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Survey accuracy. A sample survey contacted an SRS of 2854 registered voters shortly before the 2012 presidential election and asked respondents whom they planned to vote for. Election results show that \(51 \%\) of registered voters voted for Barack Obama. We will see later that in this situation the proportion of the sample who planned to vote for Barack Obama (call this proportion \(V\) ) has approximately the Normal distribution with mean \(\mu=0.51\) and standard deviation \(\sigma=0.009\). (a) If the respondents answer truthfully, what is \(P(0.49 \leq V \leq 0.53)\) ? This is the probability that the sample proportion \(V\) estimates the population proportion \(0.51\) within plus or minus \(0.02\). (b) In fact, \(49 \%\) of the respondents said they planned to vote for Barack Obama \((V=0.49)\). If respondents answer truthfully, what is \(P(V \leq 0.49)\) ?

Short Answer

Expert verified
(a) Approximately 97.27%. (b) Approximately 1.32%.

Step by step solution

01

Understand the Distribution

The problem states that the sample proportion \( V \) follows a normal distribution with mean \( \mu = 0.51 \) and standard deviation \( \sigma = 0.009 \). We need to find probabilities involving this normal distribution.
02

Calculate Probability for Interval (a)

We want to calculate the probability that the proportion \( V \) is within the range \( 0.49 \leq V \leq 0.53 \). For this, we will use the standardized normal variable \( Z \), where \[ Z = \frac{V - \mu}{\sigma} \].Find \( Z \) for both ends of the interval:- Lower bound: \( Z = \frac{0.49 - 0.51}{0.009} = -2.22 \)- Upper bound: \( Z = \frac{0.53 - 0.51}{0.009} = 2.22 \)Using normal distribution tables or a calculator, find the probability:\[ P(-2.22 \leq Z \leq 2.22) \approx 0.9727 \]
03

Calculate Probability for Point (b)

For part (b), we need to find \( P(V \leq 0.49) \). Again use the standard normal variable:- \( Z = \frac{0.49 - 0.51}{0.009} = -2.22 \)Find the probability for this \( Z \) value:\[ P(Z \leq -2.22) \approx 0.0132 \]
04

Conclusion

The calculated probabilities provide insight into how likely the observed sample proportions are assuming the respondents answered truthfully. In part (a), there is a high probability (about 97.27%) that the sample proportion falls within the desired range. In part (b), the low probability (about 1.32%) of observing \( V \leq 0.49 \) suggests that the sample result might be surprising if respondents were answering truthfully.

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Key Concepts

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

Understanding Sample Surveys
A sample survey is a method of collecting data from a portion of a larger population with the aim of gaining insights into the entire population's characteristics. Surveys are important tools in statistics because they provide a way to infer population parameters, which are often impossible or impractical to measure directly due to resource or time constraints.
Sample surveys must be carefully designed to ensure that they accurately reflect the population. This involves choosing an appropriate sample size and method of selection. The results of the survey can greatly vary depending on these factors.
In our context, a sample survey was conducted with 2,854 registered voters before an election to estimate voter intentions. By understanding the responses from this sample, one can infer details about the entire voter population, such as who they intend to vote for, as well as estimate how accurately the survey reflects the actual voting population.
The Significance of SRS (Simple Random Sample)
SRS, or Simple Random Sample, is a foundational concept in statistics which ensures that every individual in the population has an equal chance of being included in the sample. This randomization process is crucial for minimizing bias and ensuring that the sample represents the population accurately.
When conducting a survey, such as the one from our example with the voters, employing an SRS means that every voter had an equal opportunity to be surveyed. This equal opportunity reduces selection bias and allows us to generalize the survey results to the entire population with greater reliability.
  • Ensures fairness in selection
  • Minimizes systematic errors
  • Enhances the validity of conclusions drawn from the data
Understanding and correctly implementing SRS is one of the fundamental steps in statistical sampling to ensure data integrity and reliable results.
Exploring Population Proportion
Population proportion is a statistical measure that represents the fraction of the population exhibiting a specific characteristic. In the context of a survey, it reflects the percentage of respondents who have a particular opinion or trait, projected onto the entire population.
In this exercise, the population proportion under scrutiny is the percentage of voters planning to vote for Barack Obama, known to be 51%. This proportion serves as the benchmark against which survey results are compared.
The population proportion allows statisticians to evaluate whether the sample results are a valid reflection of the broader population. This comparison helps to determine the likelihood of outcomes observed within the sample actually occurring in the population, and thus is crucial for the interpretation of survey data.
Unraveling Standard Deviation
Standard deviation is a measure of variability that describes how much individual data points deviate from the mean of a dataset. In the context of normal distribution, it helps to understand the spread of the sample proportions around the mean.
In our problem, the standard deviation is 0.009, indicating that voter intentions recorded in the survey vary quite slightly from the expected population mean. This small standard deviation suggests consistent survey data close to the true mean proportion (51%).
The significance of standard deviation extends to probability calculations, such as determining the likelihood of the survey results aligning with the true population mean. By using standard deviation within the normal distribution framework, statisticians can assess how typical or atypical a particular survey result might be.

