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According to a Pew Research survey, about \(27 \%\) of American adults are pessimistic about the future of marriage and the family. This is based on a sample, but assume that this percentage is correct for all American adults. a) Using a binomial model, what is the probability that, in a sample of 20 American adults, \(25 \%\) or fewer of the people in the sample are pessimistic about the future of marriage and family? b) Now use a Normal model to compute that probability. How does this compare to your answer from part a? c) Using a Binomial model, what is the probability that, in a sample of 700 American adults, \(25 \%\) or fewer of the people of the people in the sample are pessimistic about the future of marriage and family? d) Now use a Normal model to compute that probability. How does this compare to your answer from part c? e) What do these answers tell you about the importance of checking that \(n p\) and \(n q\) are both at least \(10 ?\)

Short Answer

Expert verified
Normal approximation is less accurate for smaller samples; it improves with larger sample sizes and when both \( np \) and \( nq \) are at least 10.

Step by step solution

01

Define Variables for Binomial Model

The probability of an adult being pessimistic, \( p \), is 0.27. For part (a), we have a sample of 20 American adults, so \( n = 20 \). We want to find the probability that 25% or fewer are pessimistic. Since 25% of 20 is 5, we need \( P(X \leq 5) \), where \( X \) follows Binomial distribution \( \text{Binomial}(n = 20, p = 0.27) \).
02

Calculate Binomial Probability (a)

Use the binomial probability formula and cumulative probability. \[P(X \leq 5) = \sum_{k=0}^{5} \binom{20}{k} (0.27)^k (0.73)^{(20-k)}\]Calculate this sum to find the probability.
03

Normal Approximation for (b)

For Normal approximation, calculate mean \( \mu = n \times p = 20 \times 0.27 = 5.4 \) and standard deviation \( \sigma = \sqrt{npq} = \sqrt{20 \times 0.27 \times 0.73} \approx 2.038 \). Use continuity correction: \( P(X \leq 5.5) \). Convert to a standard normal variable, \( Z = \frac{5.5 - 5.4}{2.038} \approx 0.049 \), and use the standard normal table to find \( P(Z \leq 0.049) \approx 0.519 \).
04

Binomial Model for Larger Sample (c)

For \( n = 700 \), calculate \( np = 700 \times 0.25 = 175 \). Use \( X \sim \text{Binomial}(700, 0.27) \), and find \( P(X \leq 175) \) using binomial tables or software for such a large \( n \).
05

Normal Approximation for Larger Sample (d)

Mean for \( n = 700 \): \( \mu = 700 \times 0.27 = 189 \), and standard deviation \( \sigma = \sqrt{700 \times 0.27 \times 0.73} \approx 11.97 \). Use continuity correction: \( P(X \leq 175.5) \). Calculate \( Z = \frac{175.5 - 189}{11.97} \approx -1.126 \), then find \( P(Z \leq -1.126) \approx 0.130 \).
06

Analysis of Approximation (e)

Compared results from binomial and normal models: as sample size increases, normal approximation becomes more accurate because \( np = 189 \) and \( nq = 511 \) both satisfy \( n \times p \geq 10 \) and \( n \times q \geq 10 \). Normal approximation is less accurate in smaller samples without meeting these conditions.

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Key Concepts

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

Normal Approximation
In the world of probability theory and statistics, the Normal approximation to the Binomial distribution is a powerful tool often used to simplify calculations, especially when dealing with large sample sizes. When a Binomial distribution with parameters \(n\) (sample size) and \(p\) (probability of success) meets certain conditions, it can be approximated by a Normal distribution, making probability calculations easier.
One very important condition for using Normal approximation is that both \(np\) and \(nq\) must be greater than or equal to 10, where \(q = 1-p\). This ensures that the distribution is roughly symmetric, which is crucial for the Normal distribution to model it accurately.
The process involves calculating the mean \( \mu = np\) and standard deviation \(\sigma = \sqrt{npq}\). When applying this approximation, use a continuity correction by adjusting the binomial variable by 0.5 in the direction that accounts for the discrete nature of binomial outcomes. For example, if you need \(P(X \leq k)\) in a Normal model, calculate it as \(P(X \leq k+0.5)\).
This method simplifies probability calculations as finding areas under the Normal curve can be computed using standard statistical tables or software.
Central Limit Theorem
The Central Limit Theorem (CLT) is a foundational principle in statistics that greatly aids in understanding how sampling distribution behaves. It states that when you have a sufficiently large sample size, the sampling distribution of the sample mean will be approximately Normally distributed, regardless of the shape of the population distribution.
In the context of applying the Normal approximation to Binomial distributions, CLT justifies this conversion. As the number of trials in a binomial experiment increases, the distribution of the number of successes begins to resemble a Normal distribution. This is why large sample sizes allow for the Normal approximation to be used as a stand-in for Binomial calculations.
The CLT is particularly useful because:
  • It allows for the use of Normal tables to estimate probabilities which are otherwise complex to calculate exactly.
  • It provides a basis for creating confidence intervals and hypothesis tests using the Normal distribution.
Essentially, the larger the sample size \(n\), the more accurate the Normal approximation becomes due to the CLT.
Probability Calculation
Calculating probabilities in a Binomial distribution involves determining the likelihood of a given number of successes in a fixed number of trials. For a Binomial distribution with parameters \(n\) and \(p\), the probability of observing \(k\) successes is given by the formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \(\binom{n}{k}\) is a binomial coefficient. This tells us how many ways we can choose \(k\) successes in \(n\) trials.
For cumulative probability, when calculating \(P(X \leq k)\), you sum the individual probabilities from 0 to \(k\). This involves calculating the above formula for each value from 0 to \(k\) and summing the results.
When the sample size is large, this calculation can become complicated and computationally strenuous, which is why approximation methods such as the Normal distribution are often utilized. Using the Normal approximation streamlines these calculations significantly by allowing mathematical software or simple tables to quickly provide results.
Sample Size Importance
The sample size plays a critical role in statistical analysis, especially when applying approximation methods like the Normal approximation for Binomial distributions. A larger sample size not only increases the accuracy of the approximation but also affects the reliability and interpretability of results.
With a small sample size, the binomial distribution is typically skewed, especially when \(p\) is far from 0.5. This skewness can lead to inaccurate results if a Normal approximation is used. However, as sample size increases, the shape of the Binomial distribution becomes more symmetric and bell-shaped, making the Normal approximation more reliable.
The conditions \(np \geq 10\) and \(nq \geq 10\) ensure that both tails of the distribution can be adequately approximated. This makes interpretations based on Normal distribution more valid. Increasing sample size also generally increases the precision of estimates and the power of statistical tests.
Moreover, a sufficiently large sample size helps in reducing sampling error, thereby making the results from analyses more reflect the actual population parameters.

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