helps us understand how spread out the values in a data set or distribution are. In the context of sampling distributions, it tells us how much the sample proportions are expected to vary around the true population proportion.

For the exercise above, the standard deviation \(\sigma = 0.019\) shows the expected average distance between individual sample proportions (like how many respondents reported they voted) and the actual population proportion (the true percent of people who voted). A smaller standard deviation implies the sample proportions are closely clustered around the population proportion, while a larger one indicates greater variability.

Understanding the standard deviation is crucial for interpreting how much we can trust our sample data as a representation of the population.

Normal Distribution

The normal distribution is a continuous probability distribution that is symmetrical. It's often represented as a "bell curve" and is a fundamental concept in statistics due to its natural occurrence in many scenarios.

In the exercise, we assume the sample proportions are approximately normally distributed. This assumption allows us to use properties of the normal distribution to calculate probabilities about the sample proportion.

Because the normal distribution is defined by its mean and standard deviation, any data can be transformed into a standard normal distribution—an essential process in understanding probabilities and drawing conclusions based on sample data. This transformation is achieved via the z-score.

Z-score

A z-score is a way of standardizing an individual data point, expressing it in terms of standard deviations from the mean. This transformation helps us compare data points from different distributions or surveys.

Calculating a z-score uses the formula: \[ z = \frac{x - \mu}{\sigma} \]where \(x\) is the data point, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation. By transforming data into z-scores, we can reference standard normal distribution tables, which provide a fast way to obtain probabilities related to any specific z-score.

For example, in the exercise when finding \(P(0.52 \leq V \leq 0.60)\), z-scores were calculated to align these proportions with the standardized normal distribution, thereby allowing probability calculations.

Confidence Intervals

Confidence intervals provide a range of values, derived from a sample, that is likely to contain the population parameter with a certain level of confidence. Essentially, it gives a "guesstimation" about where the true parameter lies.

Though not directly mentioned in the task, understanding confidence intervals ties into the broader context. The exercise basically explores potential sample outcomes (the "gap" between predicted and true values via probabilities and z-scores). This approach shares a core concept with confidence intervals, which also predict a range around a statistic (like a proportion) with known reliability—usually denoted by a percentage like 95% confidence.

Using concepts like standard deviation and normal distribution, confidence intervals give insights into the degree of uncertainty we have concerning our sample data as a representation of the entire population.