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Most popular questions from this chapter

How Many Cups of Coffee? Choose an adult age 18 or over in the Unaited States at random and ask, "How many cups of coffee do you drink on average per day?" Call the response \(X\) for short. Based on a large sample survey, here is a probability model for the answer you will get: \({ }^{7}\) \begin{tabular}{l|ccccc} \hline Number & 0 & 1 & 2 & 3 & 4 or more \\ \hline Probability & \(0.36\) & \(0.26\) & \(0.19\) & \(0.08\) & \(0.11\) \\ \hline \end{tabular} (a) Verify that this is a valid finite probability model. (b) Describe the event \(X<4\) in words. What is \(P(X<4)\) ? (c) Express the event "have at least one cup of coffee on an average day" in terms of \(X\). What is the probability of this event?

In a table of random digits such as Table B, each digit is equally likely to be any of \(0,1,2,3,4,5,6,7,8\), or 9. What is the probability that a digit in the table is a \(7 ?\) (a) \(1 / 9\) (b) \(1 / 10\) (c) \(9 / 10\)

Will You Be in a Crash? The probability that a randomly chosen driver will be involved in a car crash in the next year is about \(0.05\). \(^{12}\) This is based on the proportion of millions of drivers who have crashes. (a) What do you think is your own probability of being in a crash in the next year? This is a personal probability. (b) Give some reasons why your personal probability might be a more accurate prediction of your "true chance" of being in a crash than the probability for a random driver. (c) Almost everyone says their personal probability is lower than the random driver probability. Why do you think this is true?

Birth order. A couple plans to have three children. There are eight possible arrangements of girls and boys. For example, GGB means the first two children are girls and the third child is a boy. All eight arrangements are (approximately) equally likely. (a) Write down all eight arrangements of the sexes of three children. What is the probability of any one of these arrangements? (b) Let \(X\) be the number of girls the couple has. What is the probability that \(X=2 ?\) (c) Starting from your work in part (a), find the distribution of \(X\). That is, what values \(\operatorname{can} X\) take, and what are the probabilities for each value?

Probability models? In each of the following sítuations, state whether or not the given assignment of probabilities to individual outcomes is legitimate, that is, satisfies the rules of probability. Remember, a legitimate model need not be a practically reasonable model. If the assignment of probabilities is not legitimate, give specific reasons for your answer. (a) Roll a six-sided die, and record the count of spots on the up-face: $$ \begin{gathered} \mathbf{P}(1)=0 \mathrm{P}(2)=1 / 6 P(3)=1 / 3 P(4)=1 / 3 P(5)=1 / 6 P(6)=0 \\\ P(1)=0 \quad P(2)=1 / 6 \quad P(3)=1 / 3 \\ P(4)=1 / 3 \quad P(5)=1 / 6 \quad P(6)=0 \end{gathered} $$ (b) Deal a card from a shuffled deck: $$ \begin{gathered} \mathrm{P}(\text { clubs })=12 / 52 P(\text { diamonds })=12 / 52 P(\text { hearts })=12 / 52 P(\text { spades })=16 / 52 \\ P(\text { clabs })=12 / 52 \quad P(\text { diamonis })=12 / 52 \\ P(\text { bearts })=12 / 52 \quad P(\text { spades })=16 / 52 \end{gathered} $$ (c) Choose a college student at random and record sex and enrollment status: \(P(\) female full-time \()=0.56 \mathrm{P}(\) male full-time \()=0.44 \mathrm{P}(\) female part- time \()=0.24 \mathrm{P}(\) male part-lime \()=0.17\)

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